Find all the relative extrema of $f(x)=x^4-4x^3$

Question

Find all the relative extrema of $f(x)=x^4-4x^3$

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$Solution:$

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

Answer 1

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 $$\lim\limits_{x\rightarrow+\infty}(x^4-4x^3)=\lim_{x\rightarrow+\infty}x^4\left(1-\frac{4}{x}\right)=+\infty.$$

Answer 2

More sophisticated is to take a second derivative which indicates the rate of change of the derivative when the derivative is zero. (Assuming the second derivative exists)

A derivative $f(x)=0$ means to function has "flatlined" as $x$. Four things can/will occur. [For purpose of analogy, I was consider the graphs moving in time. And values $x'< 0$ I will use past tense. $x' = x$ is "now" and I'll use present tense. And $x' > 0$ I'll use future tense.

1) The function could have been climbing, is now reaching a peak, and then will decline. (In which case this is a local maximum.)

That means that the derivative was positive, hits zero, and will then be negative.

The will be the case if $f''(x) < 0$. That would mean the derivative represented a positive steepness in the function but the steepness was decreasing, it decreases to $0$ steepness at which point the function plateaus, and then will become negatively step and and function will begin to go down.

However this could also be the case if $f''(x)=0$. The function could have been raises, the steepness (derivative) decreasing but decreasing less and less, the function then plateaus while the rate of change in steepness sloughs to zero, then the rate of change in steepness becomes active again and the steepness becomes negative and the function will begin to drop.

So $f'(x)=0$ and $f'''(x) < 0$ means $x$ is a local maximum. $f'(x)=0$ and $f''(x)=0$ means it could be, but isn't necessarily, a maximum. If $f'(x)= f''(x)=0$ and $x$ is a local maximum that is called an undulation point.

2) The function could have been dropping, is now flatlining, and will then raise.

This is the exact opposite as above for the same reasons. And is a local minimum.

The will always by the case if $f''(x) > 0$. If $f''(x)=0$ this could, but isn't nescessarily the case. If $f'(x)=f''(x) = 0$ and $f(x)$ is a local minimum this is another example of an undulation point.

3) The function was raising/dropping, is now flatling, but will then continue to raise and drop.

In this case the steepness (derivative) of the function was positive/negative, has leveled to zero, but rather than continuing to become negative/positive and will return to be positive/negative. This would be a local min/max for the derivative.

IF the second derivative exist it will have to be that $f''(x) =0$ and this point. (But it need not exist. $f'$ could be making a "sharp turn" while $f$ is still differentiable).

But $f''(x) = 0$ will not assure this must be the case.

In this case this is called a saddle point.

4) $f(x)$ have been, or will be, a constant for an interval.

If which case $f', f'', f'''$ etc so long as they actually are defined will be $0$.

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$f''(x) = 12x^2-24x$. $f''(3)=12*9-24*3 > 0$ so $x = 3; f(3)= -27$ is a local minimum.

$f''(0) = 0$ so it isn't clear what type of critical point $(0,0)$ is. Taking point sample of $x=-1,2$ show you it is a saddle point.

If we have the patience and clear head we could note that:

$f'''(x)=12x^2 - 24$ and $f'''(0)=-24 < 0$ So that means that $f'(0)=0$ was a local maximum for $f'$. So the derivative had been less than $0$, so the function had been decreasing, the derivative is $0$ and $x=0$ and the function is flatlining, but the will drop down to being negative again, so the function will start to decrease againd. SO $x=0; f(x)= 0$ is a saddle point.

Answer 3

Once you have identified $x_0=0$ and $x_0=3$ you can also evaluate the sign of $f(x)-f(x_0)$.

• $x_0=0$ where $f(0)=0$

$$f(x)=x^3\underbrace{(x-4)}_{<0}$$

Thus the sign of $f(x)$ is determined by the sign of $x^3$ around $0$ which is not constant so $x_0=0$ is an inflexion point.

• $x_0=3$ where $f(3)=-27$

Let set $u=x-3$ then $$f(x)-f(3)=(u+3)^3(u-1)+27=18u^2+8u^3+u^4=\underbrace{u^2}_{\ge 0}\underbrace{(u^2+8u+18)}_{\Delta=-8\implies >0}\ge 0$$

Thus $f(3)$ is a global minimum.