Find the absolute maximum and minimum values of $f (x) = 2 x^{4} - 8 x^{3} + 9 x^{2}$ on the closed interval $[0, 3]$ .

hard 3 marks Class 12 · Mathematics · Application of Derivatives

Answer

Minimum 0 at

x = 0

; Maximum 27 at

x = 3

The correct answer is Minimum 0 at $x = 0$ ; Maximum 27 at $x = 3$ — Hence absolute minimum 0 at $x = 0$ , absolute maximum 27 at $x = 3$ .

Step-by-step solution

Compute the derivative $f^{'} (x) = 8 x^{3} - 24 x^{2} + 18 x = 2 x (4 x^{2} - 12 x + 9)$ .
Set $f^{'} (x) = 0$ to find critical points: $x = 0$ or $4 x^{2} - 12 x + 9 = 0$ .
Solve $4 x^{2} - 12 x + 9 = 0$ gives a double root $x = 1.5$ .
Evaluate $f$ at the critical points and endpoints: $f (0) = 0$ , $f (1.5) = 3.375$ , $f (3) = 27$ .
The smallest value is $0$ at $x = 0$ ; the largest value is $27$ at $x = 3$ .
Hence absolute minimum 0 at $x = 0$ , absolute maximum 27 at $x = 3$ .

Common mistake: Ignoring the endpoints of the interval

Concept tested: Absolute extrema on closed interval