Question

3.Find the local maxima and local minima, if any, of the following functions. Findalso the local maximum and the local minimum values, as the case may be(i) f(x) = x2(iii) h (x) = sin x + cos x, 0 < x <(iv) f(x)-sin-cos x, 0 < x < 2π(v) f(x)=x3-6x2+9+15 (vi)(vii) g(x)=x2+2(ii) g(x)=x3-3xg(x)=-+-,x>0(viii)f(x)=W1-х, О < x <1Vill

Answer 1

(i)

Given function is,

f( x )= x 2

Differentiate the given function with respect to x,

f ′ ( x )=2x(1)

Put f ′ ( x )=0.

2x=0 x=0

This shows that 0 is the only critical point of the local maxima or local minima of the function.

Differentiate equation (1) with respect to x,

f ″ ( x )=2 f ″ ( 0 )=2 >0

Therefore, x=0 is the point of local minima and the local minimum value of the function is f( 0 )=0.

(ii)

Given function is,

g( x )= x 3 −3x

Differentiate the function with respect to x,

g ′ ( x )= d( x 3 −3x ) dx =3 x 2 −3 (1)

Put g'( x )=0,

3 x 2 −3=0 x 2 =1 x=±1

So, x=1and x=−1 are the two critical points of the given function.

Differentiate equation (1) with respect to x,

g ″ ( x )=6x g ″ ( 1 )=6 g ″ ( −1 )=−6

This shows that at x=1, the function is positive, so x=1 is a local minima where the local minimum value of the given function is,

g( 1 )= ( 1 ) 3 −3( 1 ) =1−3 =−2

And at x=−1, the function is negative, so x=−1 is a local maxima where the local maximum value of the given function is,

g( −1 )= ( −1 ) 3 −3( −1 ) =−1+3 =2

(iii)

Given function is,

h( x )=sinx+cosx  ,  0<x< π 2

Differentiate the function with respect to x,

h ′ ( x )=cosx−sinx(1)

Put h ′ ( x )=0,

cosx−sinx=0 cosx=sinx tanx=1 x= π 4

So, x= π 4 is the critical point of the given function.

Differentiate equation (1) with respect to x,

h ″ ( x )=−sinx−cosx h ″ ( π 4 )=−sin( π 4 )−cos( π 4 ) h ″ ( π 4 )=− 1 2 − 1 2 h ″ ( π 4 )=− 2 2

This shows that at x= π 4 , the function is negative, so x= π 4 is a local maxima where the local maximum value of the given function is,

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

(iv)

Given function is,

f( x )=sinx−cosx  ,  0<x<2π

Differentiate the function with respect to x,

f ′ ( x )=cosx+sinx(1)

Put f ′ ( x )=0,

cosx+sinx=0 cosx=−sinx tanx=−1 x= 3π 4 , 7π 4

Differentiate equation (1) with respect to x,

f ″ ( x )= d( cosx+sinx ) dx =−sinx+cosx

Substitute x= 3π 4 ,

f ″ ( x )=−sin( 3π 4 )+cos( 3π 4 ) =− 1 2 − 1 2 =− 2

For, x= 3π 4 , f ″ ( x ) is negative, so, x= 3π 4 is the point of local maxima where the local maximum value of the given function is,

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

At x= 7π 4 ,

f ″ ( x )=−sin 7π 4 +cos 7π 4 = 1 2 + 1 2 = 2

At x= 7π 4 , f ″ ( x ) is positive, so, x= 7π 4 is the point of local minima where the local minimum value of the given function is,

f( 7π 4 )=sin( 7π 4 )−cos( 7π 4 ) =− 1 2 − 1 2 =− 2

(v)

Given function is,

f( x )= x 3 −6 x 2 +9x+15

Differentiate the function with respect to x,

f ′ ( x )=3 x 2 −12x+91)

Put f ′ ( x )=0,

3 x 2 −12x+9=0 3 x 2 −3x−9x+9=0 ( 3x−9 )( x−1 )=0 x=1,3

Differentiate equation (1) with respect to x,

f ″ ( x )=6x−12 f ″ ( 1 )=−6 f ″ ( 3 )=6

This shows that at x=1, the function is negative, so x=1 is a local maxima where the local maximum value of the given function is,

f( 1 )= ( 1 ) 3 −6 ( 1 ) 2 +9( 1 )+15 =1−6+9+15 =19

And at x=3, the function is positive, so x=3 is a local minima where the local minimum value of the given function is,

f( 3 )= ( 3 ) 3 −6 ( 3 ) 2 +9( 3 )+15 =27−54+27+15 =15

(vi)

Given function is,

g( x )= x 2 + 2 x   ,  x>0

Differentiate the function with respect to x,

g ′ ( x )= 1 2 − 2 x 2 (1)

Put g ′ ( x )=0,

1 2 − 2 x 2 =0 1 2 = 2 x 2 x 2 =4 x=±2

Since x>0, so, x=2 is the only critical point of the given function.

Differentiate equation (1) with respect to x,

g ″ ( x )= 4 x 3 g ″ ( 2 )= 4 ( 2 ) 3 g ″ ( 2 )= 1 2 >0

This shows that x=2 is the local minima of the function where the local minimum value of the function is,

g( 2 )= 2 2 + 2 2 =1+1 =2

(vii)

Given function is,

g( x )= 1 x 2 +2

Differentiate the function with respect to x,

g ′ ( x )= d( 1 x 2 +2 ) dx = ( x 2 +2 )×0−2x×1 ( x 2 +2 ) 2 = −2x ( x 2 +2 ) 2 (1)

Put g ′ ( x )=0,

−2x ( x 2 +2 ) 2 =0 x=0

Differentiate equation (1) with respect to x,

g ″ ( x )= d( −2x ( x+2 ) 2 ) dx = ( x+2 ) 2 ( −2 )−( −2x )2( x+2 ) ( x+2 ) 4 = 2( x−2 ) ( x+2 ) 3 g ″ ( 0 )=− 1 2

This shows that x=0 is the point of local maxima where local maximum value of the given function is g( 0 )= 1 2 .

(viii)

Given function is,

f( x )=x 1−x

Differentiate the function with respect to x,

f ′ ( x )= 1−x +x 1 2 1−x ×( −1 ) = 2( 1−x )−x 2 1−x = 2−3x 2 1−x (1)

Put f ′ ( x )=0,

2−3x 2 1−x =0 2−3x=0 x= 2 3

Differentiate equation (1) with respect to x,

f ″ ( x )= 1 2 [ 1−x ( −3 )−( 2−3x )( − 1 2 1−x ) 1−x ] = −6( 1−x )+( 2−3x ) 4 ( 1−x ) 3 2 = 3x−4 4 ( 1−x ) 3 2 f ″ ( 2 3 )= −1 2 ( 1 3 ) 3 2

This shows that x= 2 3 is the point of local maxima where the local maximum value of the function is,

f( 2 3 )= 2 3 1− 2 3 = 2 3 1 3 = 2 3 3 = 2 3 9