Application of Derivatives: Speed-Solving Techniques (10)

Question

Find the maximum and minimum values of $f(x) = 2x^3 - 15x^2 + 36x + 1$ on the interval $[0, 5]$.

Solution — Step by Step

Final answer: Max $= 56$ at $x = 5$, Min $= 1$ at $x = 0$.

Why This Works

On a closed interval, the absolute maximum and minimum of a continuous function occur either at critical points (where $f' = 0$ or DNE) or at the endpoints. We just check all candidates and pick the largest and smallest.

The local max at $x = 2$ ($f = 29$) and local min at $x = 3$ ($f = 28$) are not the absolute extrema here — the endpoint $x = 5$ wins. This is a classic exam trap.

Alternative Method

Use the second derivative test to classify the critical points (max or min) and combine with endpoint values. $f''(x) = 12x - 30$. At $x = 2$: $f''(2) = -6 < 0$ → local max. At $x = 3$: $f''(3) = 6 > 0$ → local min. Then check endpoints. Same answer.

Common Mistake

Forgetting endpoints. Students find $f(2) = 29$, declare it the maximum, and miss that $f(5) = 56$. On closed intervals, always check both endpoints.