# Extremas

• Oct 10th 2012, 08:10 AM
Greymalkin
Extremas
Find extrema $\displaystyle (x^2-8)^{2/3}$
Answers are $\displaystyle \pm 2\sqrt{2}$ and 0 for x values,
Local minimas are $\displaystyle (\pm 2\sqrt{2},0)$
Which makes complete sense if you look at the graph, however algebraically I solve for local maxima (0,4) which makes no sense to me as its undefined on the graph, why does my book say it counts as a local maxima if its "undefined"?
• Oct 10th 2012, 09:34 AM
MarkFL
Re: Extremas
We are given:

$\displaystyle f(x)=(x^2-8)^{\frac{2}{3}}$

and we find:

$\displaystyle f'(x)=\frac{2}{3}(x^2-8)^{-\frac{1}{3}}(2x)=\frac{4x}{3(x^2-8)^{\frac{1}{3}}}$

So, we see we have the 3 critical values:

$\displaystyle x=-2\sqrt{2},0,2\sqrt{2}$

While the derivative is undefined for $\displaystyle x=\pm2\sqrt{2}$, the function is defined there, indicating we have cusps at these points.

We then find on the intervals:

$\displaystyle (-\infty,-2\sqrt{2})$ derivative is negative, function is decreasing.

$\displaystyle (-2\sqrt{2},0)$ derivative is positive, function is increasing.

$\displaystyle (0,2\sqrt{2})$ derivative is negative, function is decreasing.

$\displaystyle (2\sqrt{2},\infty)$ derivative is positive, function is increasing.

So, by the first derivative test for extrema, we find minima at:

$\displaystyle (\pm\sqrt{2},0)$

and a maximum at $\displaystyle (0,4)$.