Question

1. Find the maximum and minimum values,if any, of the following functionsgiven byii) f(x)- 9x2 12x 2(ii) fx)-- (r - 1)2+10 v) g)31

Answer 1

(i)

The function is given as,

f( x )= ( 2x−1 ) 2 +3

Differentiate above equation with respect to x,

f'( x )=2( 2x−1 )2+0 =4( 2x−1 ) =8x−4

Put f ′ ( x )=0,

8x−4=0 x= 1 2

When x< 1 2 ,f'( x ) is negative and when x> 1 2 ,f'( x ) is positive.

Since the derivative of the given function changes its sign from negative to positive at x= 1 2 , thus x= 1 2 is the point of minima.

Since the value of the function increases with increase in value of x therefore, it doesn’t have any point of maxima.

Substitute x= 1 2 in the given function,

f( 1 2 )= ( 2( 1 2 )−1 ) 2 +3 =( 0 )+3 =3

Therefore, the minimum value of the given function is 3.

(ii)

The function is given as,

f( x )=9 x 2 +12x+2

Differentiating the given function with respect to x,

f'( x )=18x+12

Put f ′ ( x )=0,

18x+12=0 x=− 12 18 =− 2 3

When x<− 2 3 ,f'( x ) is negative and when x>− 2 3 ,f'( x ) is positive.

Since the derivative of the function changes its sign from negative to positive at x=− 2 3 , thus x=− 2 3 is the point of minima.

There is no maximum of the above function as the value of function goes increasing with increase in the value of x.

Substitute x=− 2 3 in the given function,

f( − 2 3 )=9 ( − 2 3 ) 2 +12( − 2 3 )+2 =9( 4 9 )−4( 2 )+2 =−2

Therefore, the minimum value of the given function is −2.

(iii)

The function is given as,

f( x )=− ( x−1 ) 2 +10

Differentiate the given function with respect to x,

f'( x )=−2( x−1 )

Put f ′ ( x )=0,

−2( x−1 )=0 x=1

When x<1, f'( x ) is positive and when x>1, f'( x ) is negative.

Since, the derivative of the given function changes its sign from positive to negative at x=1, thus x=1 is the point of maxima.

Since, the value of derivative of the function goes decreasing for higher value of x, thus there is no point of minima.

Substitute x=1 in the given function,

f( 1 )=− ( 1−1 ) 2 +10 =10

Therefore, the maximum value of the given function is 10.

(iv)

The function is given as,

g( x )= x 3 +1

Differentiate the given function with respect to x,

g'( x )=3 x 2

Put f ′ ( x )=0,

3 x 2 =0 x=0

When x<0,g'( x ) is positive and when x>0,g'( x ) is also positive.

Since, the derivative of the given function neither changes its sign from positive to negative, nor negative to positive at x=0, thus x=0 is neither point of local maxima nor point of local minima.

So, x=0 is point of inflection.