# 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:
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 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 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 such that  $f(a)=f(b)=0.$ Then, by the Mean Value Theorem, there exists  $c$ with  $a 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.