Exercise 5.17

Question

Exercise 17:
    Suppose $f$ is a real, three-times differentiable function on $[-1,1]$, such that $$f(-1)=0,\quad f(0)=0,\quad f(1)=1,\quad f'(0)=0.$$ Prove that $f^{(3)}(x)\ge 3$ for some
    $x\in(-1,1)$.

analambanomenos · Accepted Answer

Following the hint, applying Theorem 5.15 with $\alpha=0$ and $\beta=\pm1$, we get
    \begin{align*}
        f(1) &= 1 = \frac{f''(0)}{2}+\frac{f^{(3)}(s)}{6} \
        f(-1) &= 0 = \frac{f''(0)}{2}-\frac{f^{(3)}(t)}{6}
    \end{align*}
    for some $s\in(0,1)$ and $t\in(-1,0)$. Subtracting the second equation from the first, we get $6=f^{(3)}(s)+f^{(3)}(t)$, so at least one of the two terms is $\ge 3$.

Exercise 5.17

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Exercise 5.17

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