Find the interval in which the following functions are strictly incerasing or decreasing
x2+2x−5
10−6x−2x2
−2x3−9x2−12x+1
6−9x−x2
(x+1)3(x−3)3
Find the interval in which the following functions are strictly incerasing or decreasing
x2+2x−5
10−6x−2x2
−2x3−9x2−12x+1
6−9x−x2
(x+1)3(x−3)3
Let f(x)=x2+2x−5⇒f′(x)=2x+2 (By differentiating w.r.t. x)
Putting f'(x)=0, we get 2x+2=0⇒2x=−2⇒x=−1
x=-2 divides real line into two intervals
namely (−∞,−1) and (−2,∞)
IntervalsSign of f′(x)Nature of f(x)(−∞,−1)−veStrictly deceresing(−1,∞)+veStrictly increasing
Therefore, f(x) is strictly increasing when x>-2 and strictly decreasing when x<-1
Let f(x)=10−6x−2x2⇒f′(x)=0−6−22x=−6−4x
On putting f'(x)=0, we get −6−4x=0⇒x=−32
Which divides real line into two intervals
namely (−∞,−32) and (−32,∞)
IntervalsSign of f′(x)Nature of f(x)(−∞,−32)+veStrictly increasing(−32,∞)−veStrictly deceresing
Hence, f is strictly increasing for x<−32 and strictly decreasing for x>−32
Given, f(x)=−2x3−9x2−12x+1
⇒f′(x)=−2.3x2−92x−12=−6x2−18x−12
On putting f'(x)=0, we get −6x2−18x−12=0
⇒−6(x+2)(x+1)=0⇒x=−2,−1
which divides real line into three intervals (−∞,−2),(−2,−1) and (−1,∞)
IntervalsSign of f′(x)Nature of f(x)(−∞−2)(−)(−)(−)=−veStrictly decreasing(−2,−1)(−)(−)(+)=+veStrictly increasing(−1,∞)(−)(+)(+)=−veStrictly decreasing
Therefore, f(x) is strictly increasing in −2<x<−1 and strictly decreasing for x<−2 and x>−1.
Given,f(x)=6−9x−x2⇒f′(x)=−9−2x
On putting, f(x)=6−9x−x2⇒f′(x)=−9−2x
On putting f'(x)=0, we get −9−2x=0⇒x=−92
Which divides the real line in two disjoint
intervals (−∞,−92) and (−92,∞)
IntervalsSign of f′(x)Nature of f(x)(−∞,−92)+veStrictly increasing(−92,∞)−veStrictly decreasing
Therefore, f(x) is strictly increasing when x<−92 and strictly decreasing when x>−92.
Given, f(x)=(x+1)3(x−3)3
On differentiating, we get
f′(x)=(x+1)3.3(x−3)21+(x−3)3.3(x+1)2.1=3(x−3)2(x+1)2{(x+1)+(x−3)}=3(x−3)2(x+1)2(2x−2)=6(x−3)2(x+1)2(x−1)
On putting f'(x)=0, we get x=-1, 1, 3
Which divides real line into four disjoint
intervals namely (−∞,−1)(−1,1)(1,3) and (3,∞).
IntervalsSign of f′(x)Nature of f(x)−∞<x<−1(+)(+)(−)=−veStrictly decresing−1<x<1(+)(+)(−)=−veStrictly decreasing1<x<3(+)(+)(+)=+veStrictly increasing3<x<∞(+)(+)(+)=+vestrictly increasing
Therefore, f(x) is strictly increasing in (1,3) and (3,∞) and strictly decreasing in (−∞,−1) and (−1,1)
Let f(x)=x2+2x−5⇒f′(x)=2x+2 (By differentiating w.r.t. x)
Putting f'(x)=0, we get 2x+2=0⇒2x=−2⇒x=−1
x=-2 divides real line into two intervals
namely (−∞,−1) and (−2,∞)
IntervalsSign of f′(x)Nature of f(x)(−∞,−1)−veStrictly deceresing(−1,∞)+veStrictly increasing
Therefore, f(x) is strictly increasing when x>-2 and strictly decreasing when x<-1
Let f(x)=10−6x−2x2⇒f′(x)=0−6−22x=−6−4x
On putting f'(x)=0, we get −6−4x=0⇒x=−32
Which divides real line into two intervals
namely (−∞,−32) and (−32,∞)
IntervalsSign of f′(x)Nature of f(x)(−∞,−32)+veStrictly increasing(−32,∞)−veStrictly deceresing
Hence, f is strictly increasing for x<−32 and strictly decreasing for x>−32
Given, f(x)=−2x3−9x2−12x+1
⇒f′(x)=−2.3x2−92x−12=−6x2−18x−12
On putting f'(x)=0, we get −6x2−18x−12=0
⇒−6(x+2)(x+1)=0⇒x=−2,−1
which divides real line into three intervals (−∞,−2),(−2,−1) and (−1,∞)
IntervalsSign of f′(x)Nature of f(x)(−∞−2)(−)(−)(−)=−veStrictly decreasing(−2,−1)(−)(−)(+)=+veStrictly increasing(−1,∞)(−)(+)(+)=−veStrictly decreasing
Therefore, f(x) is strictly increasing in −2<x<−1 and strictly decreasing for x<−2 and x>−1.
Given,f(x)=6−9x−x2⇒f′(x)=−9−2x
On putting, f(x)=6−9x−x2⇒f′(x)=−9−2x
On putting f'(x)=0, we get −9−2x=0⇒x=−92
Which divides the real line in two disjoint
intervals (−∞,−92) and (−92,∞)
IntervalsSign of f′(x)Nature of f(x)(−∞,−92)+veStrictly increasing(−92,∞)−veStrictly decreasing
Therefore, f(x) is strictly increasing when x<−92 and strictly decreasing when x>−92.
Given, f(x)=(x+1)3(x−3)3
On differentiating, we get
f′(x)=(x+1)3.3(x−3)21+(x−3)3.3(x+1)2.1=3(x−3)2(x+1)2{(x+1)+(x−3)}=3(x−3)2(x+1)2(2x−2)=6(x−3)2(x+1)2(x−1)
On putting f'(x)=0, we get x=-1, 1, 3
Which divides real line into four disjoint
intervals namely (−∞,−1)(−1,1)(1,3) and (3,∞).
IntervalsSign of f′(x)Nature of f(x)−∞<x<−1(+)(+)(−)=−veStrictly decresing−1<x<1(+)(+)(−)=−veStrictly decreasing1<x<3(+)(+)(+)=+veStrictly increasing3<x<∞(+)(+)(+)=+vestrictly increasing
Therefore, f(x) is strictly increasing in (1,3) and (3,∞) and strictly decreasing in (−∞,−1) and (−1,1)
