2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ A zero in what interval? $\endgroup$ – Alex Becker May 5 '13 at 7:33
  • $\begingroup$ Sorry [1,3] I forgot to state in question. $\endgroup$ – Bobby May 5 '13 at 7:38
  • 5
    $\begingroup$ 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! $\endgroup$ – Gerry Myerson May 5 '13 at 7:39
  • $\begingroup$ Do you see $x=2$ gives you zero? $\endgroup$ – Inceptio May 5 '13 at 7:41
  • $\begingroup$ I could have done that or used IVT? But the question explicitly states to use it. $\endgroup$ – Bobby May 5 '13 at 7:47
2
$\begingroup$

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.)

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Not really, this is what is called the Intermediate Value Theorem. The MVT is different. $\endgroup$ – 40 votes Jul 22 '13 at 2:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.