Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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### 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: (vii) g (π₯) = 1/(π₯^2 + 2)Finding gβ(π) gβ(π₯)=π/ππ₯ (1/(π₯^2 + 2)) gβ(π₯)=(π(π₯^2 + 2)^(β1))/ππ₯ gβ(π₯)=β1(π₯^2+2)^(β1β1) Γ (2π₯+0) gβ(π₯)=β2π₯(π₯^2+2)^(β2) gβ²(π₯)=( β2π₯ )/(π₯^2 + 2)^2 Putting gβ(π)=π ( β2π₯ )/(π₯^2+2)^2 =0 β2π₯=0 Γ(π₯^2+2)^2 β2π₯=0 π₯=0 Finding gββ(π) gβ(π₯)=(β2π₯)/(π₯^2 + 2)^2 gββ(π₯)=(π(β2π₯)/ππ₯ . γ (π₯^2 + 2)γ^2 β (π(π₯^2 + 2)^2)/ππ₯ . (β2π₯))/((π₯^2 + 2)^2 )^2 Using quotient rule as (π’/π£)^β²=(π’^β² π£ β π£^β² π’)/π£^2 =(β2 (π₯^2 + 2)^2β2 (π₯^2 + 2)^(2β1).π(π₯^2 + 2)/ππ₯ . (β2π₯))/((π₯^2 + 2)^2 )^2 =(β2 (π₯^2 + 2)^2β2 (π₯^2 + 2)(2π₯ + 0) (β2π₯))/(π₯^2 + 2)^4 =(β2 (π₯^2 + 2)^2β2 (π₯^2 + 2)(2π₯) (β2π₯))/(π₯^(2 )+ 2)^4 =(β2 (π₯^2 + 2)^2+ 8π₯^2 (π₯^2 + 2))/(π₯^(2 )+ 2)^4 =(β2 (π₯^2 + 2)[(π₯^(2 )+ 2) β 4π₯^2 ])/(π₯^2 + 2)^4 =(β2 (π₯^2 + 2)(β3π₯^2 + 2))/(π₯^2 + 2)^4 =(β2(β3π₯^2 + 2))/(π₯^(2 )+ 2)^3 Putting x = 0 in gββ(x) gββ(0)=(β2(β3(0) + 2))/(0^2 + 2)^3 =(β2(0 + 2))/(2)^3 =(β4)/8=(β1)/2 Hence gββ(π₯)<0 when π₯ = 0 β΄ π₯ = 0 is point of local maxima Thus, g(π₯) is maximum at π = 0 Maximum value of g(π) at x = 0 g(π₯)=1/(π₯^(2 )+ 2) g(0)=1/(0^2 + 2) = 1/2 Maximum value is π/π