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

Question

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

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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 $(8,-32)$

Answer 1

You can simplify a little by noticing $f$ is odd, so you can study on $[0,+\infty)$ only.

Also by setting $u=x^{\frac 13}$ you get $g(u)=2u^3-24u$ with $g'(u)=6u^2-24=6(u^2-4)$.

Thus $u=\pm 2\iff x=\pm 8$ are critical points.

Note: critical points between $f$ and $g$ are identical because the variable change is strictly monotonous i.e. $(a\circ b)'=(a'\circ b)\cdot b'$ and $b'>0$

$$g(u)-g(2)=2u^3-24u+32=2(u+4)(u-2)^2\ge 0$$ since on $[0,+\infty)$ we have $u+4>0$.

So $u=2$ is a local minimum and by symmetry $u=-2$ is a local maximum.

Answer 2

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 $f(x)=\mp32$.