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

{{inconclusive case for|first derivative test}}
==Statement==

===What fails===

Suppose <math>f</math> is a function and <math>c</math> is a point in the interior of the domain of <math>f</math>. Suppose <math>f</math> is not continuous at <math>c</math>. 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:

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

We see that:

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

Note that <math>f'</math> is not defined at 0 because <math>f</math> is not continuous at zero.

We thus see that <math>f'(x) < 0</math> for <math>x</math> to the immediate left of 0 and <math>f'(x) > 0</math> for <math>x</math> to the immediate right of 0. Thus, <math>f'</math> is changing sign from negative to positive. A naive application of the first derivative test suggests that <math>f</math> 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.