1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Your sign chart is off (you also missed the critical point $x=0$). $\endgroup$ – David Mitra Jul 21 '15 at 18:30
2
$\begingroup$

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.

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.