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  4. The function f(x) =(9-x2)2 increases in

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Q. The function $ f(x) ={{(9-{{x}^{2}})}^{2}} $ increases in

KEAMKEAM 2008Application of Derivatives

Solution:

Given, $ f(x)={{(9-{{x}^{2}})}^{2}} $
On differentiating w.r.t. $ x, $ we get $ f(x)=2(9-{{x}^{2}})(-2x) $
Now, put $ f(x)=0 $
$ \Rightarrow $ $ 2(9-{{x}^{2}})(-2x)=0 $
$ \Rightarrow $ $ x=0,\pm 3 $

$ \therefore $ $ f(x) $ is increasing in $ (-3,0)\cup (3,\infty ) $ .

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