alright, this week we studied for our test, and we did our packets; however, last week I didn't know how to do optimization, so I am going to explain that and other stuff that I know.
EVT: the EVT states that a continuous function on a close interval [a,b], must have both a minimum and a maximum on the interval. However, the max and min can occur at the endpoints.
Rolles theorem gives the conditions that guarantee the existance of an extrema in the interior of a closed inteval.
Rolles: Let f be continuous on a closed interval [a,b] and differentiable on the open interval (a,b). If f(a)= f(b) then there is at least one number, “c” in (a,b) such the f '(c)=0
MVT: If f is continuous on the closed interval [a,b] and differentiable on the interval (a,b) then there exists a number c in (a,b) such that f '(c)= (f(b)-f(a))/(b-a).
Steps for optimization
1.Identify primary and secondary equations. The primary will be the one you are maximizing or minimizing, and the secondary will be the other one.
2.Solve secondary equation for one variable, and plug into the primary. (if the primary only has one variable, this step is not necessary.)
3.Take the derivative of the primary, and set it equal to zero; solve for x.
4.Plug into secondary equation to find the other value, check end points if necessary.
Something I don't understand is the tangent line stuff, I do not even know how to start a question that asks about that. And it's all over the packets we just got!
Sunday, October 4, 2009
Ash's 7th post
Wow...is it merely the 7th week of school? It feels as if I should be going into the Christmas Holidays in about 2 weeks......not in the middle of October.
Alright, this week....Studying and Study Guides. But I do remember Mrs. Robinson saying something about us being able to explain something from past weeks right? Good!
How about I get what I don't understand out of the way right now?
1. STILL!! Optimization
2. Tangent Lines [[I get the general gist of it, just...not all of it]]
First Derivative Test (I'm sorry if everyone understands it, but I don't want to look dumb by explaining something I don't get at all wrong)
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
Also, for the powerpoint tomorrow:
1. If I downloaded fonts from the Internet, will they show up on Mrs. Robinson's computer?
2. Will someone be clicking the next slide button or does it have to be set on a timer?
Thanks!! =]
Alright, this week....Studying and Study Guides. But I do remember Mrs. Robinson saying something about us being able to explain something from past weeks right? Good!
How about I get what I don't understand out of the way right now?
1. STILL!! Optimization
2. Tangent Lines [[I get the general gist of it, just...not all of it]]
First Derivative Test (I'm sorry if everyone understands it, but I don't want to look dumb by explaining something I don't get at all wrong)
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
Also, for the powerpoint tomorrow:
1. If I downloaded fonts from the Internet, will they show up on Mrs. Robinson's computer?
2. Will someone be clicking the next slide button or does it have to be set on a timer?
Thanks!! =]
Post 7
This week in Calculus we pretty much just reviewed and had a quiz. At the end of last week and the beginning of this week, I was still confused with optimization. By the time of the quiz, I finally caught on to optimization. This week we were also given studyguides to work on right up until our exam. They're pretty big, but we'll get through them.
So first I didn't understand optimization and where the equasions came from, or even how to decide which equation was to be maximized. I kept on confusing myself and going in circles with the problems. After we were given the list for optimization "in English" I understood it a lot better. It's really simple when you know what you're looking for
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)
So far looking through the packets, I've only gotten to the limits. I don't understand how to look at a graph and find what the limit is. Can someone please explain?
So first I didn't understand optimization and where the equasions came from, or even how to decide which equation was to be maximized. I kept on confusing myself and going in circles with the problems. After we were given the list for optimization "in English" I understood it a lot better. It's really simple when you know what you're looking for
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)
So far looking through the packets, I've only gotten to the limits. I don't understand how to look at a graph and find what the limit is. Can someone please explain?
post 7
most of the week that just passed was studying for the quiz on wednesday and on wednesday we got the study guides for the first nine weeks exam. the study guides include optimization, first and second der tests, and tangent lines.
A Tangent Line is a line which touches a curve at one and only one point. The slope-intercept formula for a line is y = mx + b,where m is the slope of the line and b is the y-intercept.
The point-slope formula for a line is y – y1 = m (x – x1).This formula uses a point on the line, denoted by (x1, y1),and the slope of the line, denoted by m, to calculate the slope-intercept formula for the line.
The first derivative is an equation for the slope of a tangent line to a curve at an indicated point.
to find max and mins you use first der test then plug the critical values into original functions to get y values then plug endpoints in to original function to get y values the highest y value is the absolute max and the lowest y value is the absolute min.
i know both theorems we learned a while ago. Mean value theorem and rolle's theorem.
im somewat confused on optimization and knowing which eqn is the secondary and which one is thee primary. other than that im good
A Tangent Line is a line which touches a curve at one and only one point. The slope-intercept formula for a line is y = mx + b,where m is the slope of the line and b is the y-intercept.
The point-slope formula for a line is y – y1 = m (x – x1).This formula uses a point on the line, denoted by (x1, y1),and the slope of the line, denoted by m, to calculate the slope-intercept formula for the line.
The first derivative is an equation for the slope of a tangent line to a curve at an indicated point.
to find max and mins you use first der test then plug the critical values into original functions to get y values then plug endpoints in to original function to get y values the highest y value is the absolute max and the lowest y value is the absolute min.
i know both theorems we learned a while ago. Mean value theorem and rolle's theorem.
im somewat confused on optimization and knowing which eqn is the secondary and which one is thee primary. other than that im good
this week was mostly going over optimization in preparation for the quiz on wednesday..friday we got our study guides for the exam.
Some of the stuff on the study guides are tangent lines and using the first/second derivative test.
A Tangent Line is a line which locally touchesa curve at one and only one point.• The slope-intercept formula for a line is y = mx + b,where m is the slope of the line and b is the y-intercept.
• The point-slope formula for a line is y – y1 = m (x – x1).This formula uses a point on the line, denoted by (x1, y1),and the slope of the line, denoted by m, tocalculate the slope-intercept formula for the line.
• The first derivative is an equation for the slope of a tangentline to a curve at an indicated point.The equation for the slope of the tangent line tof(x) = x2 is f '(x), the derivative of f(x).f(x) = x2f '(x) = 2x (1)Therefore, at x = 2, the slope of the tangent line is f '(2).f '(2) = 2(2)= 4
Now , you know the slope of the tangent line, which is 4.All that you need now is a point on the tangent line to beable to formulate the equation. To find that point, simply plugthe coordinate of the shared point into the original equation, this gives you (2,4)The only step left is to use the point (2, 4) and slope, 4,in the point-slope formula for a line. Therefore: Y-4=4(x-2)
TO FIND MAX AND MINS
1. first derivative test
2. plug the critical values into origional function to get y-values.
3. plug endpoints in to origional function to get y-values.
4. highest y-value is absolute max.
5. lowest y-value is absolute min.
-absolute maxs or mins or written as a point or simply as the y-value.
For example:find the absolute max or min of f(x)=3x^4-4x^ on [-1,2].f1(x)=12x^3-12x^2=012x^2(x-1)=0x=1,0(-1, 0)U(0,1)U(1,2)f1(-.5)=-ve f1(.5)=-ve f1(1.5)=+ve
min @x=13(1)^4-4(1)^3=-1 (1,-1)3(-1)^4-4(-1)^3=-7 (-1,-7)3(2)^4-4(2)^3 (2,16)-32=16
abs min:(1,-1)or -1 -1 at x=1
i still don't fully understand optimization as a whole, anyone want to explain from the beginning?
Some of the stuff on the study guides are tangent lines and using the first/second derivative test.
A Tangent Line is a line which locally touchesa curve at one and only one point.• The slope-intercept formula for a line is y = mx + b,where m is the slope of the line and b is the y-intercept.
• The point-slope formula for a line is y – y1 = m (x – x1).This formula uses a point on the line, denoted by (x1, y1),and the slope of the line, denoted by m, tocalculate the slope-intercept formula for the line.
• The first derivative is an equation for the slope of a tangentline to a curve at an indicated point.The equation for the slope of the tangent line tof(x) = x2 is f '(x), the derivative of f(x).f(x) = x2f '(x) = 2x (1)Therefore, at x = 2, the slope of the tangent line is f '(2).f '(2) = 2(2)= 4
Now , you know the slope of the tangent line, which is 4.All that you need now is a point on the tangent line to beable to formulate the equation. To find that point, simply plugthe coordinate of the shared point into the original equation, this gives you (2,4)The only step left is to use the point (2, 4) and slope, 4,in the point-slope formula for a line. Therefore: Y-4=4(x-2)
TO FIND MAX AND MINS
1. first derivative test
2. plug the critical values into origional function to get y-values.
3. plug endpoints in to origional function to get y-values.
4. highest y-value is absolute max.
5. lowest y-value is absolute min.
-absolute maxs or mins or written as a point or simply as the y-value.
For example:find the absolute max or min of f(x)=3x^4-4x^ on [-1,2].f1(x)=12x^3-12x^2=012x^2(x-1)=0x=1,0(-1, 0)U(0,1)U(1,2)f1(-.5)=-ve f1(.5)=-ve f1(1.5)=+ve
min @x=13(1)^4-4(1)^3=-1 (1,-1)3(-1)^4-4(-1)^3=-7 (-1,-7)3(2)^4-4(2)^3 (2,16)-32=16
abs min:(1,-1)or -1 -1 at x=1
i still don't fully understand optimization as a whole, anyone want to explain from the beginning?
Week 7
Wow, I cannot believe we are already in the seventh week of school. It seems like just yesterday we were walking into the first day of class scared of how hard calculus may be. Anyways, this week we reviewed optimization on Monday and Tuesday. On Wednesday we had a quiz on optimization and recieved our review packets for exams. On Thursday we had a field trip, and on Friday we worked on our packets.
So far I have understood most of the things in our packets.
I understand how to take derivatives.
Example:
9xsin+4cox
you would use product and rule and the addition property to solve this problem.
= [9x(cosx) + sinx(9)] + 4(-sinx)
= 9xcosx + 5sinx
I also understand that slope is the same thing as derivative
I have recently discoved that a higher-order derivative is just that you keep taking derivatives until you are told to stop.
Example:
f ''(x) = 2x^(7/5), find f^(iv)(x)
first you would take the derivative of f ''(x) and get (14/5)x^(2/5). Then you would take the derivative of that to get (28/25)x^(-3/5).
Also, if it tells you to find a derivative a certain point, then take the derivative and plug the point in for x.
We copied down the Intermediate Value Theorem on Friday. I do not know how to solve a problem with it.
Example:
Find the value of c guaranteed by the Intermediate Value Theorem.
f(x) = x^2 - 2x - 3, [4,8], f(c) = 12.
Can anybody help me with that? Thanks.
So far I have understood most of the things in our packets.
I understand how to take derivatives.
Example:
9xsin+4cox
you would use product and rule and the addition property to solve this problem.
= [9x(cosx) + sinx(9)] + 4(-sinx)
= 9xcosx + 5sinx
I also understand that slope is the same thing as derivative
I have recently discoved that a higher-order derivative is just that you keep taking derivatives until you are told to stop.
Example:
f ''(x) = 2x^(7/5), find f^(iv)(x)
first you would take the derivative of f ''(x) and get (14/5)x^(2/5). Then you would take the derivative of that to get (28/25)x^(-3/5).
Also, if it tells you to find a derivative a certain point, then take the derivative and plug the point in for x.
We copied down the Intermediate Value Theorem on Friday. I do not know how to solve a problem with it.
Example:
Find the value of c guaranteed by the Intermediate Value Theorem.
f(x) = x^2 - 2x - 3, [4,8], f(c) = 12.
Can anybody help me with that? Thanks.
post 7
this week in calculus we reviewed and are getting ready for our exam by doing our beastfull study guide.
Things on the study guide that i understand include the first and second derivitive test and optimization.
First Derivative test-determine if it is continious and differentiable on the interval. If it is then you take the first derivitve and set equal to zero. Once set equal to zero solve for x and set your x into intervals. Then from there you plug in and find your max and mins.
Second derivative Test- determine if it is continious and defferentiable, if not then it doesnt work. take first derivative then take the second derivative and set the second equal to zero. Solve for x. Set up your x intervals and then plug in from within those intervals into the equation. Then you find your concavity, concave up and concave down.
Optimization - determine primary equation and then determine secondary and solve the secondary for an equation. Once solved plug it into the primary equation and then solve for that variable and then take the derivative and set equal to zero. Once set equal to zero you then plug back into the primary equatin and get your answers.
I still dont fully understand how to look at a graph and determine the derivative and the lim. The graph stuff messes me up.
Things on the study guide that i understand include the first and second derivitive test and optimization.
First Derivative test-determine if it is continious and differentiable on the interval. If it is then you take the first derivitve and set equal to zero. Once set equal to zero solve for x and set your x into intervals. Then from there you plug in and find your max and mins.
Second derivative Test- determine if it is continious and defferentiable, if not then it doesnt work. take first derivative then take the second derivative and set the second equal to zero. Solve for x. Set up your x intervals and then plug in from within those intervals into the equation. Then you find your concavity, concave up and concave down.
Optimization - determine primary equation and then determine secondary and solve the secondary for an equation. Once solved plug it into the primary equation and then solve for that variable and then take the derivative and set equal to zero. Once set equal to zero you then plug back into the primary equatin and get your answers.
I still dont fully understand how to look at a graph and determine the derivative and the lim. The graph stuff messes me up.
post #7
This week we didn't really learn anything new. On Monday and Tuesday we reviewed optimization. Then on Wednesday we took a quiz on it, then after the quiz we got our study guide for the exam which we will work on until the exam day. Thursday we went on a field trip, and Friday we worked on our packets.
lets go over the rules for finding a limit that approaches inifinty :)
1. if the top degree is equal to the bottom, divide leading coefficients.
2. if top degree is greater than bottom degree, it's infinity
3. if the top top degree is less than bottom degree, it's zero.
Taking derivatives are also all over the calc packet. like product rule, quotient rule, and all the other 36 rules we had to copy.
product rule:
f(x)= x^3cos(3x+4)
f'(x)=3x^2)(cos(3x+4)+(x^3)(-sin(3x+4)(3)
then simplify
one thing i still don't understand is optimization. if they don't give you any problems, how do you know what the primary and secondary equations are! im so confused. i also need help with tangent lines
lets go over the rules for finding a limit that approaches inifinty :)
1. if the top degree is equal to the bottom, divide leading coefficients.
2. if top degree is greater than bottom degree, it's infinity
3. if the top top degree is less than bottom degree, it's zero.
Taking derivatives are also all over the calc packet. like product rule, quotient rule, and all the other 36 rules we had to copy.
product rule:
f(x)= x^3cos(3x+4)
f'(x)=3x^2)(cos(3x+4)+(x^3)(-sin(3x+4)(3)
then simplify
one thing i still don't understand is optimization. if they don't give you any problems, how do you know what the primary and secondary equations are! im so confused. i also need help with tangent lines
Post 7
Another week down and first nine weeks exams are creeping up on us. This means it is study guide time.
In the study guide there are limits. The limit rules are: 1) if the highest exponent is the same on the top and bottom then the limit is the top coefficient over the bottom coefficient of the highest exponents. 2) If the highest exponent is on the top then the limit is infinity. 3) But if the highest exponent is on the bottom then the limit is 0.
Some examples are:
lim->infinity (3x^2-2x)/(2x^2+4x-1) Then the limit is 3/2 because it follows rule 1.
lim->infinity (8x^4+6)/(3x-9) Then the limit is infinity because it follows rule 2.
lim->infinity (x-7)/(5x^3+4x+2) Then the limit is 0 because it follows rule 3.
Also the study guide reviews taking derivivatives. That means using the product rule, quotient rule, and the regular way of taking the derivative of an equation.
Product rule is copy the first times the derivative of the last plus copy the first times the derivative of the second.
Quotient rule is copy the bottom times derivative of the top minus copy the top times the derivative of the bottom all divided by the bottom squared.
But there are some things I do not remember how to do or never did. I do not remember how to do the tangent line things. And I still do not fully understand optimization. Help with these 2 things would be great so that I will be ready for the exam.
In the study guide there are limits. The limit rules are: 1) if the highest exponent is the same on the top and bottom then the limit is the top coefficient over the bottom coefficient of the highest exponents. 2) If the highest exponent is on the top then the limit is infinity. 3) But if the highest exponent is on the bottom then the limit is 0.
Some examples are:
lim->infinity (3x^2-2x)/(2x^2+4x-1) Then the limit is 3/2 because it follows rule 1.
lim->infinity (8x^4+6)/(3x-9) Then the limit is infinity because it follows rule 2.
lim->infinity (x-7)/(5x^3+4x+2) Then the limit is 0 because it follows rule 3.
Also the study guide reviews taking derivivatives. That means using the product rule, quotient rule, and the regular way of taking the derivative of an equation.
Product rule is copy the first times the derivative of the last plus copy the first times the derivative of the second.
Quotient rule is copy the bottom times derivative of the top minus copy the top times the derivative of the bottom all divided by the bottom squared.
But there are some things I do not remember how to do or never did. I do not remember how to do the tangent line things. And I still do not fully understand optimization. Help with these 2 things would be great so that I will be ready for the exam.
Post #7
This week in Calculus we did optimization and reviewed. So, that’s what I’m going to do..explain things I now understand from the beginning of the year, and say I still don’t understand how you know what is the primary and secondary functions for optimization.
So, FIRST DERIVATIVE TEST: let c be a critical point of a function, f, that is continuous on an open interval containing c. If it is differentiable on the interval, except possibly at c, then f[c] can be classified as either:relative min [negative to positive]relative max [positive to negative]When using the first derivative test, you take the derivative of the function and set it equal to zero and solve for x. Then you set up your x values into intervals to see which ones are max and mins
SECOND DERIVATIVE TEST: concavity & points of refection. When using the second derivative test, you take the derivative of a function two times and set equal to zero and solve for x. You put these x values into intervals as well to find if a graph is concave up or down, point of inflection, etc.
Also, I understand the LIMIT RULES…Remember, you find vertical asymptotes when you set the bottom of a fraction equal to zero, then solve. After you factor the top and bottom of the fraction, if there is anything you can cancel from a function, it is a removable. Another thing i understand better is how to use the first and second derivative test.
I need help with tangent lines and optimization when figuring out which formula is what…can anyone help?
So, FIRST DERIVATIVE TEST: let c be a critical point of a function, f, that is continuous on an open interval containing c. If it is differentiable on the interval, except possibly at c, then f[c] can be classified as either:relative min [negative to positive]relative max [positive to negative]When using the first derivative test, you take the derivative of the function and set it equal to zero and solve for x. Then you set up your x values into intervals to see which ones are max and mins
SECOND DERIVATIVE TEST: concavity & points of refection. When using the second derivative test, you take the derivative of a function two times and set equal to zero and solve for x. You put these x values into intervals as well to find if a graph is concave up or down, point of inflection, etc.
Also, I understand the LIMIT RULES…Remember, you find vertical asymptotes when you set the bottom of a fraction equal to zero, then solve. After you factor the top and bottom of the fraction, if there is anything you can cancel from a function, it is a removable. Another thing i understand better is how to use the first and second derivative test.
I need help with tangent lines and optimization when figuring out which formula is what…can anyone help?
Post #7
This week in calculus we did optimization for the first three days, had a field trip Thursday, and worked on our review packets Friday so we really did not learn anything new. I am going to review how to find the maxs and mins using the shortcut method used for multiple choice questions only because I need a review on how to do it so yall might too.
The first step is to take the derivative of the equation given. Then you set the derivative equal to zero and solve for x. After, you take the second derivative and plug in each of your x's into it. If your answer turns out to be a positive number then it means the second derivative is concave up and it is a min. If it is a negative number, it means it is concave down and it is a max. If it equals zero, it means it is neither a max nor min.
EXAMPLE:
f(x) = -3x^5 + 5x^3
f'(x) = -15x^4 + 15x^2
= -15x^2 (x^2-1)
= -15x^2 (x+1) (x-1) = 0
x= -1, 0, 1
f''(x)= -60x^3 + 30x
-1: -60 (-1)^3 + 30 (-1) = positive; concave up
0: -60 (0)^3 + 30 (0) = 0
1: -60 (1)^3 + 30 (1) = negative; concave down
Therefore, there is a min at x= -1 and a max at x=1
I am having trouble with how to find a tangent line to a graph. I get stuck after I take the derivative. I used to know how to do it, but I forgot. If anybody can help that would be great.
The first step is to take the derivative of the equation given. Then you set the derivative equal to zero and solve for x. After, you take the second derivative and plug in each of your x's into it. If your answer turns out to be a positive number then it means the second derivative is concave up and it is a min. If it is a negative number, it means it is concave down and it is a max. If it equals zero, it means it is neither a max nor min.
EXAMPLE:
f(x) = -3x^5 + 5x^3
f'(x) = -15x^4 + 15x^2
= -15x^2 (x^2-1)
= -15x^2 (x+1) (x-1) = 0
x= -1, 0, 1
f''(x)= -60x^3 + 30x
-1: -60 (-1)^3 + 30 (-1) = positive; concave up
0: -60 (0)^3 + 30 (0) = 0
1: -60 (1)^3 + 30 (1) = negative; concave down
Therefore, there is a min at x= -1 and a max at x=1
I am having trouble with how to find a tangent line to a graph. I get stuck after I take the derivative. I used to know how to do it, but I forgot. If anybody can help that would be great.
Post Number Seven
So there goes the seventh week of calculus.
This past week after taking the quiz on optimization we were given our study guides for the exam. By the way I still need some help on optimization so if anyone has gotten the hang of that please let me know if you are willing to teach me. I'm finally getting the hang of everything, especially derivatives and tangent lines.
I'll do an example of a tangent line for you:
Write the equation of the tangent line to the graph of g at x=3. g(x)=x^2 - 4x + 9
First of all, you are given an x value. To find the y value, you simply just plug in the x into the original equation.
g(3)=(3)^2 -4(3) + 9
g(3) = 6
This gives you the point (3,6) which you will use later to write the equation.
Now you have to find the slope. This just means to take the derivative and then plug in the given x.
g'(x)=2x-4
g'(3)=2(3) - 4
g'(3)=2
Now that you have a point and a slope, guess what you do now..
Put it in point slope form :)
y-y1=m(x - x1)
so you're tangent line would be y-6=2(x-3)
Another problem I have is limits (aka chapter 1). If anyone is able to help guide me with that packet it'd be greatly appreciated. I know all the rules so don't comment me the rules, i just can't put them into play..
Oh and also remember that when it asks you at what value is the slope of the graph equal to a number, all you do is take the derivative and set it equal to that number then solve for x.
Anyways, I still have a lot of work to do and always accepting help.
This past week after taking the quiz on optimization we were given our study guides for the exam. By the way I still need some help on optimization so if anyone has gotten the hang of that please let me know if you are willing to teach me. I'm finally getting the hang of everything, especially derivatives and tangent lines.
I'll do an example of a tangent line for you:
Write the equation of the tangent line to the graph of g at x=3. g(x)=x^2 - 4x + 9
First of all, you are given an x value. To find the y value, you simply just plug in the x into the original equation.
g(3)=(3)^2 -4(3) + 9
g(3) = 6
This gives you the point (3,6) which you will use later to write the equation.
Now you have to find the slope. This just means to take the derivative and then plug in the given x.
g'(x)=2x-4
g'(3)=2(3) - 4
g'(3)=2
Now that you have a point and a slope, guess what you do now..
Put it in point slope form :)
y-y1=m(x - x1)
so you're tangent line would be y-6=2(x-3)
Another problem I have is limits (aka chapter 1). If anyone is able to help guide me with that packet it'd be greatly appreciated. I know all the rules so don't comment me the rules, i just can't put them into play..
Oh and also remember that when it asks you at what value is the slope of the graph equal to a number, all you do is take the derivative and set it equal to that number then solve for x.
Anyways, I still have a lot of work to do and always accepting help.
Post #7
Soo...what to say for the seventh week of calculus. I guess this week was a review week. Monday and Tuesday were basically review of optimization, then Wednesday we took a quiz on optimazation [which hopefully i did okay on] and thursday everyone went to the river center to look at colleges, then friday we worked on the packets that will help us for our nine week exams! [91 pages!]
so lets review shall we..
for the FIRST DERIVATIVE TEST you let c be a critical point of a function f taht is continuous on an open interval containing c. if it is differentiable on the interval, except possibly at c, then f[c] can be classified as either:
relative min [negative to positive]
relative max [positive to negative]
for the SECOND DERIVATIVE TEST you are dealing with concavity, & points of refection. remember though that there is a shortcut for the max/min. Also the points of inflection only happen if there is a change in concavity.
for ABSOLUTE MAX OR MIN you will do the first derivative test, plug in the critical valuse you get when you solve the first derivative, then plug in the endpoints into the original.
and now for OPTIMIZATION..umm i understand how to do the steps: you take the equations given and solve for one value in the secondary equation, then your answer will be pluged in the primary equation to bo solved for the variable and whatever you get you plug it into the secondary equation in order to get the other value needed. the values you get..are your answer! HOWEVER, WHAT I DON'T GET is how exactly to figure out what is your primary and secondary equation!
other than that..i guess it's just knowing how to take derivatives and sloving which i'm pretty good at i just mess up simple mistakes
so lets review shall we..
for the FIRST DERIVATIVE TEST you let c be a critical point of a function f taht is continuous on an open interval containing c. if it is differentiable on the interval, except possibly at c, then f[c] can be classified as either:
relative min [negative to positive]
relative max [positive to negative]
for the SECOND DERIVATIVE TEST you are dealing with concavity, & points of refection. remember though that there is a shortcut for the max/min. Also the points of inflection only happen if there is a change in concavity.
for ABSOLUTE MAX OR MIN you will do the first derivative test, plug in the critical valuse you get when you solve the first derivative, then plug in the endpoints into the original.
and now for OPTIMIZATION..umm i understand how to do the steps: you take the equations given and solve for one value in the secondary equation, then your answer will be pluged in the primary equation to bo solved for the variable and whatever you get you plug it into the secondary equation in order to get the other value needed. the values you get..are your answer! HOWEVER, WHAT I DON'T GET is how exactly to figure out what is your primary and secondary equation!
other than that..i guess it's just knowing how to take derivatives and sloving which i'm pretty good at i just mess up simple mistakes
Post #7
So, week seven is complete. I'm finally getting the hang of stuff I had trouble with at the begining of the year. I'm ready to review and do allll those packets in class! haha
Derivatives are finally easy. So, I'm going to show a few examples of those:
1. f(x)=x^5tanx (this would be product rule)
x^5sec^2x+5x^4tanx
2. y=cos(2x^5+3) (this would NOT be product rule)
-sin(2x^5+3)*10x^4
-10x^4sin(2x^5+3)
Also, anything that says find the slope or derivative of the function, at a point or x-value is so easy too. Just take the derivative and then plug in the x-value. Now for an example:
1. Find the slope of the graph of the function f(x)=x(3x^4+2) at x=5.
x(3^4+2)
x(12x^3)+(3x^4+2)(1)
12x^4+3x^4+2
12(5)^4+3(5)^4+2= 9377
I'm still having trouble on optimization. I plan on getting someone to help/teach me with that. But my real question is I forgot how to find an equation of the tangent line. Can someone help me on that?
Derivatives are finally easy. So, I'm going to show a few examples of those:
1. f(x)=x^5tanx (this would be product rule)
x^5sec^2x+5x^4tanx
2. y=cos(2x^5+3) (this would NOT be product rule)
-sin(2x^5+3)*10x^4
-10x^4sin(2x^5+3)
Also, anything that says find the slope or derivative of the function, at a point or x-value is so easy too. Just take the derivative and then plug in the x-value. Now for an example:
1. Find the slope of the graph of the function f(x)=x(3x^4+2) at x=5.
x(3^4+2)
x(12x^3)+(3x^4+2)(1)
12x^4+3x^4+2
12(5)^4+3(5)^4+2= 9377
I'm still having trouble on optimization. I plan on getting someone to help/teach me with that. But my real question is I forgot how to find an equation of the tangent line. Can someone help me on that?
Posting...#7
At the begging of the week we went over Optimization to get ready for our test that’s basically all we did for the first three days of the week But after the test Wednesday that I did bad on I decided to crack down on everything now I will do all my homework and everything so I will have to know the stuff. But anyways after the test I bombed I was relieved that the pain was finally over until I got about 6 packets of 10 pages a piece. Im kind of not mad about these packets because I know after I do them which I will actually do them this time I will know everything for the exam. So all these packets don’t bother me to much.
How to find Maxes an MIns first you do the first derivative test then plug the critical values into origional function to get y-values.next you want to plug endpoints in to origional function to get y-values.which gives you highest y-value is absolute max. lowest y-value is absolute min.
I’m still not a expert on optiomization so I need a little practice as well as a little help I will probally do that part of the study guide first cause its fresh in my head so if any of yall have any tips that will help me learn it better; please leave a comment J thanks everyone
Im posting this late because RTC is the worst telephone/ ezer-net provide ever.
How to find Maxes an MIns first you do the first derivative test then plug the critical values into origional function to get y-values.next you want to plug endpoints in to origional function to get y-values.which gives you highest y-value is absolute max. lowest y-value is absolute min.
I’m still not a expert on optiomization so I need a little practice as well as a little help I will probally do that part of the study guide first cause its fresh in my head so if any of yall have any tips that will help me learn it better; please leave a comment J thanks everyone
Im posting this late because RTC is the worst telephone/ ezer-net provide ever.
7th post
okay so the seventh week of calculus has passed and im stressing out so bad. this week we just reviewed and got packets to do.
REVIEW:
1. FIRST DERIVATIVE TEST: let c be a critical point of a function f that is continuous on a open interval containing c. If it is differentiable on the interval, except possibly at c, then f(c) van be classified as follows:
a.relative minimum if f1(x)changes from negative to positive at c.
b.relative maximum if f1(x)changes from positive to negative at c.
2. SECOND DERIVATIVE TEST: 1.points of reflection 2. intervals of concavity 3. short cut:max/min (multiple choice)4. beware points of inflection only happen if there is a change in concavity.
3. TO FIND ABSOLUTE MAX OR MIN: 1.first derivative test 2.plug in critical values to get y-values 3.plug endpoints into origional function 4.highest y-value 5.lowest y-value.
4. OPTIMIZATION: if your looking for the closest your looking for the minimum, cannot take the sq.rt of addition, ignore the sq.rt of the smallest number, for this you dont need to find points, distance formula, problem>secondary.
a. identify primary and secondary (primary the one your maximizing or minimizing)(secodary the other one)
b. solve secondary for 1 variable and plug into primary
c.take derivative
d. plug into secondary equation.
okay im having problem on the packets especially the chapter one so if anyone can kind of explain that like whats on it that would help me.
REVIEW:
1. FIRST DERIVATIVE TEST: let c be a critical point of a function f that is continuous on a open interval containing c. If it is differentiable on the interval, except possibly at c, then f(c) van be classified as follows:
a.relative minimum if f1(x)changes from negative to positive at c.
b.relative maximum if f1(x)changes from positive to negative at c.
2. SECOND DERIVATIVE TEST: 1.points of reflection 2. intervals of concavity 3. short cut:max/min (multiple choice)4. beware points of inflection only happen if there is a change in concavity.
3. TO FIND ABSOLUTE MAX OR MIN: 1.first derivative test 2.plug in critical values to get y-values 3.plug endpoints into origional function 4.highest y-value 5.lowest y-value.
4. OPTIMIZATION: if your looking for the closest your looking for the minimum, cannot take the sq.rt of addition, ignore the sq.rt of the smallest number, for this you dont need to find points, distance formula, problem>secondary.
a. identify primary and secondary (primary the one your maximizing or minimizing)(secodary the other one)
b. solve secondary for 1 variable and plug into primary
c.take derivative
d. plug into secondary equation.
okay im having problem on the packets especially the chapter one so if anyone can kind of explain that like whats on it that would help me.
Post #7
I didn't get to post my last one because the internet people fail. Specifically the AT&T people internet. At the moment, I am at PJs doing my blog when I could be spending time with my MiMi. But that's besides the point.
If I do recall correctly, we basically worked on optimization (prepping for the quiz) and that big study guide (which I'm getting along with at the moment). After stressing for, I don't know, 2 days for the OPTIMIZATION quiz (and putting "WTF?" by a lot of the attempted problems), I decided I was just going to stare at the paper until the information manifested in my head. That may have not been the exact case, but I don't think I did terribly bad on the quiz after all was said and done. I believe I actually did pretty good (remains to be seen...). Oh and by the way, I intend on asking every single little tiny question I have, so be prepared because I am so not failing the exam.
So, let's put my decent optimization skills to the test...yea right...and do an example problemo!!!! (Excessive Use of Exclamation Points without Real Enthusiasm).
Alrighty, I'm not going to do one. I changed my mind because I have to have the picture to work a problem like that, and I'd have to actually paint up one. Negative.
So, how about we do an example from the study guide...good idea? Maybe!??
Find an equation to the tangent line to the graph of the function f(x)=tan^8(x) at the point (7pi/5, 8049.922). The coefficients below are given to two decimal places.
So first to find the slope you take the derivative of the the tan thingy. You get:
8tan^7(x) * sec^2(x)----------take care of the exp. first then multiply by the der. of tan.
Now you must plug in your x value from the point given in order to find the actually numerical value slope. Which gives you 219,125.53. Now since you have a slope and a point, what do you do? Oh, that's right! Slope intercept formula!
y - 8049.922 = 219125.53 ( x - 252)-------the 252 is 7pi/5 in degrees.
and then you solf for y giving you:
y = 219125.53x - 955714.52
which is choice C!
If I do recall correctly, we basically worked on optimization (prepping for the quiz) and that big study guide (which I'm getting along with at the moment). After stressing for, I don't know, 2 days for the OPTIMIZATION quiz (and putting "WTF?" by a lot of the attempted problems), I decided I was just going to stare at the paper until the information manifested in my head. That may have not been the exact case, but I don't think I did terribly bad on the quiz after all was said and done. I believe I actually did pretty good (remains to be seen...). Oh and by the way, I intend on asking every single little tiny question I have, so be prepared because I am so not failing the exam.
So, let's put my decent optimization skills to the test...yea right...and do an example problemo!!!! (Excessive Use of Exclamation Points without Real Enthusiasm).
Alrighty, I'm not going to do one. I changed my mind because I have to have the picture to work a problem like that, and I'd have to actually paint up one. Negative.
So, how about we do an example from the study guide...good idea? Maybe!??
Find an equation to the tangent line to the graph of the function f(x)=tan^8(x) at the point (7pi/5, 8049.922). The coefficients below are given to two decimal places.
So first to find the slope you take the derivative of the the tan thingy. You get:
8tan^7(x) * sec^2(x)----------take care of the exp. first then multiply by the der. of tan.
Now you must plug in your x value from the point given in order to find the actually numerical value slope. Which gives you 219,125.53. Now since you have a slope and a point, what do you do? Oh, that's right! Slope intercept formula!
y - 8049.922 = 219125.53 ( x - 252)-------the 252 is 7pi/5 in degrees.
and then you solf for y giving you:
y = 219125.53x - 955714.52
which is choice C!
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