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6. Find the intervals in which the following functions are strictly increasing ordecreasing:(a) x2 + 2r- 5(c) -2r3- 9x2- 12r +1e) (x 1)3 (x - 3(b) 10 6r 2r2(d) 6-9x -x'

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Solution

(a)

The given function is x 2 +2x5=0.

Let f( x ) be the given function.

f( x )= x 2 +2x+5

Differentiate f( x ) with respect to x.

f ( x )= d dx f( x ) =2x+2

Substitute f( x )=0

2x+2=0 2x=2 x=1

Now, the point x=1 divides the real line into two disjoint intervals given by,

<x<1 1<x<

In the interval <x<1,

f ( x )<0 2x+2<0

The function f( x ) is strictly decreasing in the interval x( ,1 )

In the interval 1<x<,

f ( x )>0 2x+2>0

The function f( x ) is strictly increasing in the interval x( 1, )

Thus, the function is strictly increasing in the interval x>1 and strictly decreasing in the interval x<1

(b)

The given function is 106x2 x 2 .

Let f( x ) be the given function.

f( x )=106x2 x 2

Differentiate f( x ) with respect to x.

f ( x )= d dx f( x ) =64x

Substitute f( x )=0

64x=0 4x=6 x= 6 4 = 3 2

Now, the point x= 3 2 divides the real line into two disjoint intervals given by,

<x< 3 2 3 2 <x<

In the interval <x< 3 2 ,

f ( x )<0 64x<0

The function f( x ) is strictly increasing in the interval x( , 3 2 ) .

In the interval 3 2 <x<,

f ( x )<0 64x<0

The function f( x ) is strictly decreasing in the interval x( 3 2 , ) .

Thus, the function is strictly increasing in the interval x< 3 2 and strictly decreasing in the interval x> 3 2 .

(c)

The given function is 2 x 3 9 x 2 12x+1.

Let f( x ) be the given function.

f( x )=2 x 3 9 x 2 12x+1

Differentiate f( x ) with respect to x.

f ( x )= d dx f( x ) =6 x 2 18x12 =6( x 2 +3x+2 ) =6( x+1 )( x+2 )

Substitute f( x )=0

6( x+1 )( x+2 )=0 x=1,2

Now, the points x=1 and x=2 divide the real line into three disjoint intervals given by,

<x<2 2<x<1 1<x<

In the interval <x<2 and 1<x< ,

f ( x )<0 6( x+1 )( x+2 )<0

The function f( x ) is strictly decreasing in the interval x( ,2 ) and x( 1, ).

In the interval 2<x<1,

f ( x )>0 6( x+1 )( x+2 )>0

The function f( x ) is strictly increasing in the interval x( 2,1 ) .

Thus, the function is strictly increasing in the interval ( 2,1 ) and strictly decreasing in the interval ( ,2 ) and ( 1, ).

(d)

The given function is 69x x 2 .

Let f( x ) be the given function.

f( x )=69x x 2

Differentiate f( x ) with respect to x.

f ( x )= d dx f( x ) =92x

Substitute f( x )=0

92x=0 2x=9 x= 9 2

Now, the point x= 9 2 divides the real line into two disjoint intervals given by,

<x< 9 2 9 2 <x<

In the interval <x< 9 2 ,

f ( x )>0 92x>0

The function f( x ) is strictly increasing in the interval x( , 9 2 ) .

In the interval 9 2 <x<,

f ( x )<0 92x<0

The function f( x ) is strictly decreasing in the interval x( 9 2 , ) .

Thus, the function is strictly increasing in the interval x< 9 2 and strictly decreasing in the interval x> 9 2 .

(e)

The given function is ( x+1 ) 3 ( x3 ) 3 .

Let f( x ) be the given function.

f( x )= ( x+1 ) 3 ( x3 ) 3

Differentiate f( x ) with respect to x.

f ( x )= d dx f( x ) =3 ( x+1 ) 2 ( x3 ) 3 +3 ( x+1 ) 3 ( x3 ) 2 =3 ( x+1 ) 2 ( x3 ) 2 [ x3+x+1 ] =3 ( x+1 ) 2 ( x3 ) 2 ( x1 )

Substitute f( x )=0

3 ( x+1 ) 2 ( x3 ) 2 ( x1 )=0 x=1,1,3

Now, the points x=1, x=1 and x=3 divide the real line into four disjoint intervals given by,

<x<1 1<x<1 1<x<3 3<x<

In the interval <x<1 and 1<x<1 ,

f ( x )<0 3 ( x+1 ) 2 ( x3 ) 2 ( x1 )<0

The function f( x ) is strictly decreasing in the interval x( ,1 ) and x( 1,1 ).

In the interval 1<x<3 and 3<x<.

f ( x )>0 3 ( x+1 ) 2 ( x3 ) 2 ( x1 )>0

The function f( x ) is strictly increasing in the interval x( 1,3 ) and x( 3, ) .

Thus, the function is strictly increasing in the interval ( 1,3 ) and ( 3, )and strictly decreasing in the interval ( ,1 ) and ( 1,1 ).


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