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 Further note that f ′(x) is not defined at x = 0 So the
critical points are  x = 0 and 
x =81 Now evaluating the value of f at critical points
x = 0, 1
8  and at end points of the interval x = –1 and x = 1, we have
f (–1) =
4
1
3
3
12( 1)
6( 1)
18
−
−
−
=
f (0) = 12 (0) – 6(0) = 0
Rationalised 2023-24
 MATHEMATICS
174
f 81




  =
4
1
3
3
1
1
9
12
6
8
8
−4




−
=








f (1) =
4
1
3
3
12(1)
6(1)
6
−
=
Hence, we conclude that absolute maximum value of  f  is 18 that occurs at x = –1
and absolute minimum value of f is 
49
−  that occurs at 
x =81