Let $f$ be a function defined on $[a, b]$ such that $f^{\prime}(x) > 0$ for all $x \in (a, b)$. Prove that $f$ is an increasing function on $(a, b)$.

To prove that $f$ is an increasing function on $(a, b)$,we use the Mean Value Theorem $(MVT)$.
Let $x_1$ and $x_2$ be any two points in $(a, b)$ such that $x_1 < x_2$.
Since $f$ is defined on $[a, b]$ and $f^{\prime}(x) > 0$ for all $x \in (a, b)$,$f$ is continuous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$.
By the Mean Value Theorem,there exists a point $c \in (x_1, x_2)$ such that $f^{\prime}(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$.
Since $f^{\prime}(x) > 0$ for all $x \in (a, b)$,we have $f^{\prime}(c) > 0$.
Since $x_1 < x_2$,we have $x_2 - x_1 > 0$.
Therefore,$\frac{f(x_2) - f(x_1)}{x_2 - x_1} > 0$,which implies $f(x_2) - f(x_1) > 0$,or $f(x_2) > f(x_1)$.
Since $x_1 < x_2$ implies $f(x_1) < f(x_2)$ for any $x_1, x_2 \in (a, b)$,the function $f$ is strictly increasing on $(a, b)$.