# Concavity and Inflection problem, with Trig Functions.

• February 26th 2013, 01:20 PM
KhanDisciple
Concavity and Inflection problem, with Trig Functions.
Namaste!

Let $f(x)=8x^2+sin^2(4x)$ where $x \in \left[0,\frac{\pi}{4} \right]$

Find the interval where $f$ is concave up, and where $f$ is concave down. Finally, list any points of inflection (if there are any).

I began by taking the first derivative of $f(x)$:
$f'(x)=16x+8cos(4x)$ (not 100% sure this is correct)
Then I found the second derivative of $f(x)$ and got:
$f''(x)=16-32sin(4x)$

To find the critical points I (attempt to) set the second derivative to zero to find where the graph of $f(x)$ is concave up or concave down.
Broken into steps:
$0=16-32sin(4x)$

$-16=-32sin(4x)$

$\frac{-16}{-32}=sin(4x)$

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

I try to think of where the graph of $sin(x)=\frac{1}{2}$

$\frac{\pi}{6}=4x$

Finally: $x=\frac{\pi}{24}$

So this is the critical point, and possibly a point of inflection.

I test points on a number line on either side of this to find where the second derivative is positive or negative to find where the original function is concave up, or concave down.
A hint was given to use a double angle formula, however I am not very familiar with how to implement these.

However, all of my answers so far have been incorrect. I don't want the solution, I just want to know where I went wrong through this process. Any help would be greatly appreciated. Thanks in advance.
• February 26th 2013, 04:03 PM
Esteban
Re: Concavity and Inflection problem, with Trig Functions.
The first derivate is not correct
• February 26th 2013, 04:10 PM
BobP
Re: Concavity and Inflection problem, with Trig Functions.
As you suspect as a possibility, your first derivative is incorrect.

The derivative of $\sin^{2}4x,$ is $8\sin4x \cos 4x.$
• February 26th 2013, 10:31 PM
KhanDisciple
Re: Concavity and Inflection problem, with Trig Functions.
A 20 minute post, answered in 20 seconds...(sigh).
But in all seriousness, thanks.