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Q.Find the interval in which the function f given by f(x) = 2x^3-3x^2-36x+7 is strictly decreasing.

Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2019Subjective· 2mImportance★★★★★
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f′(x)=6(x−3)(x+2)f'(x)=6(x-3)(x+2), negative on (−2,3)(-2,3), so ff is strictly decreasing on (−2,3)(-2,3).

Concept. f′(x)<0f'(x)<0 on an interval ⇒f\Rightarrow f strictly decreasing there.

Steps.

f(x)=2x3−3x2−36x+7  ⟹  f′(x)=6x2−6x−36=6(x2−x−6)=6(x−3)(x+2).f(x)=2x^3-3x^2-36x+7\implies f'(x)=6x^2-6x-36=6(x^2-x-6)=6(x-3)(x+2). …

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