# How do you find the critical points for y = x^(2/3)(x^2-16) ?

Mar 27, 2015

For $f \left(x\right) = y = {x}^{\frac{2}{3}} \left({x}^{2} - 16\right)$, the critical points are $0 , 2 , - 2$

Solution
$c$ is a critcal point for $f$ if $c$ is in the domain of $f$ and either $f ' \left(c\right) = 0$ or $f ' \left(c\right)$ does not exist.

Find $f ' \left(x\right)$
We have choices for finding $f ' \left(x\right)$.
We could leave the function as written and use the product rule, or we could distribute the ${x}^{\frac{2}{3}}$ and avoid the product rule.

Using the product rule $\left(\text{I use:} \left(F S\right) ' = F ' S + F S '\right)$
$f ' \left(x\right) = \frac{2}{3} {x}^{- \frac{1}{3}} \left({x}^{2} - 16\right) + {x}^{\frac{2}{3}} \left(2 x\right)$

$= \frac{2 \left({x}^{2} - 16\right)}{3 \sqrt[3]{x}} + \frac{2 x \cdot {x}^{\frac{2}{3}}}{1}$

$= \frac{2 \left({x}^{2} - 16\right)}{3 \sqrt[3]{x}} + \frac{6 {x}^{2}}{3 \sqrt[3]{x}} = \frac{8 {x}^{2} - 32}{3 \sqrt[3]{x}}$

.Re-writing $f$ before differentiating
$f \left(x\right) = {x}^{\frac{8}{3}} - 16 {x}^{\frac{2}{3}}$

$f ' \left(x\right) = \frac{8}{3} {x}^{\frac{5}{3}} - 16 \cdot \frac{2}{3} {x}^{- \frac{1}{3}} = \frac{8 {x}^{\frac{5}{3}}}{3} - \frac{32}{3 \sqrt[3]{x}}$

$= \frac{8 {x}^{\frac{5}{3}} \cdot {x}^{\frac{1}{3}}}{3 \sqrt[3]{x}} - \frac{32}{3 \sqrt[3]{x}} = \frac{8 {x}^{2} - 32}{3 \sqrt[3]{x}}$
.
Using either method, we get $f ' \left(x\right) = \frac{8 {x}^{2} - 32}{3 \sqrt[3]{x}}$

Critical points:
$c$ is a critcal point for $f$ if $c$ is in the domain of $f$ and either $f ' \left(c\right) = 0$ or $f ' \left(c\right)$ does not exist.

$f \left(x\right) = y = {x}^{\frac{2}{3}} \left({x}^{2} - 16\right)$
For this function, the domain is $\left(- \infty , \infty\right)$
So the critical points for this $f$ will be the zeros of $f '$. and all zeros of the denominator of $f '$

$f ' \left(x\right) = \frac{8 {x}^{2} - 32}{3 \sqrt[3]{x}} = 0$ when $8 \left({x}^{2} - 4\right) = 0$
at $x = \pm 2$,
and $f ' \left(x\right)$ fails to exist at $x = 0$
Because these are all in the domain of $f$, these are all critical points for this $f$.