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Suppose $f$ is a real, three times differentiable function on $[-1,1]$, such that
$f(-1)=0, \quad f(0)=0, \quad f(1)=1, \quad f^{\prime}(0)=0 .$
Prove that $f^{(3)}(x) \geq 3$ for some $x \in(-1,1)$.
Note that equality holds for $\frac{1}{2}\left(x^{3}+x^{2}\right)$.
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\begin{aligned} f(1) &=f(0)+f^{\prime}(0)+\frac{1}{2} f^{\prime \prime}(0)+\frac{1}{6} f^{(3)}(s) \\ f(-1) &=f(0)-f^{\prime}(0)+\frac{1}{2} f^{\prime \prime}(0)-\frac{1}{6} f^{(3)}(t) \end{aligned}
for some $s \in(0,1), t \in(-1,0)$. By subtracting the second equation from the first and using the given values of $f(1), f(-1)$, and $f^{\prime}(0)$, we obtain
$1=\frac{1}{6}\left(f^{(3)}(s)+f^{(3)}(t)\right),$
which is the desired result. Note that we made no use of the hypothesis $f(0)=0$.
by Diamond (66,975 points)

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