# 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?

• Your sign chart is off (you also missed the critical point $x=0$). – David Mitra Jul 21 '15 at 18:30

## 2 Answers

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.

• @dpmcmlxxvi: Pugging in $x=7/16$ I get $y=(8(\frac{7}{16})^2-7\frac7{16})^{1/3} = \sqrt{-\frac{49}{32}} \approx -1.15261$. – Henning Makholm Jul 22 '15 at 12:10

There is another critical point at $x=0$, so you need to test values between $0$ and $7/16$ for you to get a correct result. Let's try $1/4$, with that you get $$y(1/4) = (8(1/4)^2-7(1/4))^\frac {1}{3} = \left(-\frac{5}{4}\right)^{1/3} = \text{ some negative number }$$

So your sign chart becomes

$$+ \, 0 \, - \, \frac{7}{16} - \, \frac{7}{8} - \,$$

So, really, $x = 7/16$ is a minimum.