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Question

# 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

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Solution

## 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)

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