Namaste!

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

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

I began by taking the first derivative of $\displaystyle f(x)$:

$\displaystyle f'(x)=16x+8cos(4x)$ (not 100% sure this is correct)

Then I found the second derivative of $\displaystyle f(x)$ and got:

$\displaystyle f''(x)=16-32sin(4x)$

To find the critical points I (attempt to) set the second derivative to zero to find where the graph of $\displaystyle f(x)$ is concave up or concave down.

Broken into steps:

$\displaystyle 0=16-32sin(4x)$

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

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

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

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

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

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

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.