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MathsGee Android Q&A

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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 \(\frac{1}{2}\left(x^{3}+x^{2}\right)\).
in Mathematics by Diamond (66,975 points) | 63 views

1 Answer

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Best answer
\[
\begin{aligned}
f(1) &=f(0)+f^{\prime}(0)+\frac{1}{2} f^{\prime \prime}(0)+\frac{1}{6} f^{(3)}(s) \\
f(-1) &=f(0)-f^{\prime}(0)+\frac{1}{2} f^{\prime \prime}(0)-\frac{1}{6} f^{(3)}(t)
\end{aligned}
\]
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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MathsGee Android Q&A

MathsGee Android Q&A