


Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Ex 6.5
Ex 6.5, 1 (ii)
Ex 6.5, 1 (iii) Important
Ex 6.5, 1 (iv)
Ex 6.5, 2 (i)
Ex 6.5, 2 (ii) Important
Ex 6.5, 2 (iii)
Ex 6.5, 2 (iv) Important
Ex 6.5, 2 (v) Important
Ex 6.5, 3 (i)
Ex 6.5, 3 (ii)
Ex 6.5, 3 (iii)
Ex 6.5, 3 (iv) Important
Ex 6.5, 3 (v)
Ex 6.5, 3 (vi)
Ex 6.5, 3 (vii) Important You are here
Ex 6.5, 3 (viii)
Ex 6.5, 4 (i)
Ex 6.5, 4 (ii) Important
Ex 6.5, 4 (iii)
Ex 6.5, 5 (i)
Ex 6.5, 5 (ii)
Ex 6.5, 5 (iii) Important
Ex 6.5, 5 (iv)
Ex 6.5,6
Ex 6.5,7 Important
Ex 6.5,8
Ex 6.5,9 Important
Ex 6.5,10
Ex 6.5,11 Important
Ex 6.5,12 Important
Ex 6.5,13
Ex 6.5,14 Important
Ex 6.5,15 Important
Ex 6.5,16
Ex 6.5,17
Ex 6.5,18 Important
Ex 6.5,19 Important
Ex 6.5, 20 Important
Ex 6.5,21
Ex 6.5,22 Important
Ex 6.5,23 Important
Ex 6.5,24 Important
Ex 6.5,25 Important
Ex 6.5, 26 Important
Ex 6.5, 27 (MCQ)
Ex 6.5,28 (MCQ) Important
Ex 6.5,29 (MCQ)
Last updated at Aug. 19, 2021 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
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: (vii) g (π₯) = 1/(π₯^2 + 2)Finding gβ(π) gβ(π₯)=π/ππ₯ (1/(π₯^2 + 2)) gβ(π₯)=(π(π₯^2 + 2)^(β1))/ππ₯ gβ(π₯)=β1(π₯^2+2)^(β1β1) Γ (2π₯+0) gβ(π₯)=β2π₯(π₯^2+2)^(β2) gβ²(π₯)=( β2π₯ )/(π₯^2 + 2)^2 Putting gβ(π)=π ( β2π₯ )/(π₯^2+2)^2 =0 β2π₯=0 Γ(π₯^2+2)^2 β2π₯=0 π₯=0 Finding gββ(π) gβ(π₯)=(β2π₯)/(π₯^2 + 2)^2 gββ(π₯)=(π(β2π₯)/ππ₯ . γ (π₯^2 + 2)γ^2 β (π(π₯^2 + 2)^2)/ππ₯ . (β2π₯))/((π₯^2 + 2)^2 )^2 Using quotient rule as (π’/π£)^β²=(π’^β² π£ β π£^β² π’)/π£^2 =(β2 (π₯^2 + 2)^2β2 (π₯^2 + 2)^(2β1).π(π₯^2 + 2)/ππ₯ . (β2π₯))/((π₯^2 + 2)^2 )^2 =(β2 (π₯^2 + 2)^2β2 (π₯^2 + 2)(2π₯ + 0) (β2π₯))/(π₯^2 + 2)^4 =(β2 (π₯^2 + 2)^2β2 (π₯^2 + 2)(2π₯) (β2π₯))/(π₯^(2 )+ 2)^4 =(β2 (π₯^2 + 2)^2+ 8π₯^2 (π₯^2 + 2))/(π₯^(2 )+ 2)^4 =(β2 (π₯^2 + 2)[(π₯^(2 )+ 2) β 4π₯^2 ])/(π₯^2 + 2)^4 =(β2 (π₯^2 + 2)(β3π₯^2 + 2))/(π₯^2 + 2)^4 =(β2(β3π₯^2 + 2))/(π₯^(2 )+ 2)^3 Putting x = 0 in gββ(x) gββ(0)=(β2(β3(0) + 2))/(0^2 + 2)^3 =(β2(0 + 2))/(2)^3 =(β4)/8=(β1)/2 Hence gββ(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