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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)
Ex 6.3, 3 (vi) You are here
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 June 12, 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: (vi) g (š„) = š„/2 + 2/š„ , š„ > 0g (š„) = š„/2 + 2/š„, š„ > 0 Finding gā(š) gā(š„)=š/šš„ (š„/2+2/š„) =š/šš„ (š„/2)+š/šš„ (2š„^(ā1) ) =1/2ā2š„^(ā2) =1/2ā2/š„^2 Putting gā(š)=š 1/2ā2/š„^2 =0 (š„^2ā 4)/(2š„^2 )=0 š„^2ā4=0 Ć2š„^2 (š„ā2)(š„+2)=0 So, š„=2 & š„=ā2 Since š„>0 is given, we consider only š„=2 Finding gāā(š) gā(š„)=1/2ā2/š„^2 gāā(š„)=š/šš„ (1/2ā2/š„^2 ) = 0 ā 2 . (ā2) š„^(ā2ā1) = 4š„^(ā3) = 4/š„^3 Putting value š„=2 in gāā(š„) gāā(š„)=4/(2)^3 = 4/8= 1/2 Since gāā (x) > 0, for x = 2 g(x) is minimum at x = 2 Minimum value of g(š„) is g(š„)=š„/2+2/š„ Putting š„=2 g(2)= 2/2+2/2 = 1 + 1 = 2