ALLLLRRRIGHTT!!..I survived another week of calculus! =)
and the best part is that I'm managing to pull off a 93 all of a sudden and believe me I'm NOT complaining..that's amazingg!!! =)
so it goes to show yall that you can make 0s on your AP but as long as you continue to practice and do homework and do the wiki and of course THIS then you can pass desently.
so I have A LOT of questions:
~I thought I knew how to find the Equation of a Tangent line, Turns out, I DON'T..
~I thought I knew linearization pretty well also, once again, I DON'T...
~ln and e intergration ... helpp!!
~on the non-calculator portion: number 7..relative max of ln[x] divided by x..
do you take the derivative?..giving you [x(1/x)-ln(x)] over (x)^2..
then do you set it equal to zero?..giving you x=ln(1)...
~on the non-calculator protion: number 13..integrate e^u divided by 1+e^2u
I know you have to integrate..but how do you integrate e?
and some sort of way it gives you tan inverse..ik that the bottom could give me that but i still don't see how the answer manages to be tan inverse e^u plus c
~ARE WE GOING OVER THE CALCULATOR PORTION WITH MRS. ROBINSON LIKE WE DID TODAY FOR THE NON-CALCULATOR PORTION??...I hope so, because I really need that!..if not then here are a few more questions i need answered:
~numbers 30,31, and 32..I don't know where to begin!
~number 34...what do I do when I see that they want to know what the domain of something is?
~number 36...why is it E?
~number 40...I know I have to integrate what they give me...twice in order to see the orginal equation, but the square root of x and the cosine cubed messes me up every time!!...HELP!
~WHAT'S THE ANSWER TO 41 ... I guess I missed that when she was calling it out to us!
sooo...
let's go over what to remember:
~limits are always on the AP so know that if the exponents are the same you look at it's coeffients, if they are different then you look at where the highest one is: if it's on the top then the limit is infinity, and if it's on the bottom the limit is zero.
~know the definition of a derivitive means to take the derivative of the part that has the h in it and then plug in whatever number is with the h into the derivitive to get your answer
~know that when you see [x] meaning the greatest integers or w/e..it's the ladder graph!
~to find where something increases take the first derivative, solve for x, make intervals with what you have, plug in to the derivative and see where they increase by whether or not they are positive or negagive.
~remember that for ln they try to trick you by putting the front term as the exponent!!
~slope fields...I dont know why but those are my favorite!..=)
~know that in order to find slope you do y2 minus y1 divided by x2 minus x1...
EXAMPLES:
~LRAM is to make a point on the graph and go to the left to make a rectangle
~RRAM is to make a point on the graph and go to the right to make a rectangle
~TRAM is to make a point on the graph and connect the points by making a rectangle
~MRAM is to make a point on the graph and go half to the left and half to the right and form a rectangle
DEPENDING ON WHERE ALL OF THESE ARE ON THE GRAPH, you'll know if it's under/over and by knowing that you can see what's greater than what
~find the limit of x=3 on (x+3)/((x)^2-9)
you know that you can take the bottom and factor it out giving you (x+3)(x-3)
the (x+3)'s will cancel leaving you with 1 over (x-3)
plug in 3 since that's what the limit is, giving you:
1 over (3-3) which is 1/0
THEREFORE...IT'S NONEXISTENT, UNDEFINED, OR DOES NOT EXIST
see yall monday...& DON'T FORGET TO WHERE SAINTS CLOTHES ALL WEEK =)
~ElliE~
Friday, January 29, 2010
post #24
absolute maxs and mins:
1. Find critical values by using the first derivative test: take the derivative set equal to zero and solve for the variable.
2. Plug both intervals as well as the critical values into your original equation
3. Be sure to keep the value after substitution with each appropriate number.
4. To determine which constitute a max and which constitute a min, you must look and point out those of which have the lowest values-min- and those of the highest value-max.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
e integration:
whatever is raised to the e power will be your u and du will be the derivative of u. For example:
e^2x-1dx
u=2x-1 du=2
rewrite the function as:
1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
Equation of a tangent line:
Take the derivative and plug in the x value.
If you are not given a y value, plug into the original equation to get the y value.
then plug those numbers into point slope form: y − y1 = m(x − x1)
The terms for the First Derivative Test:
1. Increasing
2. Decreasing
3. Horizontal Tangent
4. Min/Max
Steps:
1. Take the derivative
2. Set it equal to zero
3. Solve for x to get the possible critical points [[can someone clarify this part???? My notebook says it equals critical points AND mins/maxs/horizontal tangents]]
4. Set up intervals with your x value(s)
5. Plug into your first derivative
6. To find an absolute extrema, plug in the values from step 5 into your original function
linearization.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unknown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
okay so another thing im having trouble with is linearization and also implicit derivatives.
1. Find critical values by using the first derivative test: take the derivative set equal to zero and solve for the variable.
2. Plug both intervals as well as the critical values into your original equation
3. Be sure to keep the value after substitution with each appropriate number.
4. To determine which constitute a max and which constitute a min, you must look and point out those of which have the lowest values-min- and those of the highest value-max.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
e integration:
whatever is raised to the e power will be your u and du will be the derivative of u. For example:
e^2x-1dx
u=2x-1 du=2
rewrite the function as:
1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
Equation of a tangent line:
Take the derivative and plug in the x value.
If you are not given a y value, plug into the original equation to get the y value.
then plug those numbers into point slope form: y − y1 = m(x − x1)
The terms for the First Derivative Test:
1. Increasing
2. Decreasing
3. Horizontal Tangent
4. Min/Max
Steps:
1. Take the derivative
2. Set it equal to zero
3. Solve for x to get the possible critical points [[can someone clarify this part???? My notebook says it equals critical points AND mins/maxs/horizontal tangents]]
4. Set up intervals with your x value(s)
5. Plug into your first derivative
6. To find an absolute extrema, plug in the values from step 5 into your original function
linearization.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unknown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
okay so another thing im having trouble with is linearization and also implicit derivatives.
Monday, January 25, 2010
Post 23...blogger fails..miserably
I am having major problems with this blogger thing. KEEPS KICKING ME OFF!!!!!!
Ok...Math!!!!!
To begin with, there are certain steps by which you must follow in order to attain the absolute maximum and minimum of a graph.
1. Find critical values by using the first derivative test: take the derivative set equal to zero and solve for the variable.
2. Plug both intervals as well as the critical values into your original equation
3. Be sure to keep the value after substitution with each appropriate number.
4. To determine which constitute a max and which constitute a min, you must look and point out those of which have the lowest values-min- and those of the highest value-max.
In an effort to further explain the above, I will now complete an example problem similar to that on an AP exam which we take later in the year.
Example:
Given the original function f(x) = x^2 -16 on the interval [1,2], find any absolute maxima and minima.
1. First Derivative Test:
f’(x) = 2x
2x = 0
X = 0 <<
2/3. (0)^2 + 16= 16>>[0,,16]
(1)^2 + 16= 17>>[1, 17]
(2)^2 + 16= 20>>[2, 20]
4. In accordance with the above data, as well as the aforementioned steps, one can infer that there is an absolute minimum at [1, 16] and an absolute maximum at [2, 20].
Hopefully the above explanation has clarified a few things and/or helped you in some shape or form.
In yet another effort to clarify a specific topic or idea in Calculus, I shall further explain the concept from blogs one and two.
While all of the information in the previous reflections were indeed correct, after supplementary review, I have found that I could have given an example seen on some free response questions on the AP.
For example, they may ask you if a certain type of sum, such as LRAM, is an overestimate or an underestimate. My method to finding the answer to this question is relatively simple, however, it requires a substantial understanding of derivative graphs.
But in simple terms:
IF:
f’(x) > 0--increasing
f’’(x) <0--concave down
THEN:
LRAM-underestimate
RRAM-overestimate
MRAM-overestimate
TRAM-underestimate (barely-closest estimate)
IF:
f’(x) <0--decreasing
f’’(x).0--concave up
THEN:
LRAM-overestimate
RRAM-underestimate
TRAM-overestimate
MRAM-investigate
Hope that helps some!
**This comes in handy on the problems that have "T
Ok...Math!!!!!
To begin with, there are certain steps by which you must follow in order to attain the absolute maximum and minimum of a graph.
1. Find critical values by using the first derivative test: take the derivative set equal to zero and solve for the variable.
2. Plug both intervals as well as the critical values into your original equation
3. Be sure to keep the value after substitution with each appropriate number.
4. To determine which constitute a max and which constitute a min, you must look and point out those of which have the lowest values-min- and those of the highest value-max.
In an effort to further explain the above, I will now complete an example problem similar to that on an AP exam which we take later in the year.
Example:
Given the original function f(x) = x^2 -16 on the interval [1,2], find any absolute maxima and minima.
1. First Derivative Test:
f’(x) = 2x
2x = 0
X = 0 <<
2/3. (0)^2 + 16= 16>>[0,,16]
(1)^2 + 16= 17>>[1, 17]
(2)^2 + 16= 20>>[2, 20]
4. In accordance with the above data, as well as the aforementioned steps, one can infer that there is an absolute minimum at [1, 16] and an absolute maximum at [2, 20].
Hopefully the above explanation has clarified a few things and/or helped you in some shape or form.
In yet another effort to clarify a specific topic or idea in Calculus, I shall further explain the concept from blogs one and two.
While all of the information in the previous reflections were indeed correct, after supplementary review, I have found that I could have given an example seen on some free response questions on the AP.
For example, they may ask you if a certain type of sum, such as LRAM, is an overestimate or an underestimate. My method to finding the answer to this question is relatively simple, however, it requires a substantial understanding of derivative graphs.
But in simple terms:
IF:
f’(x) > 0--increasing
f’’(x) <0--concave down
THEN:
LRAM-underestimate
RRAM-overestimate
MRAM-overestimate
TRAM-underestimate (barely-closest estimate)
IF:
f’(x) <0--decreasing
f’’(x).0--concave up
THEN:
LRAM-overestimate
RRAM-underestimate
TRAM-overestimate
MRAM-investigate
Hope that helps some!
**This comes in handy on the problems that have "T
Sunday, January 24, 2010
Posting...#23
So the Saints one which is awesome they're going to the superbowl woot woot
NOW MATH. yuck
integration which is something i love
LRAM-left hand approximationx[f(a)+f(a+x)+...f(b)]
RRAM-right hand approximationx[f(a+x)+...f(b)]
MRAM-midpoint approximationx[f(mid)+f(mid)+...]
Trapezoid-approximates using a trapezoidx/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]
An example of integration is:
x^2-3 [1,4] n=3
First you find your delta x which is 1
LRAM 1[f(1)+f(2)+f(3)]1[-2+1+6]1[5]=5
RRAM 1[f(2)+f(3)+f(4)]1[1+6+13]=20
MRAM 1,2,3,42+1/2=3/23+2/2=5/24+3/2=7/21[-3/4+13/4=37/4]=47/4
Trapezoid1/2[f(1)+2f(2)+2f(3)+f(4)]1/2[-2+2+12+13]=25/2
WEll thats basically it im really having alot of trouble with the non calculator portions of the ap things that i love butt if someone can spicifically help me with related rates that would be amazing and i might actually like you since i hate all
NOW MATH. yuck
integration which is something i love
LRAM-left hand approximationx[f(a)+f(a+x)+...f(b)]
RRAM-right hand approximationx[f(a+x)+...f(b)]
MRAM-midpoint approximationx[f(mid)+f(mid)+...]
Trapezoid-approximates using a trapezoidx/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]
An example of integration is:
x^2-3 [1,4] n=3
First you find your delta x which is 1
LRAM 1[f(1)+f(2)+f(3)]1[-2+1+6]1[5]=5
RRAM 1[f(2)+f(3)+f(4)]1[1+6+13]=20
MRAM 1,2,3,42+1/2=3/23+2/2=5/24+3/2=7/21[-3/4+13/4=37/4]=47/4
Trapezoid1/2[f(1)+2f(2)+2f(3)+f(4)]1/2[-2+2+12+13]=25/2
WEll thats basically it im really having alot of trouble with the non calculator portions of the ap things that i love butt if someone can spicifically help me with related rates that would be amazing and i might actually like you since i hate all
Post #23
We meet again bloggers.
Well it is time for another blog. I always feel like I'm talking about the same thing over and over again. You think it would help me remember it but sadly no. Tomorrow, I think, we take another ap test. I'm so not looking foward to it because of my past scores. I am trying though..I wish they gave points for that in multiple choice. Okay now I should probably talk about the stuff I learned this week:
We have been talking about key words lately.
key word stuff:
1. linearization-->equation of a tangent line
2. if given a table(the sets of points)-->dealing with slope or rram/lram/ect.
3. how many-->integrate
4.(in calculator part)volume problems-->find intersection
We have learned some new formulas.
some formula stuff:
1. Volume of a sphere: V=4/3pi(r^3)
2. dealing with inverses: g^1(y)=1/f^1(x)
Example: Let f(x)=x^5+1 and let g be the inverse function of f. What is the value of g^1(0)?
g^1(y)=1/f^1(x)
Solve for x: x^5+1=0
x^5=-1
x=-1
plug in x to problem: 5(-1)^4=5
take reciprocal: 1/5
3. words like bacteria, birth rate, contaminates, you will probably use the exponential growth formula: A(subscript 0)e^kt
For what I don't know:
Substitution, once again. I think I need an example problem shown. Any takers?
Well it is time for another blog. I always feel like I'm talking about the same thing over and over again. You think it would help me remember it but sadly no. Tomorrow, I think, we take another ap test. I'm so not looking foward to it because of my past scores. I am trying though..I wish they gave points for that in multiple choice. Okay now I should probably talk about the stuff I learned this week:
We have been talking about key words lately.
key word stuff:
1. linearization-->equation of a tangent line
2. if given a table(the sets of points)-->dealing with slope or rram/lram/ect.
3. how many-->integrate
4.(in calculator part)volume problems-->find intersection
We have learned some new formulas.
some formula stuff:
1. Volume of a sphere: V=4/3pi(r^3)
2. dealing with inverses: g^1(y)=1/f^1(x)
Example: Let f(x)=x^5+1 and let g be the inverse function of f. What is the value of g^1(0)?
g^1(y)=1/f^1(x)
Solve for x: x^5+1=0
x^5=-1
x=-1
plug in x to problem: 5(-1)^4=5
take reciprocal: 1/5
3. words like bacteria, birth rate, contaminates, you will probably use the exponential growth formula: A(subscript 0)e^kt
For what I don't know:
Substitution, once again. I think I need an example problem shown. Any takers?
Ash's 23rd Post
Soo, I've decided to post this little bit of fun info:
I was reading Nightlight (parody of Twilight) and there were references to Calculus! I felt smart because I actually knew what they were talking about!! (hehe, Riemann Sums)
Anyway!!
Thursday...
It clicked!! I'm finally starting to understand these problems!! I think it's because Mrs. Robinson is going soo slow and explaining everything very thoroughly! Yay!!
So, let me explain a few things ^^
1. Limit from h->0 (DON'T LOOK AT THE COEFFS!!); it's a definition of a derivative
2. When you're asked about the Average Rate Of Change, you simply take the derivative.
3. When they ask you about relative maximum, you take the derivative, solve for x, and check intervals.
4. When you're asked about maximum value, take the derivative, solve for x, then plug in to get y.
5. Maximum acceleration = slope
6. ?? Can someone confirm this for me? When asked for the position, find the area between the two points??
7. When asked to find the vertical tangents, take the derivative.
8. When you see the word rate, take the derivative and solve it for...????
Also, I don't really understand the stuff we did on Friday?? Any help please?
Annndd!
Can someone explain to me the nPr and the nCr please? I need to calculate the probability that the Saints will LOSE the Super Bowl ^^
I was reading Nightlight (parody of Twilight) and there were references to Calculus! I felt smart because I actually knew what they were talking about!! (hehe, Riemann Sums)
Anyway!!
Thursday...
It clicked!! I'm finally starting to understand these problems!! I think it's because Mrs. Robinson is going soo slow and explaining everything very thoroughly! Yay!!
So, let me explain a few things ^^
1. Limit from h->0 (DON'T LOOK AT THE COEFFS!!); it's a definition of a derivative
2. When you're asked about the Average Rate Of Change, you simply take the derivative.
3. When they ask you about relative maximum, you take the derivative, solve for x, and check intervals.
4. When you're asked about maximum value, take the derivative, solve for x, then plug in to get y.
5. Maximum acceleration = slope
6. ?? Can someone confirm this for me? When asked for the position, find the area between the two points??
7. When asked to find the vertical tangents, take the derivative.
8. When you see the word rate, take the derivative and solve it for...????
Also, I don't really understand the stuff we did on Friday?? Any help please?
Annndd!
Can someone explain to me the nPr and the nCr please? I need to calculate the probability that the Saints will LOSE the Super Bowl ^^
post 23
This past week in calculus we went over AP tests and learned about slope fields.
The way you do this is to plug the points on the graph that is given into the equation that is given and your answers depend on what lines you draw.
The line you would use for postitive slopes is /
The line you would use for negative slope is \
The line you would use for a zero slope is a horizontal line --
The line you would use for an underfined slope is a vertical line
For old stuff to find critical points is what I am going to talk about. The steps are to:
1. Take the derivative of the equation
2. Set the derivative equal to zero
3. Solve step 2 for x
The way to find where the graph of the equation is concave up and concave down is to:
1. Take the derivative of the original equation
2. Take the derivative of the first derivative
3. Find points of inflection.
4. Set up intervals
5. Use a number in between the two numbers of the interval and plug into the derivative if it is a negative number and if it is a postitive number then it is concave up.
For stuff I am having problems with, is mainly tables and small mental mistakes. Going over the AP tests have helped me a lot. So once I can fix these too i should do a lot better on my tests.
The way you do this is to plug the points on the graph that is given into the equation that is given and your answers depend on what lines you draw.
The line you would use for postitive slopes is /
The line you would use for negative slope is \
The line you would use for a zero slope is a horizontal line --
The line you would use for an underfined slope is a vertical line
For old stuff to find critical points is what I am going to talk about. The steps are to:
1. Take the derivative of the equation
2. Set the derivative equal to zero
3. Solve step 2 for x
The way to find where the graph of the equation is concave up and concave down is to:
1. Take the derivative of the original equation
2. Take the derivative of the first derivative
3. Find points of inflection.
4. Set up intervals
5. Use a number in between the two numbers of the interval and plug into the derivative if it is a negative number and if it is a postitive number then it is concave up.
For stuff I am having problems with, is mainly tables and small mental mistakes. Going over the AP tests have helped me a lot. So once I can fix these too i should do a lot better on my tests.
Post Number Twenty Three
Welcome to Miami : )
Now on to Calculus if I must..
Some keywords I never understood:
Local minimum- first derivative, set equal to zero, solve, set up intervals and check
Maximum speed- absolute value of velocity
Linearization- find the equation of the tangent line
How Many? – integrate
Lim h> 0 + h – definition of derivative- take derivative of what is behind the parenthesis
Maximum value – find the y- value at the max; take derivative, set equal to 0, solve for x, plug in to find y.
Maximum acceleration- where is maximum slope
Starting point-position- related to original function; zeros of velocity will give you max and mins of position
Vertical tangents- take implicit derivative
Rate- take derivative
Somethings I know how to do:
Lim
x>infintity (20x^2 – 13x + 5)/(5 – 4x^3)
The limit is 0 because of the limit rule when the top degree is smaller than the bottom degree the limit is 0. 2<3
Limh>0 (ln(2+h) – ln2)/h is 1/2 .
You take the derivative of what is outside of the parenthesis to figure out the answer. Since it is ln 2 make that 1/x and plug in 2 therefore giving you the limit at ½.
The average rate of chage of function f on [1, 4] is..
You are given a table of the values
For this problem you use f(b) – f(a)/b – a
Therefore f(4) – f(1)/3 equals 6-2/3 which equals 4/3.
Some things I still don’t really get are e integration, linearization, switching from graphs of first to second to derivative to original or vice versa,.
Over and out, WHO DAT!?
Now on to Calculus if I must..
Some keywords I never understood:
Local minimum- first derivative, set equal to zero, solve, set up intervals and check
Maximum speed- absolute value of velocity
Linearization- find the equation of the tangent line
How Many? – integrate
Lim h> 0 + h – definition of derivative- take derivative of what is behind the parenthesis
Maximum value – find the y- value at the max; take derivative, set equal to 0, solve for x, plug in to find y.
Maximum acceleration- where is maximum slope
Starting point-position- related to original function; zeros of velocity will give you max and mins of position
Vertical tangents- take implicit derivative
Rate- take derivative
Somethings I know how to do:
Lim
x>infintity (20x^2 – 13x + 5)/(5 – 4x^3)
The limit is 0 because of the limit rule when the top degree is smaller than the bottom degree the limit is 0. 2<3
Limh>0 (ln(2+h) – ln2)/h is 1/2 .
You take the derivative of what is outside of the parenthesis to figure out the answer. Since it is ln 2 make that 1/x and plug in 2 therefore giving you the limit at ½.
The average rate of chage of function f on [1, 4] is..
You are given a table of the values
For this problem you use f(b) – f(a)/b – a
Therefore f(4) – f(1)/3 equals 6-2/3 which equals 4/3.
Some things I still don’t really get are e integration, linearization, switching from graphs of first to second to derivative to original or vice versa,.
Over and out, WHO DAT!?
Week #23?
Hope everyone is enjoying watching the Saint's game!!!
I'll do some AP uestions from the last test we review last week, AB Practice Exam 1.
32) The graph of F', which consists of a quarter-circle and two line segments, is shown above. At x = 2 which of the following statements is true?
A) f is not continuous
B) f is continuous but not differentiable
C) f has a relatve maximum
D) The graph of f has a point of inflection
E) none of these
This is a problem that is basically solved by process of elimination. You can first eliminate A and B because if they give you the graph of f', then that means f is continuous and differentiable. You then can reason that D would be the correct answer because x = 2 would be the point of inflection.
40) At which point on the graph of y = f(x) shown above is f '(x) <> 0?
A) A
B) B
C) C
D) D
E) E
They also give you a graph (which is the key word). This problem is also a process of elimination problem. When you look at the problem, f '(x) <> 0 means that the point is concave up. So you can automatically eliminate C and E because they are not increasing or decreasing. You can also eliminate B, because it is increasing rather than decreasing. You come down to A and D and you see that A is concave up and D is concave down, so you now have you answer of A) A.
See you all tomorrow.
OVERTIME
I'll do some AP uestions from the last test we review last week, AB Practice Exam 1.
32) The graph of F', which consists of a quarter-circle and two line segments, is shown above. At x = 2 which of the following statements is true?
A) f is not continuous
B) f is continuous but not differentiable
C) f has a relatve maximum
D) The graph of f has a point of inflection
E) none of these
This is a problem that is basically solved by process of elimination. You can first eliminate A and B because if they give you the graph of f', then that means f is continuous and differentiable. You then can reason that D would be the correct answer because x = 2 would be the point of inflection.
40) At which point on the graph of y = f(x) shown above is f '(x) <> 0?
A) A
B) B
C) C
D) D
E) E
They also give you a graph (which is the key word). This problem is also a process of elimination problem. When you look at the problem, f '(x) <> 0 means that the point is concave up. So you can automatically eliminate C and E because they are not increasing or decreasing. You can also eliminate B, because it is increasing rather than decreasing. You come down to A and D and you see that A is concave up and D is concave down, so you now have you answer of A) A.
See you all tomorrow.
OVERTIME
Post #23
Last week in calculus, we learned how to sketch a slope field and reviewed other topics we already learned.
To sketch a slope field, you are given an equation and a graph with points plotted.
You have to plug the points plotted into the differential equation and depending on the outcome of the slope, that is what line you draw.
If the slope is positive, the line with be /
Negative will be \
A zero slope is a horizontal line __
An undefined slope is a vertical line |
Given that dy/dx= y-1 /x^2 sketch a slope field for the nine points. Plugging into the equation you get
(-1,0) = 0-1/(-1)^2 = -1 so \
(-1,1) = 0 __
(-1,2) = 1 /
(1,0) = -1 \
(1,1) = 0 __
(1,2) = 1 /
(2,0) = -1/4 \ except not as steep
(2,1) = 0 __
(2,2) = 1/4 / except not as steep
It's pretty simple except you have to try to make the lines of the slopes that are the same as similar as possible and the different ones have to be recognized is different.
We also learned how to find distance given velocity.
You have to integrate to get position the plug in each time value to the integral.
So an example would be
The integral of 2x with a distance of 0
2x integrated is = x^2
Then plug in each time value:
x(0) = f(0) - f(1) = -1
x(1) = f(1) - f(2) = -3
x(2) = f(2) - f(3) = -5
x(3) = f(3) -f(4) = -7
x(4) = f(4) - f(5) = -9
x(5)
Lastly, you add up the totals and get -25
However, since distance cannot be negative, I think it would just be 25
I'm not sure if this is a good example or not, but I tried.
I'm having trouble moving between graphs still and problems such as number 39 on the calculator portion of the AP test we took. It says The region S in the figure showen above is bounded by y=sec x and y=4. What is the volume of the solid formed when S is rotated about the x-axis.
Good luck tomorrow!
Post #23
Tangent Lines
Tangent lines can be used for many purposes. Finding a tangent line requires only a little knowledge of calculus but a substantial amount of algebra.
Example:
Find the line tangent to the graph y=2x2+4x+6 at x=1.
1. Identify the equation and point of tangency. If not given a y value, plug the x value into the original equation.
y=2x2+4x+6 y=2(1)2
2. Differentiate you equation.
dy/dx=4x+4
3. Plug in x value then solve for dy/dx.
dy/dx=4(1)+4=8
Your dy/dx value is your slope from here on you can create your equation of the tangent line. There are three forms that your equation can be presented in: point-slope, slope-intercept, or standard form. For these purposes, I am using point-slope.
(y-12)=8(x-1) is your final answer.
Graph Interpretation
Every AP exam will include at least one graph on it's short answer section that requires the test-taker to interpret certain requirements from it in order to receive credit for that question.
Interpreting a graph is very easy but in order to do it properly one must understand certain properties.
1. If the original graph is increasing, the slope is positive.
2. If the original graph is decreasing, the slope is negative.
3. An interval with a positive slope on the first derivative means that there is a downward concavity on that interval in the second derivative.
4. An interval with a negative slope on the first derivative means that there is an upward concavity on that interval in the second derivative.
5. Upward concavity (bowl-shaped) is positive.
6. Downward concavity (umbrella-shaped) is negative.
7. There is a horizontal tangent where the slope=0 on the original graph.
8. Wherever there is a horizontal slope there is either a maximum or a minimum value.
9. A x intercept on the first derivative is either a maximum of a minimum on the original graph.
Tangent lines can be used for many purposes. Finding a tangent line requires only a little knowledge of calculus but a substantial amount of algebra.
Example:
Find the line tangent to the graph y=2x2+4x+6 at x=1.
1. Identify the equation and point of tangency. If not given a y value, plug the x value into the original equation.
y=2x2+4x+6 y=2(1)2
2. Differentiate you equation.
dy/dx=4x+4
3. Plug in x value then solve for dy/dx.
dy/dx=4(1)+4=8
Your dy/dx value is your slope from here on you can create your equation of the tangent line. There are three forms that your equation can be presented in: point-slope, slope-intercept, or standard form. For these purposes, I am using point-slope.
(y-12)=8(x-1) is your final answer.
Graph Interpretation
Every AP exam will include at least one graph on it's short answer section that requires the test-taker to interpret certain requirements from it in order to receive credit for that question.
Interpreting a graph is very easy but in order to do it properly one must understand certain properties.
1. If the original graph is increasing, the slope is positive.
2. If the original graph is decreasing, the slope is negative.
3. An interval with a positive slope on the first derivative means that there is a downward concavity on that interval in the second derivative.
4. An interval with a negative slope on the first derivative means that there is an upward concavity on that interval in the second derivative.
5. Upward concavity (bowl-shaped) is positive.
6. Downward concavity (umbrella-shaped) is negative.
7. There is a horizontal tangent where the slope=0 on the original graph.
8. Wherever there is a horizontal slope there is either a maximum or a minimum value.
9. A x intercept on the first derivative is either a maximum of a minimum on the original graph.
post #23
Trapezoidal is different because instead of multiplying by delta x, you multiply by delta x/2 and you also have on more term then your number of subintervals.
The formula is : delta x/2 [f(a) + 2f(a + delta x) + 2f(a+ 2 delta x) + ....f(b)]
For this problem: 1/2 [ f(-3) + 2 f(-2) + f( -1)] and then plug in.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
e integration:
whatever is raised to the e power will be your u and du will be the derivative of u. For example:
e^2x-1dx
u=2x-1 du=2
rewrite the function as:
1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
*Linearization
I'm not quite sure when to do it, i know i'm supposed to do linearization when i see approximate..but i'm not sure how to do it. Is it just finding a slope?
Equation of a tangent line:
Take the derivative and plug in the x value.
If you are not given a y value, plug into the original equation to get the y value.
then plug those numbers into point slope form: y − y1 = m(x − x1)
okay so im having trouble with the mram lram and trapezoidal so please help.
The formula is : delta x/2 [f(a) + 2f(a + delta x) + 2f(a+ 2 delta x) + ....f(b)]
For this problem: 1/2 [ f(-3) + 2 f(-2) + f( -1)] and then plug in.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
e integration:
whatever is raised to the e power will be your u and du will be the derivative of u. For example:
e^2x-1dx
u=2x-1 du=2
rewrite the function as:
1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
*Linearization
I'm not quite sure when to do it, i know i'm supposed to do linearization when i see approximate..but i'm not sure how to do it. Is it just finding a slope?
Equation of a tangent line:
Take the derivative and plug in the x value.
If you are not given a y value, plug into the original equation to get the y value.
then plug those numbers into point slope form: y − y1 = m(x − x1)
okay so im having trouble with the mram lram and trapezoidal so please help.
#23
wellll, for some reason, last week just seemed like a waste of a week because I wasn't really there, except for in calc because nothings ever a waste in there righttttt? right. anyways we went over the two AP exams we took the week before and identified key words which i think is going to help alot once it sinks in.
Example and explination of tangent lines:
1. Identify the equation and tangent poing. Plug the x value into the original equation if they don't already give you a y value.
2. Differentiate your equation.
3. Plug in x value then solve for dy/dx.
4. dy/dx value=slope, plug in to point slop form.
Find the line tangent to the graph y=2x2+4x+6 at x=1.
y=2x2+4x+6 y=2(1)2
dy/dx=4x+4
dy/dx=4(1)+4=8
(y-12)=8(x-1)
Slop fields are probably the easiest thing i've ever seen in my life.
* plug in the point on the graphy which are already given to you intooo the equation that is also already given and your answers depend on what what your outcome is and lines you draw for them.
The line for postitive slopes is /
The line for negative slope is \
The line for a zero slope is a horizontal line --
The line for an undefined slope is a vertical line
QUESTIONS: i've pretty much forgotten everything about related rates... help? pleaseeeeee?
Example and explination of tangent lines:
1. Identify the equation and tangent poing. Plug the x value into the original equation if they don't already give you a y value.
2. Differentiate your equation.
3. Plug in x value then solve for dy/dx.
4. dy/dx value=slope, plug in to point slop form.
Find the line tangent to the graph y=2x2+4x+6 at x=1.
y=2x2+4x+6 y=2(1)2
dy/dx=4x+4
dy/dx=4(1)+4=8
(y-12)=8(x-1)
Slop fields are probably the easiest thing i've ever seen in my life.
* plug in the point on the graphy which are already given to you intooo the equation that is also already given and your answers depend on what what your outcome is and lines you draw for them.
The line for postitive slopes is /
The line for negative slope is \
The line for a zero slope is a horizontal line --
The line for an undefined slope is a vertical line
QUESTIONS: i've pretty much forgotten everything about related rates... help? pleaseeeeee?
post 23
The steps for optimization are as follows:
1. Identify your primary and secondary equations. Primary will be the one the problem is asking you to minimize or maximize. I've also noticed that the secondary will usually be set equal to a number.
2. After finding secondary, solve it for one variable if there are two.
3. Once you have this variable, plug it into your primary equation for the variable you solved the secondary equation for
4. Take the derivative of the equation you just formulated and set it equal to zero (this zero will become one of your answers)
5. Once you come out with your zeros, plug them into your secondary equation and solve for the variable you have left (this will give you your second answer)
The terms for the First Derivative Test:
1. Increasing
2. Decreasing
3. Horizontal Tangent
4. Min/Max
Steps:
1. Take the derivative
2. Set it equal to zero
3. Solve for x to get the possible critical points [[can someone clarify this part???? My notebook says it equals critical points AND mins/maxs/horizontal tangents]]
4. Set up intervals with your x value(s)
5. Plug into your first derivative
6. To find an absolute extrema, plug in the values from step 5 into your original function
Can someone check this for me? I don't know if I'm not in my right mind right now [no comments] or if I'm really failing at this? Or if I just do it and not think about it now?
f(x) = 1/2x - sinx
Find the extrema on the interval (0, 2pi)
f'(x) = 1/2 - cosx
1/2 - cosx = 0
-cosx = -1/2
cosx = 1/2
x=cos^-1(1/2)
x=pi/3 and 5pi/3
Because
| +
-------
| +
cos is positive on the first and forth quadrants
300 degrees in radians is 5pi/3
and
60 degrees in radians is pi/3
well tom we have another ap quiz that im not looking forward to. Some of the stuff that im going to struggle with include the stuff when you have to look at the graphs and determine the first der. or second deriv of the graph. I need help with this
1. Identify your primary and secondary equations. Primary will be the one the problem is asking you to minimize or maximize. I've also noticed that the secondary will usually be set equal to a number.
2. After finding secondary, solve it for one variable if there are two.
3. Once you have this variable, plug it into your primary equation for the variable you solved the secondary equation for
4. Take the derivative of the equation you just formulated and set it equal to zero (this zero will become one of your answers)
5. Once you come out with your zeros, plug them into your secondary equation and solve for the variable you have left (this will give you your second answer)
The terms for the First Derivative Test:
1. Increasing
2. Decreasing
3. Horizontal Tangent
4. Min/Max
Steps:
1. Take the derivative
2. Set it equal to zero
3. Solve for x to get the possible critical points [[can someone clarify this part???? My notebook says it equals critical points AND mins/maxs/horizontal tangents]]
4. Set up intervals with your x value(s)
5. Plug into your first derivative
6. To find an absolute extrema, plug in the values from step 5 into your original function
Can someone check this for me? I don't know if I'm not in my right mind right now [no comments] or if I'm really failing at this? Or if I just do it and not think about it now?
f(x) = 1/2x - sinx
Find the extrema on the interval (0, 2pi)
f'(x) = 1/2 - cosx
1/2 - cosx = 0
-cosx = -1/2
cosx = 1/2
x=cos^-1(1/2)
x=pi/3 and 5pi/3
Because
| +
-------
| +
cos is positive on the first and forth quadrants
300 degrees in radians is 5pi/3
and
60 degrees in radians is pi/3
well tom we have another ap quiz that im not looking forward to. Some of the stuff that im going to struggle with include the stuff when you have to look at the graphs and determine the first der. or second deriv of the graph. I need help with this
post 23
LIMIT RULES:
1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity
2. If the degree of the bottom is larger than the degree of the tip, the limit approaches zero
3. If the degree of the bottom is equal to the degree of the top, then you make a fraction out of the coefficients in front of the largest degree.
linearization.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unknown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
one thing im not good at is lram rram and mram.
1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity
2. If the degree of the bottom is larger than the degree of the tip, the limit approaches zero
3. If the degree of the bottom is equal to the degree of the top, then you make a fraction out of the coefficients in front of the largest degree.
linearization.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unknown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
one thing im not good at is lram rram and mram.
post 23
alrighttt, so this week we did more practice ap's. yessssss. and went over both the calculator and noncalculator portions of the test. we even had these papers to write down what was the key words in the problem, and then how to work it next.
ok so, let's go over some stuff.
linearization ... key word is APPROXIMATE.
f(x) = f(c) + f'(c) (x-c)
^^^use that formula.
how to find critical values:
1. take the derivative of the function
2. set it equal to zero
3. solve for x.
taaadaaa.
Limits that go to zero:
If a limit is going to zero, check first to see if there is a +h at the end of top part of the equation.
if there is, take the derivative of the (whatever + h)
& remember disks and washers both have area and volume formula's, which are both the same except one is squared and one isn't. i don't remember which one though.
and can someone explain the shortcuts for finding max and min? i forgot it.
ok so, let's go over some stuff.
linearization ... key word is APPROXIMATE.
f(x) = f(c) + f'(c) (x-c)
^^^use that formula.
how to find critical values:
1. take the derivative of the function
2. set it equal to zero
3. solve for x.
taaadaaa.
Limits that go to zero:
If a limit is going to zero, check first to see if there is a +h at the end of top part of the equation.
if there is, take the derivative of the (whatever + h)
& remember disks and washers both have area and volume formula's, which are both the same except one is squared and one isn't. i don't remember which one though.
and can someone explain the shortcuts for finding max and min? i forgot it.
Post #whoknows
Lol, so this week in Calculus basically what we did is go over the practice AP tests that we took. We went over many old concepts and also learned a few new ones.
Some older concepts:
The limit as x goes to infinity rules...people are still getting this wrong and it's kind of ... ridiculous I guess.
1. if the degree of the top is bigger than the degree of the bottom, the answer is infinity or no horizontal asymptote
2. if the degree of the top is smaller than the degree of the bottom, then the answer is 0 or a horizontal asymptote at y=0.
3. if the degree of the top is equal to the degree of the bottom, divide the top coefficient by the bottom coefficient and that is your answer and it's also written as y=top/bottom for the horizontal asymptote.
Other things...
limit as h goes to 0. Usually you would treat this as a derivative rule. You would take the derivative of whatever is behind the minus sign. Sometimes they won't give you an x in it so basically what they did was already plug in a value. So put in x where you see the number plugged in then take the derivative. After you take the derivative...say its 1/x and it originally said ln2...you would plug in 2 for x. So now the answer would be 1/2 instead of just 1/x.
Important things to remember about velocity, position, and acceleration problems.
Velocity is the derivative of the position function.
Acceleration is the derivative of the velocity function.
Velocity is the integral of the acceleration function.
Position is the integral of the velocity function.
Use this knowledge to move between the different graphs or functions.
I can't really think of much more to add so I'm going to go get ready for the Saint's party :-)
Enjoy the rest of your day people.
Some older concepts:
The limit as x goes to infinity rules...people are still getting this wrong and it's kind of ... ridiculous I guess.
1. if the degree of the top is bigger than the degree of the bottom, the answer is infinity or no horizontal asymptote
2. if the degree of the top is smaller than the degree of the bottom, then the answer is 0 or a horizontal asymptote at y=0.
3. if the degree of the top is equal to the degree of the bottom, divide the top coefficient by the bottom coefficient and that is your answer and it's also written as y=top/bottom for the horizontal asymptote.
Other things...
limit as h goes to 0. Usually you would treat this as a derivative rule. You would take the derivative of whatever is behind the minus sign. Sometimes they won't give you an x in it so basically what they did was already plug in a value. So put in x where you see the number plugged in then take the derivative. After you take the derivative...say its 1/x and it originally said ln2...you would plug in 2 for x. So now the answer would be 1/2 instead of just 1/x.
Important things to remember about velocity, position, and acceleration problems.
Velocity is the derivative of the position function.
Acceleration is the derivative of the velocity function.
Velocity is the integral of the acceleration function.
Position is the integral of the velocity function.
Use this knowledge to move between the different graphs or functions.
I can't really think of much more to add so I'm going to go get ready for the Saint's party :-)
Enjoy the rest of your day people.
Post #23
This week we went over the practice ap test we took last week.... here are some concepts that i am starting to understand a little bit better:
lim h->0
ln(2+h)-ln2/h... the easiest way to do this problem is to just take the derivative of ln2. To do that you must treat 2 as x. So it becomes lnx. The derivative of lnx is 1/x. Now you just plug 2 back into the x value and you get 1/2.
Those table problems are not that bad anymore either:
They want to know the average rate of change of function f on [1,4]. To work this problem you have know know the average rate of change formula which is f(b)-f(a)/b-a. We know b is 4 and a is 1. So you look on the table, find 4 under the x column, look over at the f column and find what 4 equals there, it is equal to 6. Do the same for 1, it equals 2. So you set up your equation, 6-2/4-1 which equals 4/3.
There are also some problems that you will have to use the first derivative on. If they ask you for a relative max, then you have to either find the first derivative of the function or sometimes they give you the derivative of the function. You would solve for x to find your critical values and set them up into intervals. Then you would plug in numbers between those intervals into the function. To have a relative max, the intervals have to go from positive to negative. If none of the intervals do that, then the answer would be none of these.
What i'm still having trouble with:
1. Numbers 10-12 on the non-calculator portion.
2. Numbers 16,19 and 20 on the non-calculator portion.
3. Numbers 31, 32 and 36 on the calculator portion.
See you all tomorrow :)
lim h->0
ln(2+h)-ln2/h... the easiest way to do this problem is to just take the derivative of ln2. To do that you must treat 2 as x. So it becomes lnx. The derivative of lnx is 1/x. Now you just plug 2 back into the x value and you get 1/2.
Those table problems are not that bad anymore either:
They want to know the average rate of change of function f on [1,4]. To work this problem you have know know the average rate of change formula which is f(b)-f(a)/b-a. We know b is 4 and a is 1. So you look on the table, find 4 under the x column, look over at the f column and find what 4 equals there, it is equal to 6. Do the same for 1, it equals 2. So you set up your equation, 6-2/4-1 which equals 4/3.
There are also some problems that you will have to use the first derivative on. If they ask you for a relative max, then you have to either find the first derivative of the function or sometimes they give you the derivative of the function. You would solve for x to find your critical values and set them up into intervals. Then you would plug in numbers between those intervals into the function. To have a relative max, the intervals have to go from positive to negative. If none of the intervals do that, then the answer would be none of these.
What i'm still having trouble with:
1. Numbers 10-12 on the non-calculator portion.
2. Numbers 16,19 and 20 on the non-calculator portion.
3. Numbers 31, 32 and 36 on the calculator portion.
See you all tomorrow :)
Post #23
This week in calculus more than anything i just need to brush up on some subjects..so here it goes!
Since i always forget, and get this wrong...Equation of a tangent line:
Take the derivative and plug in the x value.
If you are not given a y value, plug into the original equation to get the y value.
then plug those numbers into point slope form: y − y1 = m(x − x1)
Linearization is just a step behind this…
Linearization:
f(x)=f(c)+f'(c)(x-c)
example:
Approximate the tangent line to y=x^2 at x=1
you find all the different values: dy/dx=2x dy/dx=2 y=(1)^2=1
then you plug into the formula to get: f(x)=1+2(x-1)
Also, another thing i always get wrong is exactly how to solve a definition of a derivative problem..
lim (ln(2+h)-ln2)/hh->0
All you have to do is take the derivative of ln2, or the last term in the numerator,
but for this you need to treat 2 like an x.
Then at the end plug 2 back in.
ln2= lnx
lnx= 1/x Substitute x for the number behind ln in the original equation
1/x= ½
i'm not really confused about anything except following graphs and still LRAM, MRAM, RRAM, nonsense..i just can't grasp it. i've watched every video and powerpoint..it isn't clicking. Can anyone explain it to me on paper?
Since i always forget, and get this wrong...Equation of a tangent line:
Take the derivative and plug in the x value.
If you are not given a y value, plug into the original equation to get the y value.
then plug those numbers into point slope form: y − y1 = m(x − x1)
Linearization is just a step behind this…
Linearization:
f(x)=f(c)+f'(c)(x-c)
example:
Approximate the tangent line to y=x^2 at x=1
you find all the different values: dy/dx=2x dy/dx=2 y=(1)^2=1
then you plug into the formula to get: f(x)=1+2(x-1)
Also, another thing i always get wrong is exactly how to solve a definition of a derivative problem..
lim (ln(2+h)-ln2)/hh->0
All you have to do is take the derivative of ln2, or the last term in the numerator,
but for this you need to treat 2 like an x.
Then at the end plug 2 back in.
ln2= lnx
lnx= 1/x Substitute x for the number behind ln in the original equation
1/x= ½
i'm not really confused about anything except following graphs and still LRAM, MRAM, RRAM, nonsense..i just can't grasp it. i've watched every video and powerpoint..it isn't clicking. Can anyone explain it to me on paper?
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