# Applying the Mean Value Theorem to conclude a function has a zero

Consider the function $f$ given by $f(x)=(x-2)^4\cos(x^2-4x+4)$. Use the Mean Value Theorem to show that $f'$ has a zero on the interval on $[1,3]$.

I notice that to do this we must show $f'(c)=0$ where $c$ is real number in the interval $[1,3]$. Now by the Mean Value Theorem,

$$\frac{f(3)-f(1)}{3-1} =f'(c)\,.$$

Notice that $f'(c)$ is indeed $0$ on the left hand side.

• A zero in what interval? – Alex Becker May 5 '13 at 7:33
• Sorry [1,3] I forgot to state in question. – Bobby May 5 '13 at 7:38
• I think you have to be a little crazy to use the MVT to show that that function has a zero in that interval, when there's a much much much easier way! – Gerry Myerson May 5 '13 at 7:39
• Do you see $x=2$ gives you zero? – Inceptio May 5 '13 at 7:41
• I could have done that or used IVT? But the question explicitly states to use it. – Bobby May 5 '13 at 7:47

First note that $x^2-4x+4 = (x-2)^2.$

Then the left-hand side of your MVT equation gives

\begin{align} \frac{f(3) - f(1)}{3-1} &= \frac{(3-2)^4 \cos (3-2)^2 - (1-2)^4 \cos (1-2)^2}{2} \\ &= \frac{(1)^4 \cos (1)^2 - (-1)^4 \cos (-1)^2}{2} \\ &= \frac{\cos(1) - \cos(1)}{2} \\ &= 0, \end{align}

which equals $f'(c)$ for some $c \in (1,3).$

(Notice that my interval is open. I think this is what you intended to type.)

First we get $f'(x)=4(x-2)^3\cos(x-2)^2-2(x-2)^5\sin(x-2)^2$, then we find that $f'(1)<0$ and $f'(3)>0$, so there exists $c\in[1,3]$ such that $f'(c)=0$.

The theorem I'm using here is is that if $f$ is continuous on $[a,b]$, and $f(a)<0$, and $f(b)>0$ then there exists some $c\in[a,b]$ such that $f(c)=0$.

Is this from mean value theorem?

• Not really, this is what is called the Intermediate Value Theorem. The MVT is different. – 40 votes Jul 22 '13 at 2:54