# Trig Exreme Values

• Nov 7th 2009, 05:02 PM
john-1
Trig Exreme Values
Determine the extreme values of the following function in the given interval.

f(x)=2sin4x+3 , XE [0, pi]

This is my work:

f'(x)=2cos4x (4)
f'(x)=8cos4x

Now I am trying to find the critical numbers, but I'm not sure how to solve for x.

f'(x)=0
cos4x=0
x=22.5

Is this right?
• Nov 7th 2009, 05:05 PM
Scott H
Remember that when $k$ is a constant, $\frac{d}{dx}(k)=0.$ Other than that, you are on the right track.
• Nov 7th 2009, 05:19 PM
john-1
EDIT:::::::::

I'm not getting correct answer with abs min with the back of my textbook.

I get f(0) = 3
and f(pi) = 3
f(22.5)=5

Therefore the max is 5 and the min is 3. The back of the book has min as 1. How did they get this?
• Nov 7th 2009, 05:27 PM
VonNemo19
Quote:

Originally Posted by john-1
EDIT:::::::::

I'm not getting correct answer with abs min with the back of my textbook.

I get f(0) = 3
and f(pi) = 3
f(22.5)=5

Therefore the max is 5 and the min is 3. The back of the book has min as 1. How did they get this?

$\cos{x}=0$ when $n\frac{\pi}{2}$ with $n$ odd.

Therefore

$\cos(4x)=0\Rightarrow{x}=n\frac{\pi}{8}$, $n$ odd.

Now considering the given interval...
• Nov 7th 2009, 05:47 PM
john-1
Yes I know, Thank you, but I am asking about the abs min.
How did they get 1?

I get f(0) = 3
and f(pi) = 3
f(22.5)=5

Therefore the max is 5 and the min is 3. The back of the book has min as 1.
• Nov 7th 2009, 06:10 PM
VonNemo19
Take a look at the graph.

$f(\frac{\pi}{8})=5$

$f(\frac{3\pi}{8})=1$ and etc...

The endpoints are therefore irrelevant.