1. ## Extremas

Find extrema $(x^2-8)^{2/3}$
Answers are $\pm 2\sqrt{2}$ and 0 for x values,
Local minimas are $(\pm 2\sqrt{2},0)$
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"?

2. ## Re: Extremas

We are given:

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

and we find:

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

So, we see we have the 3 critical values:

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

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

We then find on the intervals:

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

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

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

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

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

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

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