Question on Derivatives Real Analysis [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 22 '18 at 12:34

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