Maxima and Minima -- Second Derivative Test

A faster alternative at a single point. The first derivative test needs the sign of $f'$ on both sides of a critical point $c$. The second derivative test instead asks only about the behaviour of $f$ exactly at $c$, via the second derivative $f''(c)$ -- often much quicker when $f''$ is easy to compute, though (as the WBCHSE syllabus notes) it is "given as a provable tool" here rather than derived from scratch.

Why this works -- the concavity intuition. The second derivative $f''(x)$ measures how the slope $f'(x)$ itself is changing, i.e. it describes the curvature (concavity) of the graph. If $f''(c) < 0$, the slope $f'$ is decreasing as $x$ passes through $c$ -- consistent with $f'$ going from positive (rising) to negative (falling), exactly the $+\to-$ sign change the first derivative test looks for at a local maximum; the curve is "concave down" near $c$, shaped like the top of a dome. If $f''(c) > 0$, the slope is increasing through $c$, consistent with $f'$ going from negative to positive -- a local minimum, with the curve "concave up," shaped like the bottom of a bowl. This concavity picture is the intuition behind the test being a provable tool: it can be justified rigorously via a local (Taylor-type) approximation of $f$ near $c$, but the picture above is sufficient to use it correctly and is the level expected here.

Worked method. (1) Find $f'(x)$ and solve $f'(x) = 0$ for the critical points. (2) Find $f''(x)$. (3) Evaluate $f''$ at each critical point and apply the three-way rule above. (4) Where the value is asked for (not just the location), substitute the critical $x$ back into the original $f(x)$ -- see Example 8, where $f''(1) = -6 < 0$ gives a local maximum value $f(1) = 19$, and $f''(3) = 6 > 0$ gives a local minimum value $f(3) = 15$.