Ex 6.5

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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### Transcript

Ex 6.5, 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 (π₯) = π₯/2 + 2/π₯ , π₯ > 0g (π₯) = π₯/2 + 2/π₯, π₯ > 0 Finding gβ(π) gβ(π₯)=π/ππ₯ (π₯/2+2/π₯) =π/ππ₯ (π₯/2)+π/ππ₯ (2π₯^(β1) ) =1/2β2π₯^(β2) =1/2β2/π₯^2 Putting gβ(π)=π 1/2β2/π₯^2 =0 (π₯^2β 4)/(2π₯^2 )=0 π₯^2β4=0 Γ2π₯^2 (π₯β2)(π₯+2)=0 So, π₯=2 & π₯=β2 Since π₯>0 is given, we consider only π₯=2 Finding gββ(π) gβ(π₯)=1/2β2/π₯^2 gββ(π₯)=π/ππ₯ (1/2β2/π₯^2 ) = 0 β 2 . (β2) π₯^(β2β1) = 4π₯^(β3) = 4/π₯^3 Putting value π₯=2 in gββ(π₯) gββ(π₯)=4/(2)^3 = 4/8= 1/2 Since gββ (x) > 0, for x = 2 g(x) is minimum at x = 2 Minimum value of g(π₯) is g(π₯)=π₯/2+2/π₯ Putting π₯=2 g(2)= 2/2+2/2 = 1 + 1 = 2