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Suppose \(f^{\prime}(x)>0\) in \((a, b)\). Prove that \(f\) is strictly increasing in \((a, b)\), and let \(g\) be its inverse function. Prove that \(g\) is differentiable, and that
\[
g^{\prime}(f(x))=\frac{1}{f^{\prime}(x)} \quad(a<x<b)
\]
in Mathematics by Diamond (66,769 points) | 71 views

1 Answer

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For any \(c, d\) with \(a<c<d<b\) there exists a point \(p \in(c, d)\) such that \(f(d)-f(c)=f^{\prime}(p)(d-c)>0\). Hence \(f(c)<f(d)\).

We know that the inverse function \(g\) is continuous. (Its restriction to each closed subinterval \([c, d]\) is continuous, and that is sufficient.) Now observe that if \(f(x)=y\) and \(f(x+h)=y+k\), we have
\[
\frac{g(y+k)-g(y)}{k}-\frac{1}{f^{\prime}(x)}=\frac{1}{\frac{f(x+h)-f(x)}{h}}-\frac{1}{f^{\prime}(x)}
\]
Since we know \(\lim \frac{1}{\varphi(t)}=\frac{1}{\lim \varphi(t)}\) provided \(\lim \varphi(t) \neq 0\), it follows that for any \(\varepsilon>0\) there exists \(\eta>0\) such that
\[
\left|\frac{1}{\frac{f(x+h)-f(x)}{h}}-\frac{1}{f^{\prime}(x)}\right|<\varepsilon
\]
if \(0<|h|<\eta\). Since \(h=g(y+k)-g(y)\), there exists \(\delta>0\) such that \(0<|h|<\eta\) if \(0<|k|<\delta\). The proof is now complete.
by Diamond (66,769 points)

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