Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

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„^(ā1) ) =1/2ā2š„^(ā2) =1/2ā2/š„^2 Putting gā(š)=š 1/2ā2/š„^2 =0 (š„^2ā 4)/(2š„^2 )=0 š„^2ā4=0 Ć2š„^2 (š„ā2)(š„+2)=0 So, š„=2 & š„=ā2 Since š„>0 is given, we consider only š„=2 Finding gāā(š) gā(š„)=1/2ā2/š„^2 gāā(š„)=š/šš„ (1/2ā2/š„^2 ) = 0 ā 2 . (ā2) š„^(ā2ā1) = 4š„^(ā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