Ex 6.3, 3 (vi) - Chapter 6 Class 12 Application of Derivatives

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Ex 6.3, 3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (vi) g (š‘„) = š‘„/2 + 2/š‘„ , š‘„ > 0g (š‘„) = š‘„/2 + 2/š‘„, š‘„ > 0 Finding gā€™(š’™) gā€™(š‘„)=š‘‘/š‘‘š‘„ (š‘„/2+2/š‘„) =š‘‘/š‘‘š‘„ (š‘„/2)+š‘‘/š‘‘š‘„ (2š‘„^(āˆ’1) ) =1/2āˆ’2š‘„^(āˆ’2) =1/2āˆ’2/š‘„^2 Putting gā€™(š’™)=šŸŽ 1/2āˆ’2/š‘„^2 =0 (š‘„^2āˆ’ 4)/(2š‘„^2 )=0 š‘„^2āˆ’4=0 Ɨ2š‘„^2 (š‘„āˆ’2)(š‘„+2)=0 So, š‘„=2 & š‘„=āˆ’2 Since š‘„>0 is given, we consider only š‘„=2 Finding gā€™ā€™(š’™) gā€™(š‘„)=1/2āˆ’2/š‘„^2 gā€™ā€™(š‘„)=š‘‘/š‘‘š‘„ (1/2āˆ’2/š‘„^2 ) = 0 ā€“ 2 . (āˆ’2) š‘„^(āˆ’2āˆ’1) = 4š‘„^(āˆ’3) = 4/š‘„^3 Putting value š‘„=2 in gā€™ā€™(š‘„) gā€™ā€™(š‘„)=4/(2)^3 = 4/8= 1/2 Since gā€™ā€™ (x) > 0, for x = 2 g(x) is minimum at x = 2 Minimum value of g(š‘„) is g(š‘„)=š‘„/2+2/š‘„ Putting š‘„=2 g(2)= 2/2+2/2 = 1 + 1 = 2

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.