Single Variable Calculus, Chapter 4, 4.1, Section 4.1, Problem 68

Show that 5 is a critical number
Suppose that $g(x) = 2 + (x - 5)^3$, show that 5 is a critical number but $g$ does not have a local extreme value at 5.
Taking the derivative of function,

$\begin{equation} \begin{aligned} g'(x) &= \frac{d}{dx}(2) + \frac{d}{dx} (x-5)^3 \cdot \frac{d}{dx} (x-5)\\ \\ g'(x) &= 0 + 3(x-5)^2 (1)\\ \\ g'(x) &= 3(x-5)^2 \end{aligned} \end{equation}$

Checking if 5 is a critical number,


$\begin{equation} \begin{aligned} g'(5) &= 3(5-5)^2\\ \\ g'(5) &= 3(0)^2\\ \\ g'(5) &= 0 \end{aligned} \end{equation}$

It shows that 5 is a critical number of $g(x)$. But as you can see that $g'(x)$ has positive coefficient and even power. Hence, $g'(x)$ is always positive and $g(x)$ is always increasing. Therefore, $g$ does not have a local extreme value at 5.