Find the absolute maximum and minimum values of the function $f (x) = x^{3} - 6 x^{2} + 9 x + 1$ on the closed interval $[0, 4]$ .

medium 2 marks Class 12 · Mathematics · Application of Derivatives

Answer

Maximum value

5

, Minimum value

1

The correct answer is Maximum value $5$ , Minimum value $1$ — Hence the absolute minimum value is $1$ and the absolute maximum value is $5$ .

Step-by-step solution

Compute the derivative: $f^{'} (x) = 3 x^{2} - 12 x + 9$ .
Factor the derivative: $f^{'} (x) = 3 (x^{2} - 4 x + 3) = 3 (x - 1) (x - 3)$ .
Set $f^{'} (x) = 0$ to locate critical points: $x = 1$ and $x = 3$ .
Check that both critical points lie inside the interval $[0, 4]$ .
Evaluate $f (x)$ at the endpoints and at the critical points:
$f (0) = 1$ ;
$f (1) = 1 - 6 + 9 + 1 = 5$ ;
$f (3) = 27 - 54 + 27 + 1 = 1$ ;
$f (4) = 64 - 96 + 36 + 1 = 5$ .
Compare the values: the smallest is $1$ (at $x = 0$ and $x = 3$ ), the largest is $5$ (at $x = 1$ and $x = 4$ ).
Hence the absolute minimum value is $1$ and the absolute maximum value is $5$ .

Common mistake: Neglecting to evaluate the function at the interval endpoints, which can also be extrema.

Concept tested: Absolute extrema on a closed interval