Concept preview

What you will be able to do

1. 01State the second derivative test: at a critical number $c$ with $f'(c)=0$, a positive $f''(c)$ gives a local minimum and a negative $f''(c)$ gives a local maximum.
2. 02Apply the test numerically by evaluating $f''$ at each critical number, and read the result as concavity — concave up cradles a minimum, concave down caps a maximum.
3. 03Recognise that $f''(c)=0$ makes the test INCONCLUSIVE, and fall back on the first derivative test to decide whether $c$ is a maximum, a minimum, or neither.

Suppose $c$ is a critical number of a twice-differentiable function $f$, meaning $f'(c)=0$ so the tangent line at $x=c$ is horizontal. The second derivative test inspects the concavity there. If $f''(c)>0$ the graph is concave up at $c$, so the horizontal tangent sits at the bottom of a local dip: $c$ is a LOCAL MINIMUM. If $f''(c)<0$ the graph is concave down at $c$, so the horizontal tangent caps a local rise: $c$ is a LOCAL MAXIMUM. And if $f''(c)=0$, the test is silent — it is INCONCLUSIVE — and you must decide the matter another way.