anonymous
  • anonymous
I am having a problem with a this problem: Given the function: f(x) = 8(cos(x))^2 - 16sin(x) where 0 <= x <= 2pi Find the intervals where f is increasing and decreasing, find the local min and max and find the inflection points
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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yuki
  • yuki
To find where f is increasing or decreasing, first you have to find the slope of f, " f' " right? Were you able to find the derivative of f ?
anonymous
  • anonymous
Yes, the derivative of f is -16cos(x)(-sinx)- 16cosx
anonymous
  • anonymous
At least thats what I think

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yuki
  • yuki
Good. Now what we need to do is to find where the slope is increasing. In other words, where f' > 0. So we will have to solve for 16cos(x)sin(x)-16cos(x) >0 would you be able to do this ?
anonymous
  • anonymous
need help
anonymous
  • anonymous
I'm having a problem doing this ... my brain is fried
anonymous
  • anonymous
um, i think the derivative is maybe : 16cos(x)(-sin(x)-16cos(x) ...
yuki
  • yuki
Ok, when you are solving for trig then these are the usual steps that you are going to take. if you can see that 16cos(x) is common in both terms, then you should factor it. it will look like 16cos(x) {1-sin(x)} >0 . Do you know what to do from here ?
anonymous
  • anonymous
We can factor -16cosx, so -16cos(x)(sinx +1)
anonymous
  • anonymous
That means f'x = 0 when sinx=-1 or cos(x)=0, right?
yuki
  • yuki
Yes, but since this is an interval you have to figure out where f' is positive or negative. before we proceed, can you find x ?
anonymous
  • anonymous
The problem I'm having is the intervals on the X-axis ... would we set them up at the critical numbers and the endpoints? cos(x)=0 at pi/2 and 3pi/2 sin(x) = -1 then x = 3pi/2
anonymous
  • anonymous
not getting x
yuki
  • yuki
Ok, so you understand that those x's are the points where f' = 0. Now to check the interval, all you have to do is to plot those x's on the numberline and check the intervals between them. In your case the intervals you will check is [0,pi/2), (pi/2, 3pi/2), (3pi/2,2pi]. so for example, if I want to check whether f' is positive or negative on the first interval, you just plug in a convenient number in it, such as pi/4 or pi/6. if f' >0, it will be true for all x's in that interval.
anonymous
  • anonymous
mean value theorem?
anonymous
  • anonymous
I think I've got it
anonymous
  • anonymous
72500-60900=11600*7=81200 u guys r awesome
anonymous
  • anonymous
catch guys nex time Thnks again
yuki
  • yuki
hahaha. No no you don't have to do anything like that. to check the first interval [0,pi/2) you just plug in a convenient value x into f'. If I plugged in x = pi/4, I can see that 16cos(pi/4) {1-sin(pi/4)} is negative since 16(sqrt2)/2 >0 and 1- sqrt(2)/2 is negative. so I know that on the interval [0, pi/2), f' <0 therefore f is decreasing.
anonymous
  • anonymous
And if you plug in, say 5pi/4? it's increasing on the second interval and decreasing on the third
anonymous
  • anonymous
What about the inflection points?
anonymous
  • anonymous
I guess it's where f'(x) changes sign?
anonymous
  • anonymous
Or do i need to get f''(x)
yuki
  • yuki
To find the inflection points you will find f" and let it equal to 0. Then you will do the same thing again to determine if it really is an "inflection point" If f" = 0, that doesn't necessarily mean that it is an inflection point right? so you will check the interval again to see if the concavity "changed" meaning f" >0 becomes f" <0 before and after those points where f" =0. It might be an overkill for this problem, but if you want to be 100% sure then you should check it. to find out whether f has a maximum or a minimum at those x's where you had f' =0, you will plug those x's into the f". if f">0, f is concave up meaning that at that point the graph looks like a " U " so we know it will be a minimum. if f"<0 it becomes a maximum.
yuki
  • yuki
I need to go, but if you need more help, ask others about how to find an inflection point and how to use the "second derivative test." They should be able to help you out :) It was fun talking to you. Yuki
anonymous
  • anonymous
I don't know if I can do anymore and my homework is due. I just can't think. My brain is friend from 2 big software projects

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