To find absolute max/min on a closed interval, we need to evaluate f(x) at all critical points and endpoints.
Find critical points by setting f′(x)=0:
f′(x)=6x2−18x+12=0
x2−3x+2=0
(x−1)(x−2)=0
x=1 and x=2
Both lie within [0,3], so both are valid.
Evaluate f(x) at x=0,1,2,3:
f(0)=2(0)3−9(0)2+12(0)−5=−5
f(1)=2(1)3−9(1)2+12(1)−5=2−9+12−5=0
f(2)=2(8)−9(4)+12(2)−5=16−36+24−5=−1
f(3)=2(27)−9(9)+12(3)−5=54−81+36−5=4
From these values:
Absolute maximum value =4, occurring at x=3 (point of maxima)
Absolute minimum value =−5, occurring at x=0 (point of minima)
Note: Even though x=1 and x=2 are critical points, the function reaches its highest and lowest values at the endpoints of the interval. This is why we always check endpoints on a closed interval.
A → (IV), B → (III), C → (I), D → (II)