Find all the relative extrema of $f(x)=x^4-4x^3$
$Solution:$
Step 1: Solve $f'(x)=0$.
$f'(x)=4x^3-12x^2=0$
$\rightarrow$ $4x^2(x-3)=0$
$\rightarrow$ $x=0$ and $x=3$
Step 2: Draw a number line and evaluate the sign of the derivative on each section (I don't know how to draw a number line on the computer but I'll do what I can).
Lets pick a number in the region $(-\infty,0)$, how about $x=-1$:
$f'(-1)=4(-1)^2(-4)=-16<0$
Now lets pick a number in the region $(0,3)$, how about $x=2$:
$f'(2)=4(2)^2(2-3)=-1<0$
Now lets pick a number in the region $(3,\infty)$, how about $x=4$:
$f'(4)=4(4)^2(4-3)=64>0$
So our function is decreasing as $x$ increases towards $0$, then our function decreases as $x$ increases towards $3$, then our function increases as $x$ increases towards $\infty$. Therefore we have a relative minimum when $x=3$ and no relative maximum. Lets solve for the corresponding $y$ value:
$f(3)=3^4-4(3)^3 = 81-4(27)=-27$
So we have a relative minimum at $(3,-27)$ and our relative maximum DNE
Best Answer
Solution without derivative.
By AM-GM $$x^4-4x^3=3\cdot\frac{x^4}{3}+27-4x^3-27\geq4\sqrt[4]{\left(\frac{x^4}{3}\right)^3\cdot27}-4x^3-27=$$ $$=4|x^3|-4x^3-27\geq-27.$$ The equality occurs for $x=3,$ which says that we got a minimal value.
The maximum does not exist because $$\lim\limits_{x\rightarrow+\infty}(x^4-4x^3)=\lim_{x\rightarrow+\infty}x^4\left(1-\frac{4}{x}\right)=+\infty.$$