9-{{x}^{2}}}\] \[f'(x)=\frac{1}{2\sqrt{9-{{x}^{2}}}}\times (-2x)=-\frac{x}{\sqrt{9-{{x}^{2}}}}\] For function to be increasing \[-\frac{x}{\sqrt{9-{{x}^{2}}}}>0\] or \[-x>0\] or \[x<0\] but \[\sqrt{9-{{x}^{2}}}\] is DEFINED only when \[9-{{x}^{2}}>0\] or \[{{x}^{2}}-9<0\] \[(x+3)(x-3)<0\] i.e., \[-3

"> 9-{{x}^{2}}}\] \[f'(x)=\frac{1}{2\sqrt{9-{{x}^{2}}}}\times (-2x)=-\frac{x}{\sqrt{9-{{x}^{2}}}}\] For function to be increasing \[-\frac{x}{\sqrt{9-{{x}^{2}}}}>0\] or \[-x>0\] or \[x<0\] but \[\sqrt{9-{{x}^{2}}}\] is DEFINED only when \[9-{{x}^{2}}>0\] or \[{{x}^{2}}-9<0\] \[(x+3)(x-3)<0\] i.e., \[-3

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What is the interval in which the function \[f(x)=\sqrt{9-{{x}^{2}}}\] is increasing? \[(f(x)>0)\]

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) . 10 months ago

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[b] \[f(x)=\sqrt{9-{{x}^{2}}}\] \[f'(x)=\frac{1}{2\sqrt{9-{{x}^{2}}}}\times (-2x)=-\frac{x}{\sqrt{9-{{x}^{2}}}}\] For function to be increasing \[-\frac{x}{\sqrt{9-{{x}^{2}}}}>0\] or \[-x>0\] or \[x<0\] but \[\sqrt{9-{{x}^{2}}}\] is DEFINED only when \[9-{{x}^{2}}>0\] or \[{{x}^{2}}-9<0\] \[(x+3)(x-3)<0\] i.e., \[-3

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Posted on 19 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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