# 009A Sample Final 1, Problem 6

Consider the following function:

${\displaystyle f(x)=3x-2\sin x+7}$

(a) Use the Intermediate Value Theorem to show that  ${\displaystyle f(x)}$  has at least one zero.

(b) Use the Mean Value Theorem to show that  ${\displaystyle f(x)}$  has at most one zero.

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:

(a)

Step 1:
First note that   ${\displaystyle f(0)=7.}$
Also,  ${\displaystyle f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).}$
Since  ${\displaystyle -1\leq \sin(x)\leq 1,}$

${\displaystyle -2\leq -2\sin(x)\leq 2.}$

Thus,  ${\displaystyle -10\leq f(-5)\leq -6}$  and hence  ${\displaystyle f(-5)<0.}$
Step 2:
Since  ${\displaystyle f(-5)<0}$  and  ${\displaystyle f(0)>0,}$  there exists  ${\displaystyle x}$  with  ${\displaystyle -5  such that
${\displaystyle f(x)=0}$  by the Intermediate Value Theorem. Hence,  ${\displaystyle f(x)}$  has at least one zero.

(b)

Step 1:
Suppose that  ${\displaystyle f(x)}$  has more than one zero. So, there exist  ${\displaystyle a,b}$  with  ${\displaystyle a  such that  ${\displaystyle f(a)=f(b)=0.}$
Then, by the Mean Value Theorem, there exists  ${\displaystyle c}$  with  ${\displaystyle a  such that  ${\displaystyle f'(c)=0.}$
Step 2:
We have  ${\displaystyle f'(x)=3-2\cos(x).}$
Since  ${\displaystyle -1\leq \cos(x)\leq 1,}$
${\displaystyle -2\leq -2\cos(x)\leq 2.}$
So,  ${\displaystyle 1\leq f'(x)\leq 5,}$
which contradicts  ${\displaystyle f'(c)=0.}$  Thus,  ${\displaystyle f(x)}$  has at most one zero.