# Math Help - Functions

1. ## Functions

Answer the following questions for the function
defined on the interval . Enter points, such as inflection points in ascending order, i.e. smallest x values first.
Remember that you can enter "pi" for as part of your answer.

A. is concave down on the region to
B. A global minimum for this function occurs at
C. A local maximum for this function which is not a global maximum occurs at
D. The function is increasing on to and on to .

2. ## Hmm

Originally Posted by tennisgirl
Answer the following questions for the function

defined on the interval . Enter points, such as inflection points in ascending order, i.e. smallest x values first.
Remember that you can enter "pi" for as part of your answer.

A. is concave down on the region to

B. A global minimum for this function occurs at

C. A local maximum for this function which is not a global maximum occurs at

D. The function is increasing on to and on to .
This seems like a homework question or a test question...so I will not answer for you since that is frowned upon but I will give you hint find where $f'(x)=0$.... $f''(x)=0$....and use taht to set up test intervals for each...and for helping with the derivative maybe this set up will be nicer for you $\bigg[sin\bigg(\frac{x}{4}\bigg)\bigg]^2$

Actually, it's not a test question. It's a homework question (only worth one point). I'm more interested in the method of solving and I already found answers to part of it.

I have found the global minimum to be zero. I found the function to be increasing on the interval 0 to 2.6415926, but I don't know the other interval (the first part is -12.46, but I don't know the endpoint of increase).
I also am having trouble finding where the function is concave down. I took the second derivative, but can't find the local maximum. I think it might be -12.466. Can you give me some hints?

Thanks.

4. ## Ok

Originally Posted by tennisgirl
Actually, it's not a test question. It's a homework question (only worth one point). I'm more interested in the method of solving and I already found answers to part of it.

I have found the global minimum to be zero. I found the function to be increasing on the interval 0 to 2.6415926, but I don't know the other interval (the first part is -12.46, but I don't know the endpoint of increase).
I also am having trouble finding where the function is concave down. I took the second derivative, but can't find the local maximum. I think it might be -12.466. Can you give me some hints?

Thanks.
if $f(x)=sin\bigg(\frac{x}{4}\bigg)^2$...then $f'(x)=2sin\bigg(\frac{x}{4}\bigg)\cdot{cos\bigg(\f rac{x}{4}\bigg)}\cdot\frac{1}{4}=\frac{1}{4}\cdot\ sin\bigg(\frac{x}{2}\bigg)$...now setting this equal to zero we get $x=-6.283,x=0,6.28$ now setting up the test intervals we get $(-\infty,-6.283),(-6.283,0)(0,6.238)(6.238\infty)$...now just test a number in each interval in the derivative...positive values are increasing intervals and negative values are decreasing intervals

5. Originally Posted by tennisgirl
Actually, it's not a test question. It's a homework question (only worth one point). I'm more interested in the method of solving and I already found answers to part of it.

I have found the global minimum to be zero. I found the function to be increasing on the interval 0 to 2.6415926, but I don't know the other interval (the first part is -12.46, but I don't know the endpoint of increase).
I also am having trouble finding where the function is concave down. I took the second derivative, but can't find the local maximum. I think it might be -12.466. Can you give me some hints?

Thanks.

This may help:

$f(x)=\sin^{2}(\frac{x}{4})$

taking the derivative we get

$\frac{df}{dx}=\frac{1}{2}\sin(\frac{x}{4})\cos(\fr ac{x}{4})$

using the double angle formula we get

$\frac{df}{dx}=\frac{1}{4}\sin(\frac{x}{2})$

now taking the derivatvie again we get

$\frac{d^2f}{dx^2}=\frac{1}{8}\cos(\frac{x}{2})$

for max and min use the zero's of the 1st derivative and the end points of the interval

See what you can do from here.

6. ## Ok

Originally Posted by TheEmptySet
This may help:

$f(x)=\sin^{2}(\frac{x}{4})$

taking the derivative we get

$\frac{df}{dx}=\frac{1}{2}\sin(\frac{x}{4})\cos(\fr ac{x}{4})$

using the double angle formula we get

$\frac{df}{dx}=\frac{1}{4}\sin(\frac{x}{2})$

now taking the derivatvie again we get

$\frac{d^2f}{dx^2}=\frac{1}{8}\cos(\frac{x}{2})$

for max and min use the zero's of the 1st derivative and the end points of the interval

See what you can do from here.
But you should set $f''(x)=0$ and set up test intervals based on those numbers...where $f''(x)$ changes signs you have an inflection point