Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

Transcript

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/šš„ (1/(š„^2 + 2)) gā(š„)=(š(š„^2 + 2)^(ā1))/šš„ gā(š„)=ā1(š„^2+2)^(ā1ā1) Ć (2š„+0) gā(š„)=ā2š„(š„^2+2)^(ā2) gā²(š„)=( ā2š„ )/(š„^2 + 2)^2 Putting gā(š)=š ( ā2š„ )/(š„^2+2)^2 =0 ā2š„=0 Ć(š„^2+2)^2 ā2š„=0 š„=0 Finding gāā(š) gā(š„)=(ā2š„)/(š„^2 + 2)^2 gāā(š„)=(š(ā2š„)/šš„ . ć (š„^2 + 2)ć^2 ā (š(š„^2 + 2)^2)/šš„ . (ā2š„))/((š„^2 + 2)^2 )^2 =(ā2 (š„^2 + 2)^2ā2 (š„^2 + 2)^(2ā1).š(š„^2 + 2)/šš„ . (ā2š„))/((š„^2 + 2)^2 )^2 =(ā2 (š„^2 + 2)^2ā2 (š„^2 + 2)(2š„ + 0) (ā2š„))/(š„^2 + 2)^4 =(ā2 (š„^2 + 2)^2ā2 (š„^2 + 2)(2š„) (ā2š„))/(š„^(2 )+ 2)^4 =(ā2 (š„^2 + 2)^2+ 8š„^2 (š„^2 + 2))/(š„^(2 )+ 2)^4 =(ā2 (š„^2 + 2)[(š„^(2 )+ 2) ā 4š„^2 ])/(š„^2 + 2)^4 =(ā2 (š„^2 + 2)(ā3š„^2 + 2))/(š„^2 + 2)^4 =(ā2(ā3š„^2 + 2))/(š„^(2 )+ 2)^3 Putting x = 0 in gāā(x) gāā(0)=(ā2(ā3(0) + 2))/(0^2 + 2)^3 =(ā2(0 + 2))/(2)^3 =(ā4)/8=(ā1)/2 Hence gāā(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