Exercise 5.2

Question

Exercise 2: Suppose $f'(x)>0$ 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
    $$g'\bigl(f(x)\bigr)=\frac{1}{f'(x)}\quad\quad(a<x<b).$$

ghostofgarborg · Accepted Answer

\def\eps{\varepsilon} ( extit{Matt ``Frito'' Lundy}) \ If $f$ were not strictly increasing in $(a,b)$, there would exist $x,y$ with both $a < x < y < b$ and $f(x) \geq f(y)$. By ``the'' mean value theorem, there would exist a $z \in (x,y)$ such that $f'(z) = \frac{f(y) - f(x)}{y - x} \leq 0$, which contradicts $f'(x) >0$ in $(a,b)$, so $f$ must be strictly increasing. Fix $\varepsilon > 0$ and $x \in (a,b)$. $f'(x) >0$ means there exists a $\eta > 0$ and a $\delta_1 >0$ such that $0 < \left| x - t ight| < \delta_1$ implies $$\left| \frac{f(t) - f(x)}{t-x} ight| > \eta$$ So: $$\frac{1}{\left| \frac{f(t) - f(x)}{t-x} ight|} < \frac{1}{\eta}$$ From the definition of $f'(x)$, we also have for $\eta f'(x) \varepsilon > 0$, there exists a $\delta_2 > 0$ such that $0 < \left| x - t ight| < \delta_2$ implies $$\left| f'(x) - \frac{f(t) - f(x)}{t-x} ight| < \eta f'(x) \varepsilon$$ Let $y = f(x)$ and $\delta = \min \{\delta_1,\delta_2\}$. Then for any $u \in (f(x - \delta), f(x+\delta))$, let $g(u) = t$ so $t \in (x -\delta, x + \delta)$ and: \begin{align*} \left|\frac{g(u) - g(y)}{u - y} - \frac{1}{f'(x)} ight| &= \left| \frac{f'(x) - \frac{f(t) - f(x)}{t-x}}{f'(x) \left[ \frac{f(t) - f(x)}{t-x} ight] } ight| \ &< \frac{\eta f'(x) \varepsilon}{\eta f'(x)} \ &= \varepsilon \end{align*} Which shows that $$g'(f(x)) = \frac{1}{f'(x)}.$$

Exercise 5.2

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Exercise 5.2

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