Question

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'

Answer 1

(a)

The given function is x 2 +2x−5=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 10−6x−2 x 2 .

Let f( x ) be the given function.

f( x )=10−6x−2 x 2

Differentiate f( x ) with respect to x.

f ′ ( x )= d dx f( x ) =−6−4x

Substitute f( x )=0

−6−4x=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 −6−4x<0

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

In the interval − 3 2 <x<∞,

f ′ ( x )<0 −6−4x<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 −18x−12 =−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 6−9x− x 2 .

Let f( x ) be the given function.

f( x )=6−9x− x 2

Differentiate f( x ) with respect to x.

f ′ ( x )= d dx f( x ) =−9−2x

Substitute f( x )=0

−9−2x=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 −9−2x>0

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

In the interval − 9 2 <x<∞,

f ′ ( x )<0 −9−2x<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 ( x−3 ) 3 .

Let f( x ) be the given function.

f( x )= ( x+1 ) 3 ( x−3 ) 3

Differentiate f( x ) with respect to x.

f ′ ( x )= d dx f( x ) =3 ( x+1 ) 2 ( x−3 ) 3 +3 ( x+1 ) 3 ( x−3 ) 2 =3 ( x+1 ) 2 ( x−3 ) 2 [ x−3+x+1 ] =3 ( x+1 ) 2 ( x−3 ) 2 ( x−1 )

Substitute f( x )=0

3 ( x+1 ) 2 ( x−3 ) 2 ( x−1 )=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 ( x−3 ) 2 ( x−1 )<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 ( x−3 ) 2 ( x−1 )>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 ).