First derivative test fails for function that is discontinuous at the critical point

Statement

What fails

Suppose $f$ is a function and $c$ is a point in the interior of the domain of $f$. Suppose $f$ is not continuous at $c$. Then, we cannot use the first derivative test directly, and naively looking at the statement of the test could yield incorrect conclusions.

The remedy

See variation of first derivative test for discontinuous function with one-sided limits.

Example

Consider the function:

$f(x) := \lbrace \begin{array}{rl} x^2, & x < 0 \\ x^3 + 1, & x\ge 0 \\\end{array}$

We see that:

$f'(x) = \lbrace \begin{array}{rl} 2x, & x < 0 \\ 3x^2, & x > 0 \\\end{array}$

Note that $f'$ is not defined at 0 because $f$ is not continuous at zero.

We thus see that $f'(x) < 0$ for $x$ to the immediate left of 0 and $f'(x) > 0$ for $x$ to the immediate right of 0. Thus, $f'$ is changing sign from negative to positive. A naive application of the first derivative test suggests that $f$ has a local minimum at zero. However, this is not correct. The reason is that the function jumps upward at 0, hence it does not attain a local minimum from the left at zero.