Let $f$ be a differentiable function on open interval $(a,b)$. Suppose $f'(x)$ is not identically zero. Show that there exists an subinterval $(c,d)$ such that $f(x)$ is strictly increasing or strictly decreasing on $(c,d)$.
How to prove this?
I think this statement is wrong…
Best Answer
I show that it's true in case $f$ is not only differentiable but also its derivative is continuous.
Since $f'$ is non-zero function, there is a point $\tilde{x} \in (a, b)$ such that $f'(\tilde{x}) > 0$ or $f'(\tilde{x}) < 0$. Suppose $f'(\tilde{x}) > 0$. Then take $c$ and $d$ as follows: $$ \begin{align*} c &:= \inf\{\, \tilde{c} \mid a \leq \tilde{c} \leq \tilde{x}, \quad f'(c') > 0 \quad \text{for all $c' \in (\tilde{c}, \tilde{x}]$} \,\} \\ d &:= \sup\{\, \tilde{d} \mid \tilde{x} \leq \tilde{d} \leq b, \quad f'(d') > 0 \quad \text{for all $d' \in [\tilde{x}, \tilde{d})$} \,\}. \\ \end{align*}$$ From assumption that $f'$ is continuous, $c \neq \tilde{x} \neq d$. The interval $(c, d)$ is the required one (indeed, the largest interval containing $\tilde{x}$).