Given that
- $f:\Bbb{R}\longrightarrow \Bbb{R}$ is a differentiable function
- $x_0\inℝ$
- $f$ is bijective with $f^{-1}:\Bbb{R}\longrightarrow \Bbb{R}$
- $f'(x_0) \ne 0$
- $f^{-1}$ is continuous
Use Carathéodory’s rule of the derivative to prove that $f^{-1}$ is differentiable at $y_0=f(x_0)$
How would you go about starting this proof?
You are given that there exists a continuous function $q$ with the property that $$f(x) = f(x_0) + q(x)(x - x_0)$$ for all $x$ in a neighborhood of $x_0$. The continuous function $q$ satisfies $q(x_0) = f'(x_0)$ so $q(x_0) \not= 0$ and thus $q(x) \not = 0$ in another neighborhood of $x_0$. Denote the common neighborhood by $I$.
Since $f$ is continuous and one-to-one, $f(I)$ is a neighborhood of $f(x_0)$. Denote this neighborhood by $J$. Let $y_0 = f(x_0)$.
Let $y \in J$ and let $x = f^{-1}(y)$. Then $x \in I$ so that $$f(x) = f(x_0) + q(x)(x - x_0) \implies y = y_0 + q(f^{-1}(y))(f^{-1}(y) - f^{-1}(y_0))$$ which rearranges to $$f^{-1}(y) = f^{-1}(y_0) + \frac{1}{q(f^{-1}(y))} (y - y_0).$$
Since $q \circ f^{-1}$ is continuous and nonzero on $J$ so is its reciprocal. Now invoke Caratheodory.