Theorem (0.4 in notes)
If \(f\) is increasing on \((a,b)\), then \(f'(x) \ge 0\) for all \(x \in (a,b)\).
Proof
Fix any point \(x \in (a,b)\). For \(h > 0\) small enough, both \(x\) and \(x+h\) lie in \((a,b)\). Since \(f\) is increasing, we have
$$
\begin{align*}
f(x+h) - f(x) \ge 0.
\end{align*}
$$
Hence
$$
\begin{align*}
\frac{f(x+h) - f(x)}{h} \ge 0.
\end{align*}
$$
Similarly, for \(h < 0\) small enough we still have \(x+h \in (a,b)\), and because \(h < 0\), the inequality \(f(x+h) \le f(x)\) implies
$$
\begin{align*}
\frac{f(x+h) - f(x)}{h} \ge 0.
\end{align*}
$$
Taking limits as \(h \to 0\) from both sides, we obtain
$$
\begin{align*}
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \ge 0.
\end{align*}
$$
Thus, \(f'(x) \ge 0\) for all \(x \in (a,b)\). \(\ \blacksquare\)
Note
The statement “If \(f\) is strictly increasing, then \(f'(x) > 0\) for all \(x\)” is false. Consider the function
$$
\begin{align*}
f(x) = x^{3}.
\end{align*}
$$
It is strictly increasing on all of \(\mathbb{R}\), since for all \(x_1 < x_2\),
$$
\begin{align*}
x_1^{3} < x_2^{3}.
\end{align*}
$$
However, its derivative is
$$
\begin{align*}
f'(x) = 3x^{2},
\end{align*}
$$
which satisfies \(f'(0) = 0\). Thus, \(f'(x)\) is not strictly positive everywhere, even though \(f\) is strictly increasing.
References
- Introduction to Analysis, An, 4th edition by William Wade
- Lecture Notes by Professor Chun Kit Lai