Exercise 5.2.12

Question

[Inverse functions]
  If $f:[a, b] ightarrow \mathbf{R}$ is one-to-one, then there exists an inverse function $f^{-1}$ defined on the range of $f$ given by $f^{-1}(y)=$ $x$ where $y=f(x)$. In Exercise 4.5.8 we saw that if $f$ is continuous on $[a, b]$, then $f^{-1}$ is continuous on its domain. Let's add the assumption that $f$ is differentiable on $[a, b]$ with $f^{\prime}(x) 
eq 0$ for all $x \in[a, b]$. Show $f^{-1}$ is differentiable with
  $$
  \left(f^{-1}ight)^{\prime}(y)=\frac{1}{f^{\prime}(x)} \quad 	ext { where } y=f(x) .
  $$

Ulisse Mini · Accepted Answer

Typically I'd use the chain rule, but to be rigorous we must prove $f^{-1}$ is differentiable.
  Consider the limit
  $$
  \left(f^{-1}ight)'(d) = \lim_{y 	o d} \frac{f^{-1}(y) - f^{-1}(d)}{y-d}
  $$
  Since both $y$ and $d$ are in the range of $f$ we can substitute $y=f(x)$ and $d=f(c)$
  $$
  \lim_{f(x) 	o f(c)} \frac{x - c}{f(x) - f(c)} = \lim_{x 	o c} \frac{x - c}{f(x) - f(c)} = \frac{1}{f'(c)}
  $$
  Changing $f(x) 	o f(a)$ into $x 	o a$ is possible because $f$ is continuous, meaning that $x$ being close to $a$ causes $f(x)$ to be close to $f(a)$. (This can be made more rigorous, but I'm lazy)

Exercise 5.2.12

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Exercise 5.2.12

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