Thread: Three Curve Analysis Questions

I have three cuve sketching questions. It's something I'm getting better at but each one of these three questions has something which I cannot figure out.


or http://img104.imageshack.us/img104/5928/61711sn1.png if you can't see the image

For the 3rd one, fully analyzing means find critical points, interval of increase and decrease, local max and min, intercepts, symmetry, inflection points/concavity, and sketch the graph. There are a few other things but I know how to do those. Just those ones are hard if you can't figure out the function.


Here's my attempt at solution:


For the first one, 6, I know that you have to find y prime but it involves implicit differentiation and curve analysis which is throwing me off.

For the second one, I know that there aren't any global maxima but I can't find the local maxima either. I also can't factor the original question and can't figure out the intercepts which is making the rest of the "fully analyze" part hard.

For the third one, I can't figure out the maxima's probably because it's in radical form which is also giving me problems with the inflection points.

-Thanks

Originally Posted by SportfreundeKeaneKent

I have three cuve sketching questions. It's something I'm getting better at but each one of these three questions has something which I cannot figure out.


or http://img104.imageshack.us/img104/5928/61711sn1.png if you can't see the image

For the 3rd one, fully analyzing means find critical points, interval of increase and decrease, local max and min, intercepts, symmetry, inflection points/concavity, and sketch the graph. There are a few other things but I know how to do those. Just those ones are hard if you can't figure out the function.


Here's my attempt at solution:


For the first one, 6, I know that you have to find y prime but it involves implicit differentiation and curve analysis which is throwing me off.

For the second one, I know that there aren't any global maxima but I can't find the local maxima either. I also can't factor the original question and can't figure out the intercepts which is making the rest of the "fully analyze" part hard.

For the third one, I can't figure out the maxima's probably because it's in radical form which is also giving me problems with the inflection points.

-Thanks

This is easy I just need you to tell me what each of the original functions are...Then I can get the answers for you....

RE:
#17...More to come!

Re: #11

Originally Posted by SportfreundeKeaneKent

I have three cuve sketching questions. It's something I'm getting better at but each one of these three questions has something which I cannot figure out.


or http://img104.imageshack.us/img104/5928/61711sn1.png if you can't see the image

For the 3rd one, fully analyzing means find critical points, interval of increase and decrease, local max and min, intercepts, symmetry, inflection points/concavity, and sketch the graph. There are a few other things but I know how to do those. Just those ones are hard if you can't figure out the function.


Here's my attempt at solution:


For the first one, 6, I know that you have to find y prime but it involves implicit differentiation and curve analysis which is throwing me off.

For the second one, I know that there aren't any global maxima but I can't find the local maxima either. I also can't factor the original question and can't figure out the intercepts which is making the rest of the "fully analyze" part hard.

For the third one, I can't figure out the maxima's probably because it's in radical form which is also giving me problems with the inflection points.

-Thanks

here's #6

you are correct, we have to use implicit differentiation to find y'. now here's the trick, we take the derivative of x normally, but everytime we take the derivative of y, we add y' to the end of it. why? we are finding the derivative with respect to x (because we want dy/dx -- the derivative of y with respect to x). so when we take the derivative of x, what we did was take the derivative of x with respect to x, so we add dx/dx. but it turns out the derivative notations can be treated as regular fractions, so the dx/dx cancels and leaves 1, so when we differentiate x we add nothing. when we differentiate y however, we differentiate y with respect to x, so we add dy/dx (of course i don't mean literally add, i mean attach to the end, more like multiply). so here goes.

x^2 + 2xy - y^3 = 0
note that we need the product rule for the middle term. so differentiating implicitly we get (i will write out all the notations at first so you get a feel for it):

2x dx/dx + 2y dx/dx + 2x dy/dx - 3y^2 dy/dx = 0

notice i added dx/dx when i took the derivative of an x-term and dy/dx when i took the derivative of a y-term. this simplifies to:

2x + 2y + 2x dy/dx - 3y^2 dy/dx = 0 ............now solve for dy/dx
=> 2x dy/dx - 3y^2 dy/dx = -2x - 2y
=> dy/dx(2x - 3y^2) = -2x - 2y
=> dy/dx = (-2x - 2y)/(2x - 3y^2)

now we want the value of the slope when x = -1. but we have y in our function for dy/dx. so, let's find what y is when x = -1 from the original function

x^2 + 2xy - y^3 = 0

2xy - y^3 = -x^2
when x = -1
=> 1 - 2y - y^3 = 0
alas, there is no nice way to solve this, not from what i can see. the answer will be around y = 0.4534 (i used a program to find this) as the only real root. so we would plug that in for y and -1 for x in our dy/dx expression to get the answer

RE: #6.

SportfreundeKeaneKent I know I said that I hated the two other problems your teacher gave you, but honestly I love problem #6. Because it involves implicit differentiation, Solving for Y, then using Newtons Method. Watch carefully how I work this sucker...

I don’t understand part of #11 when it asks you to find the max and minimum points. How do you get the interval of decrease to be (-infinity,-1) in union with (-1,infinity)? Is there no interval of increase and is finding the actual points impossible because it doesn’t factor? It also asks for the point of inflection but will there not be a point of inflection because f''(x) is also undefined?

Examine the graph when situations such as these arise. Study the Domain and Range of this picture...

Originally Posted by SportfreundeKeaneKent

I don’t understand part of #11 when it asks you to find the max and minimum points. How do you get the interval of decrease to be (-infinity,-1) in union with (-1,infinity)? Is there no interval of increase and is finding the actual points impossible because it doesn’t factor? It also asks for the point of inflection but will there not be a point of inflection because f''(x) is also undefined?

#11

Originally Posted by SportfreundeKeaneKent

I don’t understand part of #11 when it asks you to find the max and minimum points. How do you get the interval of decrease to be (-infinity,-1) in union with (-1,infinity)? Is there no interval of increase and is finding the actual points impossible because it doesn’t factor?

correct, there is no interval of increase for this function. it is decreasing everywhere on it's domain. let's check that.

f(x) = (1 - 2x)/(x + 1)
=> f ' (x) = [-2(x + 1) - (1 - 2x)]/(x + 1)^2
.............= (-2x - 2 - 1 + 2x)/(x + 1)^2
.............= -3/(x + 1)^2
.............= -3(x + 1)^-2

Note that this number is ALWAYS negative where defined. so the slope is negative everywhere and hence the function is decreasing everywhere.

now we know it decreases everywhere it is defined, so it's interval of decrease is just it's domain, which qbkr21 found by taking everything except the vertical asymptote (the domain is the set of all x-values (or more generally, inputs) for which a function is defined).

what "actual points" were you talking about?

It also asks for the point of inflection but will there not be a point of inflection because f''(x) is also undefined?

we have the possibility of an inflection point where f '' (x) = 0, note i said possibility, it is not always the case, a max or min can give zero for the second derivative as well. to make sure something is a an inflection point, you have to check the slope on both sides of the point. if the slope has the same sign on both sides, it is an inflection.

now, f '' (x) = 6(x + 1)^-3
this is never zero, so there are no inflection points (f '' (x) is not undefined in general, only for x = -1)

I have a question about #11.

What are the x intercepts (I can't figure them out and I need them to graph the function)?

Also, how do you show if thes is symmetric about the x axis or origin using the f(-x) = f(x) or -f(x) method?

Originally Posted by SportfreundeKeaneKent

I have a question about #11.

What are the x intercepts (I can't figure them out and I need them to graph the function)?

f(x) = (1 -2x)/(x + 1)
for x-intercept, y = 0
=> 0 = (1 -2x)/(x + 1)
=> 0 = 1 - 2x ..............the bottom cannot be zero, so the top must be zero
=> x = 1/2 .........x-intercept

Also, how do you show if thes is symmetric about the x axis or origin using the f(-x) = f(x) or -f(x) method?[/quote]

if f(x) = f(-x) we are symmetric about the y-axis
i don't remember the exact formulas for the other two, but i know that if we are symmetric about the origin, then for every point (x,y) we have (-x,-y) and if we are symmetric about the x-axis, then for every point (x,y) we also have (x,-y)

you check these by just plugging in

example: is our f symmetric about the y-axis?

f(-x) = [1 - 2(-x)]/(-x + 1) = (1 + 2x)/(1 - x) not= f(x) so no

Looking at #17, I see that it is symmetric about the x-axis. What are the x intercepts for it because I can't graph it without the x-intercepts and I can't find them because of the unfactorable form the original function is in
(y=x^4-4x^3+16x)

Originally Posted by SportfreundeKeaneKent

Looking at #17, I see that it is symmetric about the x-axis. What are the x intercepts for it because I can't graph it without the x-intercepts and I can't find them because of the unfactorable form the original function is in
(y=x^4-4x^3+16x)

so it's the same procedure. to find the x-intercepts, you set y = 0

y = x^4 - 4x^3 + 16x
for x-intercepts:
0 = x^4 - 4x^3 + 16x
=> 0 = x(x^3 - 4x^2 + 16)
=> x = 0 or x^3 - 4x^2 + 16 = 0
again, the second one looks like an exercise in Newton's method, it should be around -1.6786