
Extremas
Find extrema $\displaystyle (x^28)^{2/3}$
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"?

Re: Extremas
We are given:
$\displaystyle f(x)=(x^28)^{\frac{2}{3}}$
and we find:
$\displaystyle f'(x)=\frac{2}{3}(x^28)^{\frac{1}{3}}(2x)=\frac{4x}{3(x^28)^{\frac{1}{3}}}$
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 $\displaystyle (0,4)$.