Surprised that $f(x)=x^\frac{1}{3} + 1$ has no extrema

Question

I was doing a problem involving the first-derivative test for relative extrema for $f(x)=x^\frac{1}{3} + 1$ and found the derivative: $$f'(x)=\frac{1}{3} * x^\frac{-2}{3}$$

I then discovered that all my tools for finding roots (Quadratic formula, factoring, synthetic division) seem to not fit! Obviously 0 is a root since there is only one term which is an x term. The answer in the back says "No relative extrema" which surprised me. How I might I have identified that there are no extrema?

Edit: changed on to one

Answer 1

Remember that to find extrema, set the derivative to zero and solve the equation:

$$f'(x) = \dfrac{1}{3}x^{-\frac{2}{3}} = 0$$ $$x^{-\frac{2}{3}} = 0$$ $$ \dfrac{1}{x^{\frac{2}{3}}} = 0$$

Realize that no real number reciprocal will lead to $0$, thus the equation has no solution—meaning no local extrema.

Note that this doesn't necessarily mean there are no extrema. A good example is $f(x) = x^{\frac{2}{3}}$ which has a cusp at $0$, meaning it isn't differentiable there (and the equation of the derivative $\frac{2}{3}x^{-\frac{1}{3}} = 0$ has no solutions), but the function still has an absolute minimum at $x = 0$. But in this case, $x^{\frac{1}{3}} + 1$ increases without bound in the positive and negative $y$ directions, thus there are also no absolute extrema.

$0$ is not a solution. $0^{-\frac{2}{3}}$ is undefined: $\dfrac{1}{0^{\frac{2}{3}}} = \dfrac{1}{0}$

Answer 2

note that $$\frac{1}{3}x^{-2/3}=\frac{1}{3x^{2/3}}\neq 0$$

Answer 3

For fun:

Let $x\not =0$, $x$ real;

$y= x^{1/3} +1;$

$x \not=0$ implies $y \not =1.$

$(y-1)^3 =x;$

Differentiate both sides with resp. to $x$:

$3(y-1)^2\dfrac{dy}{dx} =1$, or

$\dfrac{dy}{dx}= \dfrac{1}{3(y-1)^2}$, $y \not =1$.

$\dfrac{dy}{dx} >0$, $y \not=1$.

Hence for $x\not =0$,

$y=f(x) $ strictly increasing .

No maximum.

Answer 4

You should be careful. It is true that if a function has an extremal point in the interior of its domain then the derivative vanishes at that point, provided the derivative exists at that point to begin with.

A simple example is $g(x)=|x|$, which has obviously an absolute minimum at $0$, but its derivative vanishes nowhere (where it exists, of course).

Your function $f(x)=\sqrt[3]{x}+1$ has derivative $$ f'(x)=\frac{1}{3\sqrt[3]{x^2}} $$ which vanishes nowhere, but does not exist at $0$. This tells us that the function has no relative extremal points different from $0$, but tells us nothing about $0$.

On the other hand the derivative is everywhere positive on $(-\infty,0)\cup(0,\infty)$. Since the function is everywhere continuous, it is strictly increasing.

Hence the answer no relative extremal point is correct, but requires some more work than just using the fact that the derivative vanishes nowhere.

Beware also that if the function is defined, say, over $[a,b]$, then $a$ or $b$ can be extremal points even if the (one-sided) derivative doesn't vanish at those points. Think to $g(x)=x$ defined over $[0,1]$.

Answer 5

$0$ isn't a root of the derivative, it's a pole.

But that aside, $0$ really is a critical point of this function; the only one, in fact.

As a critical point, it's a place where a relative extremum could be located.

Not a place where one must be located.

That is your mistake; the fact $0$ is a critical point isn't a guarantee that it's actually a relative extrema; in fact, it's not.

The fact 0 is not a critical point can easily be seen from graphing the function:

(source: WolframAlpha)

or even more simply by remembering that $x^{1/3}$ is a strictly increasing function of $x$. You don't even have to bother with computing derivatives to recognize that there are no relative extrema.