Well another week down in calculus lol.. let's go over some easy stuff first:
Tangent lines:
The problem will give you a function and an x value. Sometimes they may give you a y value; if not then you plug the x value into the original function and solve for y to get the y value. Next, you take the derivative of the function and plug in x to get the slope. After that, you plug everything into point-slope form.
First derivative test:
For the first derivative test, you are solving for max and mins and may be trying to see where the graph is increasing and decreasing. You take the derivative of the function and and set it equal to zero and solve for the x values (critical points). Then you set those points up into intervals between negative infinity and infinity. Then, you plug in numbers between those intervals to see if it is positive or negative.
Second derivative test:
For the second derivative test, you are solving to see whether the graph is concave up, concave down, or where there is a point of inflection in the graph. You take the derivative of the function twice and set it equal to zero and solve for the x values. You set those values up into intervals between negative infinity and infinity. You then plug in numbers between those intervals to see if it is positive or negative. If it is positive, it is concave up. If it is negative it is concave down. Where there is a change in concavity, there is a point of inflection.
Some things i do not understand:
The problems where they give you a graph and you have to seperate them into triangles and rectangles to find the area.
Integrals with trig functions in them
Those problems where they give you two equations and you have two variables (most of the time a and b) and you have to solve for one and solve for the other and then set equal.
Hope you all have a great weekend :)
Saturday, March 6, 2010
Friday, March 5, 2010
Post something
Okay so for my blog this weekend...I don't really know what to post.. My mind is so boggled down atm from doing so many things at once today :o
Anyway, some tips and help on a few things...
If they give you a graph...that CLEARLY consists of triangle and rectangles...and it says find the area under the curve from 0 to some number...All you do is split it up into triangles and rectangle...find the area of each rectangle and triangle (remember, rectangle is length times width and triangle is 1/2 times length times width) and then add all the numbers on the top and subtract any numbers from below the axis from that...and there you go, you just got easy points ;-).
Also, don't forget to be using your trig identities...for example
The problem on the test was some integral of
sin(2x)e^(sin^2(x))
Most of you set your u to sin^2(x) and took the derivative as 2sinxcosx....but you all stopped there! Don't stop, 2sinxcosx is actually in the problem. If you will, jump back into the time space continuum and remember the trig identities test and think...okay...sin...double angle formula...OH, i know. sin(2x)=2sinxcosx. So the whole time it was already there. So you can now proceed on with the problem e^u, integrate it and do whatever definite integral it wants you to do. do not forget to use trig identities.
Also, people people people. It gives you a function...and it asks where is the slope of the tangent line equal to 6...what do you do? Just take the derivative and set equal to 6 and solve...that one was a no-brainer that a LOT of you missed.
Let's see...what else...oh, yeah.
Instantaneous speed: take a derivative, plug in the number. Simple as that. Don't miss this please--this is like zomgfreemonies except its points...and it's on the AP.
Once again...I can not reiterate enough the following:
Original - Position
1st Derivative - Velocity
2nd Derivative - Acceleration
Also, displacement refers to the area under the position curve...so if it gives you a formula for displacement, i am pretty sure you take the derivative of that to get to position...so that will change things.
Original - Displacement
1st Derivative - Position
2nd Derivative - Velocity
3rd Derivative - Acceleration
4th Derivative - Jerk
and so on and so forth....so just remember where you are in the line of things and you will be fine by integrating or deriving... (zomg we integrate and take derivatives? crazy ... i didn't think calculus did that)
Anyway, I'm tired of typing and slightly delirious if you haven't noticed...
ZzZzZZZZzzzzzzzzz............
Anyway, some tips and help on a few things...
If they give you a graph...that CLEARLY consists of triangle and rectangles...and it says find the area under the curve from 0 to some number...All you do is split it up into triangles and rectangle...find the area of each rectangle and triangle (remember, rectangle is length times width and triangle is 1/2 times length times width) and then add all the numbers on the top and subtract any numbers from below the axis from that...and there you go, you just got easy points ;-).
Also, don't forget to be using your trig identities...for example
The problem on the test was some integral of
sin(2x)e^(sin^2(x))
Most of you set your u to sin^2(x) and took the derivative as 2sinxcosx....but you all stopped there! Don't stop, 2sinxcosx is actually in the problem. If you will, jump back into the time space continuum and remember the trig identities test and think...okay...sin...double angle formula...OH, i know. sin(2x)=2sinxcosx. So the whole time it was already there. So you can now proceed on with the problem e^u, integrate it and do whatever definite integral it wants you to do. do not forget to use trig identities.
Also, people people people. It gives you a function...and it asks where is the slope of the tangent line equal to 6...what do you do? Just take the derivative and set equal to 6 and solve...that one was a no-brainer that a LOT of you missed.
Let's see...what else...oh, yeah.
Instantaneous speed: take a derivative, plug in the number. Simple as that. Don't miss this please--this is like zomgfreemonies except its points...and it's on the AP.
Once again...I can not reiterate enough the following:
Original - Position
1st Derivative - Velocity
2nd Derivative - Acceleration
Also, displacement refers to the area under the position curve...so if it gives you a formula for displacement, i am pretty sure you take the derivative of that to get to position...so that will change things.
Original - Displacement
1st Derivative - Position
2nd Derivative - Velocity
3rd Derivative - Acceleration
4th Derivative - Jerk
and so on and so forth....so just remember where you are in the line of things and you will be fine by integrating or deriving... (zomg we integrate and take derivatives? crazy ... i didn't think calculus did that)
Anyway, I'm tired of typing and slightly delirious if you haven't noticed...
ZzZzZZZZzzzzzzzzz............
Post #29
Another week down...
Let’s see what I remember shall we.. if you take the derivative of f^2(x) you treat it as if it was like sine. You do 2f(x) times f’(x) times 1. So it’s almost as if you have (f(x))^2 and you use the chain rule. Bring the 2 to the front, subtract 1, times by the derivative of the inside. And since the derivative of f(x) is f’(x) times the derivative of x, which is 1. These rules seem to help me A LOT!
Lets say you have:
x^2-2 DIVIDED BY 4-x^2 AS THE LIMIT APPROACHES 2
soo..
for this you end up getting two divided by zero. For finite limits though, if you get a number over zero, it’s always infinity. Therefore the answer is INFINITY!
Remember:
If the top exponent is greater than the bottom it’s the limit as it approaches infinity, and if the top exponent is less than the bottom it’s the limit as it approaches zero!
DON’T GET THEM CONFUSED LIKE I DID!
Here’s a trick:
Remember if you’re divided a number by another number and you have a bigger number on top it’s usually not zero. Also, if you’re dividing a number by another number and you have a smaller number on top it’s usually not a big number. Therefore, when the bigger number is on top since it’s a bigger number than zero it’s INFINITY and when the bigger number is on the bottom that means that it’s being lessened so it’s ZERO!
volume by disks
so you know that the formula is pi times the integral of the [function given] squared times dx. well then you gotta know what to do right so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one...oh, REMEMBER TO GRAPH..
just look at what they give you...they're should be numbers that they want them inbetween
IF NOT..WHAT DO YOU DO???
volume by washers
so you know that the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the inbetween number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. REMEMBER TO GRAPH!
just a reminder if it's for the x-axis then you solve for y and if it's on the y-axis then you solve for x..and if you plug the y-axis one's in your calculator to graph then you'll have to turn your calculator sideways with the screen on the right to see how it would really look.
area...it's worked the same just they'll give you two equations and if you don't have numbers for the inbetween then you'll set them equal to get them, but if you do then don't worry and keep going. for area there is no pi and no squaring. so you put the integral of the first equation minus the second equation and simplify and solve. then take the integral of it and plug in the numbers inbetween!
examples:
AREA:
y=-4x^2+41x+94 AND y=x-2 inbetween 7 and 1
so graph and then put the top equation over the bottom equation
the integral of (-4x^2+41x+94)-(x-2)dx
simplify: the integral of (-4x^2+40x+96) dx
((-4/3)(x)^3+20(x)^2+96(x)) of 1 and 7
f(7)=(3584/3) f(1)=(344/3)
ANSWER: 1080
VOLUME:
y=the sqr. root of (-2(x)^2-10(x)+48) inbetween 1 and -2
so graph and then put take the equation they give you into the integral and square it
PI times the integral of (-2(x)^2-10(x)+48) dx
PI [(-2/3)(x)^3-5(x)2+48(x)] of 1 and -2
f(1)=(127/3) f(-2)=(-332/3)
ANSWER: (153 PI)
hope this helps...
~ElliE~
Let’s see what I remember shall we.. if you take the derivative of f^2(x) you treat it as if it was like sine. You do 2f(x) times f’(x) times 1. So it’s almost as if you have (f(x))^2 and you use the chain rule. Bring the 2 to the front, subtract 1, times by the derivative of the inside. And since the derivative of f(x) is f’(x) times the derivative of x, which is 1. These rules seem to help me A LOT!
Lets say you have:
x^2-2 DIVIDED BY 4-x^2 AS THE LIMIT APPROACHES 2
soo..
for this you end up getting two divided by zero. For finite limits though, if you get a number over zero, it’s always infinity. Therefore the answer is INFINITY!
Remember:
If the top exponent is greater than the bottom it’s the limit as it approaches infinity, and if the top exponent is less than the bottom it’s the limit as it approaches zero!
DON’T GET THEM CONFUSED LIKE I DID!
Here’s a trick:
Remember if you’re divided a number by another number and you have a bigger number on top it’s usually not zero. Also, if you’re dividing a number by another number and you have a smaller number on top it’s usually not a big number. Therefore, when the bigger number is on top since it’s a bigger number than zero it’s INFINITY and when the bigger number is on the bottom that means that it’s being lessened so it’s ZERO!
volume by disks
so you know that the formula is pi times the integral of the [function given] squared times dx. well then you gotta know what to do right so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one...oh, REMEMBER TO GRAPH..
just look at what they give you...they're should be numbers that they want them inbetween
IF NOT..WHAT DO YOU DO???
volume by washers
so you know that the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the inbetween number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. REMEMBER TO GRAPH!
just a reminder if it's for the x-axis then you solve for y and if it's on the y-axis then you solve for x..and if you plug the y-axis one's in your calculator to graph then you'll have to turn your calculator sideways with the screen on the right to see how it would really look.
area...it's worked the same just they'll give you two equations and if you don't have numbers for the inbetween then you'll set them equal to get them, but if you do then don't worry and keep going. for area there is no pi and no squaring. so you put the integral of the first equation minus the second equation and simplify and solve. then take the integral of it and plug in the numbers inbetween!
examples:
AREA:
y=-4x^2+41x+94 AND y=x-2 inbetween 7 and 1
so graph and then put the top equation over the bottom equation
the integral of (-4x^2+41x+94)-(x-2)dx
simplify: the integral of (-4x^2+40x+96) dx
((-4/3)(x)^3+20(x)^2+96(x)) of 1 and 7
f(7)=(3584/3) f(1)=(344/3)
ANSWER: 1080
VOLUME:
y=the sqr. root of (-2(x)^2-10(x)+48) inbetween 1 and -2
so graph and then put take the equation they give you into the integral and square it
PI times the integral of (-2(x)^2-10(x)+48) dx
PI [(-2/3)(x)^3-5(x)2+48(x)] of 1 and -2
f(1)=(127/3) f(-2)=(-332/3)
ANSWER: (153 PI)
hope this helps...
~ElliE~
Thursday, March 4, 2010
Tangent Lines
Ok someone asked about this and I couldn't find the blog again... These are what we consider "gimme" questions. They usually don't have complicated derivatives, etc. In an non-implicit problem. You take the derivative of the equation given and then plug in the x-value given to get the slope. Then you plug the slope and x-value into point slope. You will need the y-value to finish the point-slope equation. If it is not given you find it by plugging the x-value into the ORIGINAL equation and solving for y. See several of the problems we have done in earlier APS. We had one in the diagnostic test I believe.
Integral of sin^2(x)
Ok so I am a little disappointed to hear that people are using the chain rule in integration. We have NEVER used the chain rule in integration. Your options are substitution and that is all. You can't substitute in this problem because the derivative of sin(x) is not in the integral. However, it is a trig function so you can use identities to manipulate it. For lack of being able to type mathematically here is a link to it already done.
http://wiki.answers.com/Q/Integral_of_sin_squared_x
Monday, March 1, 2010
post 28
Blogs.
The steps for related rates are:
1. Identify all of the variables and equations
2. Identify the things that you are looking for
3. Sketch a graph and then label that graph
4. Create and write an equation using all of the variables
5. Take the derivative of this equation with respect to time
6. Substitute everything back in
7. Solve the equation
The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
*delta x=b-a/number of subintervals*
SUBSTITUTION:
1. Find u & du
2. set u = whatever isn't the derivative
3. take the derivative of u
4. substitute back in
e integration:
very simple.
whatever e is raised to is your u, the derivative of u is du.
:)
IMPLICIT DERIVATIVE:
take derivative like normal, of both sides of the equation.
any derivative taken for y, mark with dy/dx behind it.
solve for dy/dx
The steps for related rates are:
1. Identify all of the variables and equations
2. Identify the things that you are looking for
3. Sketch a graph and then label that graph
4. Create and write an equation using all of the variables
5. Take the derivative of this equation with respect to time
6. Substitute everything back in
7. Solve the equation
The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
*delta x=b-a/number of subintervals*
SUBSTITUTION:
1. Find u & du
2. set u = whatever isn't the derivative
3. take the derivative of u
4. substitute back in
e integration:
very simple.
whatever e is raised to is your u, the derivative of u is du.
:)
IMPLICIT DERIVATIVE:
take derivative like normal, of both sides of the equation.
any derivative taken for y, mark with dy/dx behind it.
solve for dy/dx
post 28
This past week all we did was another set of AP tests and corrections.
The steps for related rates are:
1. Identify all of the variables and equations
2. Identify the things that you are looking for
3. Sketch a graph and then label that graph
4. Create and write an equation using all of the variables
5. Take the derivative of this equation with respect to time
6. Substitute everything back in
7. Solve the equation
The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
*delta x=b-a/number of subintervals*
Also I am going to talk about taking implicit derivatives. The steps for taking implicit derivatives are:
1. Take the derivative of both sides like you would normally do
2. Everytime the derivative of y is taken it needs to be notated with either y ' or dy/dx
3. Solve for dy/dx or y ' as if you are solving for x.
One problem am having is integrating something such as sin^2(x).
The steps for related rates are:
1. Identify all of the variables and equations
2. Identify the things that you are looking for
3. Sketch a graph and then label that graph
4. Create and write an equation using all of the variables
5. Take the derivative of this equation with respect to time
6. Substitute everything back in
7. Solve the equation
The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
*delta x=b-a/number of subintervals*
Also I am going to talk about taking implicit derivatives. The steps for taking implicit derivatives are:
1. Take the derivative of both sides like you would normally do
2. Everytime the derivative of y is taken it needs to be notated with either y ' or dy/dx
3. Solve for dy/dx or y ' as if you are solving for x.
One problem am having is integrating something such as sin^2(x).
post 28
all we did this last week was take an AP test, and then make corrections on it. and this three day weekend thing screwed me up, so i'm a day late on my blog, but anyways, i'm going to do this one on implicit derivatives because that's a super easy topic. :)
Implicit derivatives involve both x's and y's, unlike normal derivatives.
1: So, first you have to take the derivative of whatever they give you as you normally would.
2: Whenever you take the derivative of y, you have to note it with dy/dx.
3: Solve for dy/dx
(if you want to find slope plug in an x and y value)
example: y^3+y^2-5y-x^2=-4
First you just take the derivative, but don't forget to not the derivatives of the y's! So you get: 3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0
Then you have to solve for dy/dx, so you get:
dy/dx(3y^2+2y-5)=2x which then is further solved for to get dy/dx=2x/(3y^2+2y-5)
and that's it for that problem, it's done.
Another example:
Okay, let's say you want to find the slope of 3(x^2+y^2)^2=100xy at the point (3,1)
First you take the derivative, which involves all kinds of product and exponent rule...
6(x^2+y^2)(2x+2y(dy/dx))=100(y+x(dy/dx))
then, you need to foil it n stuff, so you get:
12x^3+12x^2(dy/dx)+12xy^2+12y^2(dy/dx))=100y+100x(dy/dx)
then, as usual, you would have to solve for dy/dx:
dy/dx=(-12^3-12xy^2+100y)/(12x^2+12y-100x)
after you solved for dy/dx, you plug in your x and y value from the point given to get your answer, so I think the final answer would be: 1.84 (if I put it in my calculator right)
that is it for implicit derivatives, and they are really easy to identify, it is the exact same thing as a derivative pretty much, just with x's and y's. Just don't forget to plug in the point that some problems will give you at the end, I have forgotten to do it before.
One thing that i still screw up on is MRAM, i don't know what my problem is, even when i look at the formula i still can't solve the problem, so if someone could review that, it would be good.
Implicit derivatives involve both x's and y's, unlike normal derivatives.
1: So, first you have to take the derivative of whatever they give you as you normally would.
2: Whenever you take the derivative of y, you have to note it with dy/dx.
3: Solve for dy/dx
(if you want to find slope plug in an x and y value)
example: y^3+y^2-5y-x^2=-4
First you just take the derivative, but don't forget to not the derivatives of the y's! So you get: 3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0
Then you have to solve for dy/dx, so you get:
dy/dx(3y^2+2y-5)=2x which then is further solved for to get dy/dx=2x/(3y^2+2y-5)
and that's it for that problem, it's done.
Another example:
Okay, let's say you want to find the slope of 3(x^2+y^2)^2=100xy at the point (3,1)
First you take the derivative, which involves all kinds of product and exponent rule...
6(x^2+y^2)(2x+2y(dy/dx))=100(y+x(dy/dx))
then, you need to foil it n stuff, so you get:
12x^3+12x^2(dy/dx)+12xy^2+12y^2(dy/dx))=100y+100x(dy/dx)
then, as usual, you would have to solve for dy/dx:
dy/dx=(-12^3-12xy^2+100y)/(12x^2+12y-100x)
after you solved for dy/dx, you plug in your x and y value from the point given to get your answer, so I think the final answer would be: 1.84 (if I put it in my calculator right)
that is it for implicit derivatives, and they are really easy to identify, it is the exact same thing as a derivative pretty much, just with x's and y's. Just don't forget to plug in the point that some problems will give you at the end, I have forgotten to do it before.
One thing that i still screw up on is MRAM, i don't know what my problem is, even when i look at the formula i still can't solve the problem, so if someone could review that, it would be good.
Ash's 28th Post
So it just hit me that it's Monday and not Sunday...I hate that.
Anyway, I'm going to explain some AP questions that I got right! Yippee!!
Non-Calculator Portion:
1. If g(x) = 1/32x^4-5x^2, find g'(4).
First step: take the derivative
1/8x^3-10x
Second Step: Plug in 4
1/8(4)^3-10(4)
Third Step: Solve
Answer = -32
5. Evaluate lim h->0 [5(1/2+h)^4-5(1/2)^4]/h
First Step: This is a definition of a derivative, so you take the derivative of 5(1/2)^4
4(5)(1/2)^3
Second Step: Solve
Answer = 5/2
Calculator Portion:
34. The graph of y=x^3-2x^2-5x+2 has a local maximum at:
First Step: Plug into your calculator
Second step: 2nd->Calc
Third Step: go to maximum
Forth Step: enter your bounds and then a "guess"
Answer: (-.786,4.209)
39. Find two non-negative numberfs x and y whose sum is 100 and for which (x^2)(y) is a maximum.
First step: Add up answer choices to make sure all equal 100
Second Step: Plug the x and y values into the given formula and see which is the highest number
Answer: x=66.667 and y=33.333
QUESTIONS:
How do you find the domain of a function without a calculator? I've always had trouble with domain and range even in Advanced Math.
How do you work piecewises? Number 7 is an example on the Non-calculator portion: Find k so that f(x) {(x^2-16)/x-4; x *does not equal* 4 and k; x=4 is continuous for all x. a) all real values, b)0, c)16, d)8, e) no real values
How do you take the derivative of a trig function squared? (sin^2(x) or cos^2(x))
Can someone go over the Mean Value Theorem for Derivatives? Number 12: Find a positive value c, for x, that satisfies the conclusion of the MVTD for f(x)=3x^2-5x+1 on [2,5] a)1, b)13/6, c)11/6, d)23/6, e)7/2
Hope everyone enjoys their day off!
Anyway, I'm going to explain some AP questions that I got right! Yippee!!
Non-Calculator Portion:
1. If g(x) = 1/32x^4-5x^2, find g'(4).
First step: take the derivative
1/8x^3-10x
Second Step: Plug in 4
1/8(4)^3-10(4)
Third Step: Solve
Answer = -32
5. Evaluate lim h->0 [5(1/2+h)^4-5(1/2)^4]/h
First Step: This is a definition of a derivative, so you take the derivative of 5(1/2)^4
4(5)(1/2)^3
Second Step: Solve
Answer = 5/2
Calculator Portion:
34. The graph of y=x^3-2x^2-5x+2 has a local maximum at:
First Step: Plug into your calculator
Second step: 2nd->Calc
Third Step: go to maximum
Forth Step: enter your bounds and then a "guess"
Answer: (-.786,4.209)
39. Find two non-negative numberfs x and y whose sum is 100 and for which (x^2)(y) is a maximum.
First step: Add up answer choices to make sure all equal 100
Second Step: Plug the x and y values into the given formula and see which is the highest number
Answer: x=66.667 and y=33.333
QUESTIONS:
How do you find the domain of a function without a calculator? I've always had trouble with domain and range even in Advanced Math.
How do you work piecewises? Number 7 is an example on the Non-calculator portion: Find k so that f(x) {(x^2-16)/x-4; x *does not equal* 4 and k; x=4 is continuous for all x. a) all real values, b)0, c)16, d)8, e) no real values
How do you take the derivative of a trig function squared? (sin^2(x) or cos^2(x))
Can someone go over the Mean Value Theorem for Derivatives? Number 12: Find a positive value c, for x, that satisfies the conclusion of the MVTD for f(x)=3x^2-5x+1 on [2,5] a)1, b)13/6, c)11/6, d)23/6, e)7/2
Hope everyone enjoys their day off!
post 28
all we did this week was correct our ap tests
1. The graph of y=5x^4-x^5 has an inflection point or points at
y=5x^4-x^5
20x^3-5x^3
20x^2(3-x)
3-x=0
-x=-3
divide by -1
x=3
2. d/dx(integral form 0 to x^2)sin^2t(dt)
sin^2(x^2)
2xsin^2(x^2)
3. The average value of f(x)=1/x from x=1 to x=e is
(1/e-1)[ln e -ln 1] *the e and 1 are the absoule value of
(1/e-1)[1-0]
= 1/e-1
4. Find the area under the curve y=2x-x^2 from x=1 to x=2 with n=4 left-endpoint rectangles.
use LRAM
delta x=2-1/4=1/4
1/4[f(1)+(5/4)+f(6/4)+f(7/4)]
1/4(0+.4375+.75+.9375)
1/4(2.125)= 17/32
i forgot houw to do problems (like #37 on calculator part) when a ladder slides down a wall and what is the speed of the bottom sliding out
1. The graph of y=5x^4-x^5 has an inflection point or points at
y=5x^4-x^5
20x^3-5x^3
20x^2(3-x)
3-x=0
-x=-3
divide by -1
x=3
2. d/dx(integral form 0 to x^2)sin^2t(dt)
sin^2(x^2)
2xsin^2(x^2)
3. The average value of f(x)=1/x from x=1 to x=e is
(1/e-1)[ln e -ln 1] *the e and 1 are the absoule value of
(1/e-1)[1-0]
= 1/e-1
4. Find the area under the curve y=2x-x^2 from x=1 to x=2 with n=4 left-endpoint rectangles.
use LRAM
delta x=2-1/4=1/4
1/4[f(1)+(5/4)+f(6/4)+f(7/4)]
1/4(0+.4375+.75+.9375)
1/4(2.125)= 17/32
i forgot houw to do problems (like #37 on calculator part) when a ladder slides down a wall and what is the speed of the bottom sliding out
Subscribe to:
Posts (Atom)