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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) You are here
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
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 March 16, 2023 by Teachoo
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: (ii) π(π₯)=π₯3 β3π₯π(π₯)=π₯3 β3π₯ Finding gβ(π) gβ(π₯)=π(π₯^3 β 3π₯)/ππ₯ gβ(π₯)=3π₯^2β3 Putting gβ(π)=π 3π₯^2β3=0 3π₯^2=3 π₯^2=3/3 π₯^2=1 π₯=Β±1 So, x = 1 & x = β1 Finding gββ(π) gβ(π₯)=3π₯^2β3 gββ(π₯)=π(3π₯^2β3)/ππ₯ = 6π₯β0 = 6π₯ Putting π=π in gββ(x) gββ(1)=6(1)= 6 > 0 Thus, gββ(π₯)>0 when π₯=1 β π₯=1 is point of local minima & g(π₯) is minimum at π₯=1 Local minimum value g(π₯)=π₯^3β3π₯ g(1)=(1)^3β3(1) =1β3 =βπ Putting π=βπ in gββ(x) gββ(β1)=6(β1)= β6 < 0 Thus, gββ(π₯)<0 when π₯=β1 β π₯=β1 is point of local maxima & g(π₯) is maximum at π₯=β1 Local minimum value g(π₯)=π₯^3β3π₯ g(β1)=(β1)^3β3(β1) =β1+3 =π