Question

Find the absolute maximum and minimum values of the function f given by

Answer 1

The function is given as f( x )= cos 2 x+sinx in the interval x∈[ 0,π ].

Differentiate the given function with respect to x,

f ′ ( x )=−2cosxsinx+cosx

Put f ′ ( x )=0, then,

−2cosxsinx+cosx=0 cosx( 2sinx−1 )=0

Then,

cosx=0 x= π 2

Or,

sinx= 1 2 x= π 6

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

f( π 2 )= cos 2 π 2 +sin π 2 =0+1 =1

Substitute the value x= π 6 in the given function,

f( π 6 )= cos 2 π 6 +sin π 6 = ( 3 2 ) 2 + 1 2 = 3 4 + 1 2 = 5 4

Substitute the value x=0 in the given function,

f( 0 )= cos 2 0+sin0 =1

Substitute the value x=π in the given function,

f( π )= cos 2 π+sinπ = ( −1 ) 2 +0 =1

It can be observed that the absolute maximum value of the given function is 5 4 at the point x= π 6 and the absolute minimum value is 1 at the points x= π 2 ,0,π.