# 009A Sample Final 2, Problem 7

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

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

between  ${\displaystyle f(a)}$  and  ${\displaystyle f(b),}$  then there is at least one number  ${\displaystyle x}$  in the closed interval such that  ${\displaystyle f(x)=c.}$

2. Mean Value Theorem
Suppose  ${\displaystyle f(x)}$  is a function that satisfies the following:

${\displaystyle f(x)}$  is continuous on the closed interval  ${\displaystyle [a,b].}$

${\displaystyle f(x)}$  is differentiable on the open interval  ${\displaystyle (a,b).}$

Then, there is a number  ${\displaystyle c}$  such that  ${\displaystyle a  and  ${\displaystyle f'(c)={\frac {f(b)-f(a)}{b-a}}.}$

Solution:

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