Here is the question: Let $f(x)$ be a three times differentiable function on $[−1, 1]$ such that $f(−1) = 0$, $f(0) = 0$, $f(1) = 1$ and $f′(0) = 0$. Prove that $f′′′(x) ≥ 3$ for some $x ∈ (−1, 1)$.

My attempt: Well I think I have to use either the mean-value theorem or Taylor's Theorem, or both. I have no idea. Can someone just give a hint in order for me to start on this problem? Thank you very much!!


marked as duplicate by Paramanand Singh calculus Apr 22 '18 at 12:34

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(i).If $f'''(x)\geq 3$ for some $x\in [0,1]$ then we're done.

(ii). If $f'''(x)<3$ for all $x\in [0,1]$ then for some $x\in (0,1)$ we have $$1=f(1)=f(0)+f'(0)+f''(0)/2+f'''(x)/6=$$ $$=f''(0)/2+x^3f'''(x)/6<$$ $$<f''(0)/2+3/6$$ implying $f''(0)>1.$ Then for some $y\in (-1,0)$ we have $$f(-1)=f(0)+(-1)f'(0)+(-1)^2f''(0)/2+(-1)^3f'''(y)/6$$ Now plug in the values of $f(-1), f(0),$ and $f'(0),$ and consider that $f''(0)>1.$ What does it tell you about $f'''(y)$?

  • $\begingroup$ I tried expansions about $x=0$ because we have more info there (that is, $f(0)=f'(0)=0$) than at $x=\pm 1.$ $\endgroup$ – DanielWainfleet Apr 22 '18 at 9:42

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