Find Maximum and Minimum at x I am given the following, and asked to find the local maximum and local minimum.
$$ y = (8x^2-7x)^\frac {1}{3}$$
After finding the derivative I've concluded that my critical points are $x = \frac {7}{16}$ and $x = \frac {7}{8}$ after plotting my values on a sign chart I got the following.
$$
   +   \frac{7}{16}   -   \frac{7}{8}   -
$$
 Based on this I have a local maximum at $\frac {7}{16}$ but my book says that is incorrect. Where am I messing up?

Since $x\mapsto x^{1/3}$ is strictly increasing, the maxima and minima of $f(x)^{1/3}$ are at exactly the same $x$-values as the maxima and minima of $f(x)$ itself. So you can save a lot of work by simply looking at $8x^2-7x$.
This is a parabola with the arms pointing up, so it has a global minimium at the apex, $x=7/16$, no other local minima, and no local maxima at all.
