Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a twice continuously differentiable function with $f(0)=f(1)=f'(0)=0$ what is $f''(0)$?
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3If we assume $f(x)=ax^2(x-1)$ , then the answer is $-2a$, so we need more information. – Peter Mar 30 '16 at 15:23
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well, it could also be a trick question from the beginning: in case you mean the first derivative of $f(0)$, then we clearly get $0$, since the derivative of a constant simply vanishes. And therefore you'd get while following this interpretation $f''(0)=0$. One could distinguish if we'd write $f'(x)\rvert_{x=a}$ – user190080 Mar 30 '16 at 16:24
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There's a type for the 2nd derivate: Second derivative "formula derivation" Pick: $$f''(x) = \lim_{h\to0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2}$$ Then simplify, using L'Hôspital's rule, for $x=0\ \ \ \ (x=1).$
Theodoros Mpalis
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The question is not well defined.
As an example: if you chose $y=ax^n(x-1)^m$ than for $n,m>2$ we have $y''(0)=0$, but for $n=m=2$ the sign of $y''(0)$ depends on $a$.
Emilio Novati
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In OP there are no conditions about $y'(1)$. Anyway on that $y'(1)=0$. – Emilio Novati Mar 30 '16 at 16:02
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