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.

Foundations:
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)

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 exists $a,b$ 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) 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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