Ex 6.3, 3 (ii) - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
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) You are here
Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important
Ex 6.3, 3 (v)
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 April 16, 2024 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: (ii) š(š„)=š„3 ā3š„š(š„)=š„3 ā3š„ Finding gā(š) gā(š„)=š(š„^3 ā 3š„)/šš„ gā(š„)=3š„^2ā3 Putting gā(š)=š 3š„^2ā3=0 3š„^2=3 š„^2=3/3 š„^2=1 š„=Ā±1 So, x = 1 & x = ā1 Finding gāā(š) gā(š„)=3š„^2ā3 gāā(š„)=š(3š„^2ā3)/šš„ = 6š„ā0 = 6š„ Putting š=š in gāā(x) gāā(1)=6(1)= 6 > 0 Thus, gāā(š„)>0 when š„=1 ā š„=1 is point of local minima & g(š„) is minimum at š„=1 Local minimum value g(š„)=š„^3ā3š„ g(1)=(1)^3ā3(1) =1ā3 =āš Putting š=āš in gāā(x) gāā(ā1)=6(ā1)= ā6 < 0 Thus, gāā(š„)<0 when š„=ā1 ā š„=ā1 is point of local maxima & g(š„) is maximum at š„=ā1 Local minimum value g(š„)=š„^3ā3š„ g(ā1)=(ā1)^3ā3(ā1) =ā1+3 =š