007A Sample Final 1, Problem 7 Detailed Solution

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.

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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:

(a)

Step 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  $f(-5)<0.$

Step 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)

Step 1:
Suppose that  $f(x)$  has more than one zero.
So, there exist  $a,b$  with  $a<b$  such that  $f(a)=f(b)=0.$
Then, by the Mean Value Theorem, there exists  $c$  with  $a<c<b$
such that  $f'(c)=0.$

Step 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.

Final Answer:
(a)     See solution above.
(b)     See solution above.

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