f : $I \to \Bbb R$ is differentiable 3 times in open interval $I$ which contains the closed interval [-1,1]. $f(0)=f(-1)=f'(0)=0$ and $f(1)=1$

show that exists a point $c \in (-1,1) s.t. f^{(3)} (c) \ge 3$

What I did: I used Rolle's theorem to prove that there are points in the derivatives where they equal 0. Don't really know how to get to 3... maybe using the intermediate value somehow- but don't really have an idea...

Thanks in advance


A stronger statement holds: there exists $c$ in $(-1,1)$ such that $f^{(3)}(c)=3$.

Observe that the degree $3$ polynomial $$ g(x)=\frac{1}{2}x^2(x+1) $$ has exactly the same properties as your function $f$. And note that $g^{(3)}(x)=3$.

Now $h=:f-g$ satisfies $$ h(-1)=h(0)=h(1)=h'(0)=0. $$ By Rolle, there exist $-1<a<0<b<1$ such that $h'(a)=h'(0)=h'(b)=0$.

Two more applications of Rolle yield $-1<c<1$ such that $$ 0=h^{(3)}(c)=f^{(3)}(c)-g^{(3)}(c)=f^{(3)}(c)-3. $$

  • $\begingroup$ How did the idea of using g come to your mind ? $\endgroup$ – Gabriel Romon Mar 14 '13 at 17:58
  • 1
    $\begingroup$ @GabrielRomon I started by looking for an example of a funciton $f$ with these properties. For polynomials $p$, this is equivalent to $p(x)=\frac{1}{2}x^2(x+1)(1+a_1x+\ldots+a_nx^n)$. Then I tried the simplest one, $g$, and computed the third derivative to see if it worked. After that, it was natural by linearity. $\endgroup$ – Julien Mar 14 '13 at 18:05
  • $\begingroup$ @GabrielRomon Actuallly, there is a mistake in my comment: it should be $p(x)=g(x)q(x)$ with $q$ polynomial such that $q(1)=1$. $\endgroup$ – Julien Mar 14 '13 at 18:21
  • $\begingroup$ This is really elegant $\endgroup$ – Gabriel Romon Mar 14 '13 at 18:35
  • $\begingroup$ Thanks. Actually, I can't think of any other proof. $\endgroup$ – Julien Mar 14 '13 at 18:58

I have a sketch of an answer that might help:

I believe that since $[-1,1]$ is closed, the function $f'''$ must reach a maximum $m$ on the interval. So we have: $f'''(x) \leq m$

Then you can integrate both sides over the range $[0,x]$, i.e.

$\int^x_0 f'''(t) dt \leq mx$

$f''(x) - f''(0) \leq mx$

$f''(x) \leq mx + f''(0)$

Trouble is, we are not given $f''(0)$ so leave as is for now.

Now repeat the process:

$\int^x_0 f''(t) dt \leq \frac{1}{2}mx^2 + f''(0)x$

$f'(x) - f'(0) \leq \frac{1}{2}mx^2 + f''(0)x$

$f'(x) \leq \frac{1}{2}mx^2 + f''(0)x$

and again:

$\int^x_0 f'(t) dt \leq \frac{1}{6}mx^3 + \frac{1}{2}f''(0)x^2$ $f(x) - f(0) \leq \frac{1}{6}mx^3 + \frac{1}{2}f''(0)x^2$ $f(x) \leq \frac{1}{6}mx^3 + \frac{1}{2}f''(0)x^2$

Setting $x=1$ :

$1 \leq \frac{1}{6}m + \frac{1}{2}f''(0)$

So you get an inequality involving $m$ and $f''(0)$.

Now you can repeat the whole process again but this time use the interval $[-x,0]$. You should then get another inequality involving $m$ and $f''(0)$.

Using substitution you should be able to derive an inequality involving just $m$ that should yield your answer.

  • $\begingroup$ It says $f'''$ exists, not that it is continuous. So your first claim is not valid. $\endgroup$ – Julien Mar 14 '13 at 17:31
  • $\begingroup$ I believe you have to assume continuity in this question otherwise all bets are off. I agree with your comment however. $\endgroup$ – Darren Mar 14 '13 at 17:38
  • $\begingroup$ No. See my answer. $\endgroup$ – Julien Mar 14 '13 at 17:40
  • $\begingroup$ Even without the assumption of continuity, wouldn't $f'''$ have a maximum value over the closed interval? $\endgroup$ – Darren Mar 14 '13 at 18:00
  • $\begingroup$ No there is no reason for this.. $\endgroup$ – Gabriel Romon Mar 14 '13 at 18:47

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.