Prove that $\frac{x^3}{3}-\sin x-x=0$ has only one positive root?

Question

I need to prove that this equation

$\frac{x^3}{3}-\sin x-x=0$

has only one positive root. How can I do that?

Answer 1

Let $f(x)=\frac{x^3}{3}-\sin(x)-x$.

$f$ is an odd function with $f(0)=0$.

observe that if $c>0$ is a root then $-c$ is also a root.

suppose there are two positive roots

$c_1>c_2>0$.

so,

$f(-c_1)=f(-c_2)=f(0)=f(c_2)=f(c_1)$

and by using Rolle's theorem three times, we find that

$\exists c\in (-c_1,c_1)\;\;:\; f'''(c)=0$

but $f'''(x)=2+\cos(x)>0$

and the Contradiction.

Answer 2

Note that $f(0) = 0$,$ f'(0) <0$, and $f''(x) = \sin x + 2x\ge x$ is strictly positive for $x>0$.

Hence $f'$ is strictly increasing for $x>0$ and $f'(x)-f(0) \ge \int_0^x t dt = {x^2\over 2}$.

Let $x_0>0$ be such that $f'(x_0) = 0$, then $f$ is strictly decreasing for $x \in [0,x_0)$ and strictly increasing for $x > x_0$, and $f(x_0)<0$.

Since $\lim_{x \to \infty} f(x) = \infty$, we see that there must be a positive zero (strange wording!) of $f$ in $I=(x_0 , \infty)$ and since $f$ is strictly increasing on $I$, this zero is unique.