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

Question

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.

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 05 '13 at 07:39
I could have done that or used IVT? But the question explicitly states to use it. — Bobby, May 05 '13 at 07:47
For both $x = 1$ and $x = 3$, what does $(x-2)^4$ equal? What does $\cos(x^2 - 4x + 4) \ = \ \cos([x-2]^2)$ equal? What can you conclude about $f'(c)$ ? — colormegone, May 05 '13 at 07:52
Sorry, I did not write the question properly. Apologies, it's to show f' has a zero in the interval. — Bobby, May 05 '13 at 07:52
So you do want the MVT. What does you find for $f(1)$ and $f(3)$ from the above? — colormegone, May 05 '13 at 07:54
Oops thanks haha, I mis calculated. We get f'(c)=0. Thanks for that Reckless. — Bobby, May 05 '13 at 07:55
It's still crazy to use MVT. $f'$ is of the form $(x-2)^3g$, and $g(2)$ is finite, so $f'(2)=0$. — Gerry Myerson, May 05 '13 at 12:50
@GerryMyerson Indeed. Or simply the fact that $t\mapsto f(2+t)$ is even (and differentiable). — Did, May 06 '13 at 07:43

Optional · Answer 1 · 2014-02-10T08:55:16.607

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.)

score 1 · Answer 2 · edited Mar 09 '18 at 04:13

1

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?

edited Mar 09 '18 at 04:13

Mike Pierce

18,938

answered May 05 '13 at 08:12

Jimmy Wang

629

1

Not really, this is what is called the Intermediate Value Theorem. The MVT is different. – 40 votes Jul 22 '13 at 02:54

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

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