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)
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important You are here
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: (vii) g (š„) = 1/(š„^2 + 2)Finding gā(š) gā(š„)=š/šš„ (1/(š„^2 + 2)) gā(š„)=(š(š„^2 + 2)^(ā1))/šš„ gā(š„)=ā1(š„^2+2)^(ā1ā1) Ć (2š„+0) gā(š„)=ā2š„(š„^2+2)^(ā2) gā²(š„)=( ā2š„ )/(š„^2 + 2)^2 Putting gā(š)=š ( ā2š„ )/(š„^2+2)^2 =0 ā2š„=0 Ć(š„^2+2)^2 ā2š„=0 š„=0 Finding gāā(š) gā(š„)=(ā2š„)/(š„^2 + 2)^2 gāā(š„)=(š(ā2š„)/šš„ . ć (š„^2 + 2)ć^2 ā (š(š„^2 + 2)^2)/šš„ . (ā2š„))/((š„^2 + 2)^2 )^2 =(ā2 (š„^2 + 2)^2ā2 (š„^2 + 2)^(2ā1).š(š„^2 + 2)/šš„ . (ā2š„))/((š„^2 + 2)^2 )^2 =(ā2 (š„^2 + 2)^2ā2 (š„^2 + 2)(2š„ + 0) (ā2š„))/(š„^2 + 2)^4 =(ā2 (š„^2 + 2)^2ā2 (š„^2 + 2)(2š„) (ā2š„))/(š„^(2 )+ 2)^4 =(ā2 (š„^2 + 2)^2+ 8š„^2 (š„^2 + 2))/(š„^(2 )+ 2)^4 =(ā2 (š„^2 + 2)[(š„^(2 )+ 2) ā 4š„^2 ])/(š„^2 + 2)^4 =(ā2 (š„^2 + 2)(ā3š„^2 + 2))/(š„^2 + 2)^4 =(ā2(ā3š„^2 + 2))/(š„^(2 )+ 2)^3 Putting x = 0 in gāā(x) gāā(0)=(ā2(ā3(0) + 2))/(0^2 + 2)^3 =(ā2(0 + 2))/(2)^3 =(ā4)/8=(ā1)/2 Hence gāā(š„)<0 when š„ = 0 ā“ š„ = 0 is point of local maxima Thus, g(š„) is maximum at š = 0 Maximum value of g(š) at x = 0 g(š„)=1/(š„^(2 )+ 2) g(0)=1/(0^2 + 2) = 1/2 Maximum value is š/š