007A Sample Final 2, Problem 7 Detailed Solution

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

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Background Information:
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  $a<c<b$  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  $0<x<1$  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  $a<c<b$  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.

Final Answer:
See solution above.

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