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I was given this exercise and not sure how to approach this.

Given $f:\mathbb{R}\longrightarrow\mathbb{R}$ is differentiable 3 times, $f'(0)=0, f(1)=1, f(-1)=f(0)=0$, show that there exists some point $c\in(-1,1)$ such that $f'''(c)\geq3$.

This exercise is part of homework about Taylor series, so I assumed the solution has to use it in some way.

What i've tried:

I have info about $f'(0)$ so i'll write the Taylor expansion for $f(x)$ with $n=3$ at $x=0$ (namely Maclaurin expansion).

$f(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(c)}{3!}x^3=\frac{f''(0)}{2!}x^2+\frac{f'''(c)}{3!}x^3$

Where $c$ is the point from Taylor's theorem.

Now, plugging in $1$, I get:

$1=f(1)=\frac{f''(0)}{2!}+\frac{f'''(c_1)}{3!}$ (There exists a point $c_1\in(0,1)$ such that this happens).

And $-1$:

$0=f(-1)=\frac{f''(0)}{2!}-\frac{f'''(c_2)}{3!}$ (There exists a point $c_2\in(-1,0)$ such that this happens).

Adding up both equations I can derive that:

$3!=f'''(c_1)+f'''(c_2)$

And I am not really sure how to continue, or what to infer from this, or even if I am in the right direction. Any help would be appreciated.

natitati
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  • Yes, thank you, $f(-1)=f(0)=0$ and not $1$. – natitati Mar 30 '24 at 19:45
  • That $3$ on the left-hand side of your final equation should be $3! = 6$. It then follows that the average of your two points is $3$, so one is at least $3$. (we could even show the existence of some $c$ such that $f'''(c) = 3$ from the same work, if you know Darboux's theorem) – Brian Moehring Mar 30 '24 at 19:52
  • Oh, yes... that's right. And that settles it. And by Darboux's "intermediate value" theorem $f'''(x)$ receives all points between $f'''(c_1)$ and $f'''(c_2)$, and since one of them is at least 3, and therefore the other is at most 3, we can derive that there exists a point $d$ between $c_1$ and $c_2$ such that $f'''(d)=3$. Brilliant. – natitati Mar 30 '24 at 20:04
  • I'm actually voting to re-open, since while the linked duplicate does answer the question, it doesn't do so using Taylor series. It seems rather reductive to point someone asking for a specific type of proof to an answer using a different method. The only alternative would seem to be posting your answer, corrected, as a solution to that question (at least then it could be justified as an honest answer to your question), and I don't know if that's any better. – Brian Moehring Mar 30 '24 at 20:10
  • So everything I did was correct - and what I was missing was inferring that since their sum is $6$, one of them has to be at least 3, yep?

    Okay, now how should I post the answer? Should I copy everything from my question and add that summary sentence?

     – natitati Mar 30 '24 at 20:16

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