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Question

Find the interval in which the following function is strictly incerasing or decreasing

2x39x212x+1


Solution

Given, f(x)=2x39x212x+1
f(x)=2.3x292x12=6x218x12
On putting f'(x)=0, we get 6x218x12=0
6(x+2)(x+1)=0x=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.

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