Given, f(x)=|x2−3x+2|+2x−3 is defined on [−2,1] x2−3x+2=(x−1)(x−2) So, on the interval [−2,1] If x≤1, then x−1≤0 If x≤2 then x−2≤0 ∴ On [−2,1],(x−1)(x−2)≥0 So, |x2−3x+2|=x2−3x+2 for x∈[−2,1] ∴f(x)=(x2−3x+2)+2x−3 =x2−x−1 Now, f′(x)=2x−1 For, critical points, set f′(x)=0 we get 2x−1=0 ⇒x=
1
2
∈[−2,1] Now, at x=−2f(−2)=(−2)2−(−2)−1 ⇒4+2−1=5 At x=
1
2
,f(
1
2
)=(
1
2
)2−
1
2
−1 ⇒
1
4
−
1
2
−1⇒
1−2−4
4
=−
5
4
At x=1,f(1)=12−1−1=−1 So, the absolute minimum value, m=
−5
4
and the absolute maximum value, M=5 ∴M−4m=5−4×(−
