009A Sample Final 1, Problem 6

Consider the following function:

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

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

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

ExpandFoundations:
Recall:
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:

(a)

ExpandStep 1:
First note that  $f(0)=7.$
Also,
$f(-5)=-15-2\sin(-5)+7=-8-2\sin(-5).$
Since  $-1\leq \sin(x)\leq 1,$
$-2\leq -2\sin(x)\leq 2.$
Thus,  $-10\leq f(-5)\leq -6$  and hence  $f(-5)<0.$

ExpandStep 2:
Since $f(-5)<0$  and  $f(0)>0,$  there exists $x$ with  $-5<x<0$  such that
$f(x)=0$  by the Intermediate Value Theorem. Hence, $f(x)$  has at least one zero.

(b)

ExpandStep 1:
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  $f'(c)=0.$

ExpandStep 2:
We have $f'(x)=3-2\cos(x).$  Since  $-1\leq \cos(x)\leq 1,$
$-2\leq -2\cos(x)\leq 2.$
So, $1\leq f'(x)\leq 5,$ which contradicts $f'(c)=0.$
Thus, $f(x)$  has at most one zero.

ExpandFinal Answer:
(a) Since $f(-5)<0$  and  $f(0)>0,$  there exists $x$ with  $-5<x<0$  such that
$f(x)=0$  by the Intermediate Value Theorem. Hence, $f(x)$  has at least one zero.
(b) See Step 1 and Step 2 above.

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