Question 96
Let $f : [0, 1] \to (0, \infty)$ be a continuous function on $[0, 1]$, and
$$A = \int_{0}^{1} f(t) \, dt.$$
a) Show that the function $F : [0, 1] \to [0, A]$, defined for any $x \in [0, 1]$ by
$$
F(x) = \int_{0}^{x} f(t) \, dt,
$$
is invertible, with a differentiable inverse.
b) Show that the equation
$$
\int_{0}^{x} f(t) \, dt = \int_{g(x)}^{1} f(t) \, dt \quad (1)
$$
where $g$ is a function $[0, 1] \to [0, 1]$, has a unique solution $g(x)$ for each $x \in [0, 1]$, and express $g(x)$ in terms of $F$ and $F^{-1}$.
c) Show that $g$ is differentiable, strictly decreasing, and has a unique fixed point $c \in [0, 1]$. Compute $g'(c)$.
$$A = \int_{0}^{1} f(t) \, dt.$$
a) Show that the function $F : [0, 1] \to [0, A]$, defined for any $x \in [0, 1]$ by
$$
F(x) = \int_{0}^{x} f(t) \, dt,
$$
is invertible, with a differentiable inverse.
b) Show that the equation
$$
\int_{0}^{x} f(t) \, dt = \int_{g(x)}^{1} f(t) \, dt \quad (1)
$$
where $g$ is a function $[0, 1] \to [0, 1]$, has a unique solution $g(x)$ for each $x \in [0, 1]$, and express $g(x)$ in terms of $F$ and $F^{-1}$.
c) Show that $g$ is differentiable, strictly decreasing, and has a unique fixed point $c \in [0, 1]$. Compute $g'(c)$.