Theorem (0.4 in notes)
If \(f'(x) > 0\) for all \(x \in (a,b)\), then \(f\) is strictly increasing on \((a,b)\).
Proof
Let \(x_1, x_2 \in (a,b)\) with \(x_1 < x_2\). By the Mean Value Theorem, there exists a point \(c \in (x_1, x_2)\) such that
$$
\begin{align*}
f(x_2) - f(x_1) &= f'(c)\,(x_2 - x_1).
\end{align*}
$$
Since \(f'(c) > 0\) and \(x_2 - x_1 > 0\), it follows that
$$
\begin{align*}
f(x_2) - f(x_1) &> 0,
\end{align*}
$$
and hence
$$
\begin{align*}
f(x_2) &> f(x_1).
\end{align*}
$$
Thus, \(f\) is strictly increasing on \((a,b)\). \(\ \blacksquare\)
References
- Introduction to Analysis, An, 4th edition by William Wade
- Lecture Notes by Professor Chun Kit Lai