Find all the relative extrema of the function $f(x)=2x-24x^{1/3}$ Find all the relative extrema of the function $f(x)=2x-24x^{1/3}$

Solution:
Step 1: Find the values of $x$ where $f'(x)=0$ and $f'(x)$ DNE.
$f'(x)=2-8x^{\frac{-2}{3}}=2-\frac{8}{x^{\frac{2}{3}}}$
We can see that $f'(x)$ DNE when $x=0$. Let's figure out what values of $x$ give $f'(x)=0$
$f'(x)=2-8x^{\frac{-2}{3}}=2-\frac{8}{x^{\frac{2}{3}}}=0$
$\rightarrow x^{\frac{2}{3}}=4$
$\rightarrow $x^2=64$
$\rightarrow x= \pm 8$
Thus the critical points are $x=0,x=8,x=-8$.
Step 2: Make a number line and plot the sign of $f'(x)$ in each section to figure out where $f(x)$ is increasing and decreasing
$f'(-10) \cong 2.77 >0$
$f'(-1) \cong -6 <0$
$f'(1) \cong -6 <0$
$f'(9) \cong .151 >0$
So $f(x)$ is increasing as $x$ increasing towards -8, and then $f(x)$ decreases as $x$ approaches zero. Thus $f(-8)$ corresponds to a relative maxmimum.
Also, $f(x)$ is decreasing as $x$ increasing towards 8, and then $f(x)$ is increasing as $x$ approaches $\infty$. Thus $f(8)$ corresponds to a relative minimum.
Step 3: Find the corresponding $y$ values
$f(-8) = 32$
$f(8) = -32$
Thus we have a relative maximum at $(-8,32)$ and a relative minimum at $(8,-32)$

Let $t=x^{1/3}$ so $x=t^3$.  
Then we want extrema of $2t^3-24t$.  
These occur when $6t^2-24=0$; i.e., $t=\pm2$.  
Then $x=\pm8$, and $f(x)=\mp32$.
