e , at x = 1, x = 2, x = 3 and at x = 5 So
f (1) = 2(13) – 15(12) + 36 (1) + 1 = 24
f (2) = 2(23) – 15(22) + 36 (2) + 1 = 29
f (3) = 2(33) – 15(32) + 36 (3) + 1 = 28
f (5) = 2(53) – 15(52) + 36 (5) + 1 = 56
Thus, we conclude that absolute maximum value of f on [1, 5] is 56, occurring at
x =5, and absolute  minimum value of f on [1, 5] is 24 which occurs at x = 1 Example 28 Find absolute maximum and minimum values of a function f given by
4
1
3
3
( )
12
6
,
[ 1, 1]
f x
x
x
x
=
−
∈ −
Solution We have
f (x) =
4
1
3
3
12
6
x
x
−
or
f ′(x) =
31
2
2
3
3
2
2(8
1)
16
x
x
x
x
−
−
=
Thus, f ′(x) = 0 gives 
x =81