Question

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

Answer 1

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 ?

Answer 2

Yes, the derivative of f is -16cos(x)(-sinx)- 16cosx

Answer 3

At least thats what I think

Answer 4

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 ?

Answer 5

need help

Answer 6

I'm having a problem doing this ... my brain is fried

Answer 7

um, i think the derivative is maybe : 16cos(x)(-sin(x)-16cos(x)
...

Answer 8

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 ?

Answer 9

We can factor -16cosx, so -16cos(x)(sinx +1)

Answer 10

That means f'x = 0 when sinx=-1 or cos(x)=0, right?

Answer 11

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 ?

Answer 12

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

Answer 13

not getting x

Answer 14

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.

Answer 15

mean value theorem?

Answer 16

I think I've got it

Answer 17

72500-60900=11600*7=81200
u guys r awesome

Answer 18

catch guys nex time
Thnks again

Answer 19

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.

Answer 20

And if you plug in, say 5pi/4?
it's increasing on the second interval and decreasing on the third

Answer 21

What about the inflection points?

Answer 22

I guess it's where f'(x) changes sign?

Answer 23

Or do i need to get f''(x)

Answer 24

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.

Answer 25

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

Answer 26

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