Question 38
Let $f : [0, 1] \to \mathbb{R}$ be a function which is continuous on $[0, 1]$ and differentiable in $0$. Consider the function $s : [0, 1] \to \mathbb{R}$,
$$
s(x) = \sup\{c \in [0; x] \mid \int_{0}^{x} f(t) \, dt = x f(c)\}.
$$
a) Prove that the function $f \circ s$ is differentiable in $0$ and $(f \circ s)'(0) = \frac{1}{2} f'(0)$.
b) If $f'(0) \neq 0$, prove that $s$ is differentiable in $0$ and $s'(0) = \frac{1}{2}$.
$$
s(x) = \sup\{c \in [0; x] \mid \int_{0}^{x} f(t) \, dt = x f(c)\}.
$$
a) Prove that the function $f \circ s$ is differentiable in $0$ and $(f \circ s)'(0) = \frac{1}{2} f'(0)$.
b) If $f'(0) \neq 0$, prove that $s$ is differentiable in $0$ and $s'(0) = \frac{1}{2}$.