How to prove $x^3$ is strictly increasing

Question

I am trying to use $f(x)=x^3$ as a counterexample to the following statement.

If $f(x)$ is strictly increasing over $[a,b]$ then for any $x\in (a,b), f'(x)>0$.

But how can I show that $f(x)=x^3$ is strictly increasing?

Answer 1

Consider $a^3-b^3$, where $a\gt b$. We want to show this is positive. We have $$a^3-b^3=(a-b)(a^2+ab+b^2).$$ Note that $a^2+ab+b^2$ is positive unless $a=b=0$. There are many ways to do this. For example, $$a^2+ab+b^2=\frac{1}{2}\left(a^2+b^2+(a+b)^2\right).$$ More conventionally, complete the square. We have $4(a^2+ab+b^2)=(2a+b)^2+3b^2$.

Remark: If we are in a calculus mood, note that $$\frac{a^3-b^3}{a-b} =3c^2$$ for some $c$ between $a$ and $b$. This argument breaks down if $c=0$. But that can only happen when $0$ is in the interval $(b,a)$. That means $b$ is negative and $a$ is positive, making the inequality $a^3\gt b^3$ obvious.

Answer 2

Well, you know that $f$ is strictly increasing in $[0,+\infty)$ and in $(-\infty,0]$. Morevoer, $f$ is positive in the first interval and negative in the second one!

Answer 3

You want to show that the function $f(x) = x^3$ is strictly increasing on $\mathbb{R}$.

Maybe you can just use the definition. That is, let $a < b$. Assume that $0<a$ (you can do the other cases I am sure). Let $h = b - a < 0$. You want to show that $f(a) < f(b)$.

So $$\begin{align} f(a) = a^3 &= (b - h)^3\\ &= b^3 - 3b^2h +3 bh^2 - h^3 \\ &<b^3 \\ &= f(b). \end{align} $$ This is clear since $h > 0$, so $-3b^2h + 3bh^2 = -3hb(b - h) = -3hba < 0$.

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All that said, you could, of course just use $f(x) = x$ as a counter example.

Answer 4

Suppose $a<b$. We want to prove that $a^3 < b^3$.

Well, $a<b, \ \therefore \exists \delta>0$ such that $b = a+\delta$. So:

$b^3 = (a+ \delta)^3 = a^3 + 3a^2\delta + 3\delta^2a + \delta^3 = a^3 + (3a\delta(a+\delta)+\delta^3)$.

Now we consider five different cases:

1. $\ a<-\delta$
2. $\ a = -\delta$
3. $\ -\delta<a<0$
4. $\ a=0$
5. $\ a > 0$.

In cases $1, 2, 4$ and $5,$ it is fairly easy to see that $3a\delta(a+\delta)+\delta^3 > 0,\ $ so all that remains is case $3$:

$-\delta<a<0 \implies \exists k$ with $0<k<1,\ $ such that $a = -k\delta.\ $ Then, $\ a\delta(a+\delta) = -k\delta^2(1-k)\delta = -k(1-k)\delta^3$, and using the fact that $k(1-k) \leq \frac{1}{4},\ $ we see that $-k(1-k)\delta^3 \geq -\frac{1}{4}\delta^3,\ $ and so $\ 3a\delta(a+\delta)+\delta^3 \geq (-\frac{3}{4}\delta^3 + \delta^3) = \frac{1}{4}\delta^3 > 0$.

Answer 5

$$\frac{d x^3}{dx}=3x^2\geqslant0, \forall x \in[1,2]$$

Function is strictly increasing in $[1,2]$ and...

Answer 6

Here is an intuitive proof that could be made rigorous using Lebesgue measure.

Start with $x>0$. Place a cube with edge length $x$ so that it has three faces on the coordinate planes in three-dimensional space and two corners at $(0,0,0)$ and at $(x,x,x)$. Then $f(x)=x^3$ is the volume of the cube.

Strict monotonicity of $f$ is a consequence of the volume being strictly bigger for a strictly bigger cube. To incorporate negative values of $x$, argue with the fact that $f$ is odd.