Q.Find the points of local maximum and local minimum of $f(x) = x^3 - 6x^2 + 9x + 15$, and state the corresponding maximum and minimum values.

Find the stationary points by setting $f'(x)=0$, then classify each using the second derivative test.

Step 1 — Find $f'(x)$ and the stationary points. $f'(x) = 3x^2-12x+9 = 3(x^2-4x+3) = 3(x-1)(x-3)$. Setting $f'(x)=0$: $x=1$ or $x=3$.

Step 2 — Find $f''(x)$ for the second derivative test. $f''(x) = 6x-12$.

Step 3 — Classify $x=1$. $f''(1) = 6(1)-12=-6<0$, so $x=1$ is a local maximum.

Step 4 — Classify $x=3$. $f''(3) = 6(3)-12=6>0$, so $x=3$ is a local minimum.

Step 5 — Find the function values. $f(1) = 1-6+9+15=19$. $f(3)=27-54+27+15=15$. …

The first-derivative sign-change check in the Independent check above is a reliable alternative whenever $f''(x)$ is awkward to evaluate, and alwa …

Stopping after finding the stationary points $x=1,3$ without classifying them (or assuming the smaller $x$-value is always the maximum) is a common incomplete answer — the classification must always c …