

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 6.3
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
Ex 6.3, 2 (iii)
Ex 6.3, 2 (iv) Important
Ex 6.3, 2 (v) Important
Ex 6.3, 3 (i)
Ex 6.3, 3 (ii)
Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important
Ex 6.3, 3 (v) You are here
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important
Ex 6.3, 3 (viii)
Ex 6.3, 4 (i)
Ex 6.3, 4 (ii) Important
Ex 6.3, 4 (iii)
Ex 6.3, 5 (i)
Ex 6.3, 5 (ii)
Ex 6.3, 5 (iii) Important
Ex 6.3, 5 (iv)
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10
Ex 6.3,11 Important
Ex 6.3,12 Important
Ex 6.3,13
Ex 6.3,14 Important
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18 Important
Ex 6.3,19 Important
Ex 6.3, 20 Important
Ex 6.3,21
Ex 6.3,22 Important
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25 Important
Ex 6.3, 26 Important
Ex 6.3, 27 (MCQ)
Ex 6.3,28 (MCQ) Important
Ex 6.3,29 (MCQ)
Last updated at May 29, 2023 by Teachoo
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: (v) π (π₯)=π₯3 β6π₯2+9π₯+15π (π₯)=π₯3 β6π₯2+9π₯+15 Finding fβ(π) fβ(π₯)=π(π₯3 β 6π₯2 + 9π₯ + 15" " )/ππ₯ fβ(π₯)=3π₯^2β12π₯+9 fβ(π₯)=3(π₯^2β4π₯+3) Putting fβ(π)=π 3(π₯^2β4π₯+3)=0 π₯^2β4π₯+3=0 π₯^2β3π₯βπ₯+3=0 π₯(π₯β3)β1(π₯β3)=0 (π₯β1)(π₯β3)=0 So, x = 1 & x = 3 Finding fββ(π) fβ(π₯)=3(π₯^2β4π₯+3) fββ(π₯)=π(3(π₯^2 β 4π₯+3))/ππ₯ = 3(2π₯β4+0) = 6π₯β12 Putting π=π in fββ(π) fββ(1)=6(1)β12 = 6 β 12 = β 6 < 0 Since fββ(π₯)<0 when π₯=1 β π₯=1 is point of local maxima β΄ f(π₯) is maximum at π=π Maximum value of f(π₯) at π₯ = 1 f(π₯)=π₯^3β6π₯^2+9π₯+15 f(1)=(1)^3β6(1)^2+9(1)+15 = 1 β 6 + 9 + 15 = 19 Putting π=π in fββ(x) fββ(π₯)=6π₯β12 fββ(3)=6(3)β12 = 18 β 12 = 6 > 0 Since fββ(π₯)>0 when π₯=3 β π₯=3 is point of local minima β΄ f(π₯) is minimum at π=π Minimum value of f(π₯) at π₯ = 3 f(π₯)=π₯^3β6π₯^2+9π₯+15 f(3)=(3)^3β6(3)^2+9(3)+15 = 27 β 54 + 27 + 15 = 15