# 009A Sample Final 2, Problem 7

Show that the equation  $x^{3}+2x-2=0$ has exactly one real root.

Foundations:
1. Intermediate Value Theorem
If  $f(x)$ is continuous on a closed interval  $[a,b]$ and  $c$ is any number

between  $f(a)$ and  $f(b),$ then there is at least one number  $x$ in the closed interval such that  $f(x)=c.$ 2. Mean Value Theorem
Suppose  $f(x)$ is a function that satisfies the following:

$f(x)$ is continuous on the closed interval  $[a,b].$ $f(x)$ is differentiable on the open interval  $(a,b).$ Then, there is a number  $c$ such that  $a and  $f'(c)={\frac {f(b)-f(a)}{b-a}}.$ Solution:

Step 1:
First, we note that
$f(0)=-2.$ Also,
$f(1)=1.$ Since  $f(0)<0$ and  $f(1)>0,$ there exists  $x$ with  $0 such that
$f(x)=0$ by the Intermediate Value Theorem.
Hence,  $f(x)$ has at least one zero.
Step 2:
Suppose that  $f(x)$ has more than one zero.
So, there exist  $a,b$ such that
$f(a)=f(b)=0.$ Then, by the Mean Value Theorem, there exists  $c$ with  $a such that
$f'(c)=0.$ We have  $f'(x)=3x^{2}+2.$ Since  $x^{2}\geq 0,$ $f'(x)\geq 2.$ Therefore, it is impossible for  $f'(c)=0.$ Hence,  $f(x)$ has at most one zero.