Question

5.Find the absolute maximum value and the absolute minimum value of the followingfunctions in the given interval(i) f(x)-. xe [-2, 2](ii) f(x)-sin x + cos x , x e [0, π](ii) f(x) -4xx)f(x(12+3, xel-3,1]11) JC

Answer 1

(i)

The function is given as f( x )= x 3 in the interval x∈[ −2,2 ].

Differentiate the given function with respect to x,

f ′ ( x )=3 x 2

Put f ′ ( x )=0, then,

3 x 2 =0 x=0

Substitute the value x=0 in the given function,

f( 0 )= ( 0 ) 3 =0

Substitute the value x=−2 in the given function,

f( −2 )= ( −2 ) 3 =−8

Substitute the value x=2 in the given function,

f( 2 )= ( 2 ) 3 =8

It can be observed that the absolute maximum value of the given function is 8 and the absolute minimum value is −8.

(ii)

The function is given as f( x )=sinx+cosx in the interval x∈[ 0,π ].

Differentiate the given function with respect to x,

f ′ ( x )=cosx−sinx

Put f ′ ( x )=0, then,

cosx−sinx=0 cosx=sinx tanx=1 x= π 4

Substitute the value x= π 4 in the given function,

f( π 4 )=cos π 4 +sin π 4 = 1 2 + 1 2 = 2 2 = 2

Substitute the value x=0 in the given function,

f( 0 )=cos0+sin0 =1

Substitute the value x=π in the given function,

f( π )=cosπ+sinπ =−1

It can be observed that the absolute maximum value of the given function is 2 and the absolute minimum value is −1.

(iii)

The function is given as f( x )=4x− 1 2 x 2 in the interval x∈[ −2, 9 2 ].

Differentiate the given function with respect to x,

f ′ ( x )=4−x

Put f ′ ( x )=0, then,

4−x=0 x=4

Substitute the value x=4 in the given function,

f( 4 )=4( 4 )− 1 2 ( 4 ) 2 =16−8 =8

Substitute the value x=0 in the given function,

f( −2 )=4( −2 )− 1 2 ( −2 ) 2 =−8−2 =−10

Substitute the value x= 9 2 in the given function,

f( 9 2 )=4( 9 2 )− 1 2 ( 9 2 ) 2 =18− 81 8 = 63 8 =7.87

It can be observed that the absolute maximum value of the given function is 8 and the absolute minimum value is −10.

(iv)

The function is given as f( x )= ( x−1 ) 2 +3 in the interval x∈[ −3,1 ].

Differentiate the given function with respect to x,

f ′ ( x )=2( x−1 )

Put f ′ ( x )=0, then,

2( x−1 )=0 x=1

Substitute the value x=1 in the given function,

f( 1 )= ( 1−1 ) 2 +3 =3

Substitute the value x=−3 in the given function,

f( −3 )= ( −3−1 ) 2 +3 = ( −4 ) 2 +3 =16+3 =19

It can be observed that the absolute maximum value of the given function is 19 and the absolute minimum value is 3.