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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

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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 $$ 0=h^{(3)}(c)=f^{(3)}(c)-g^{(3)}(c)=f^{(3)}(c)-3. $$

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  • $\begingroup$ How did the idea of using g come to your mind ? $\endgroup$ – Gabriel Romon Mar 14 '13 at 17:58
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    $\begingroup$ @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
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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.

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  • $\begingroup$ 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

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