Ex 6.3,10 - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Absolute minima/maxima
Last updated at April 16, 2024 by Teachoo
Ex 6.3, 10 Find the maximum value of 2đĽ3 â 24đĽ + 107 in the interval [1, 3]. Find the maximum value of the same function in [â3, â1]. Let f(đĽ)=2đĽ^3â24đĽ+107 Finding fâ(đĽ) fâ(đĽ)=đ(2đĽ^3 â 24đĽ + 107)/đđĽ = 2 Ă 3đĽ^2â24 = 6đĽ^2â24 = 6 (đĽ^2â4" " ) Putting f(đĽ)=0 6 (đĽ^2â4" " )=0 đĽ^2â4" = 0 " đĽ^2=4 đĽ=Âąâ4 đĽ=Âą2 Thus, đĽ=2 , â 2 Since x is in interval [1 , 3] đĽ = 2 is only Critical point Also, since given the interval đĽ= â [1 , 3] We calculate f(x) at đĽ= 1 , 2 & 3 Hence maximum value of f(đ)=đđ at đ = 3 in the interval [1 , 3] For the interval [âđ , âđ] đĽ = â2 is only Critical Point Also, since given the interval đĽ= â [â3,â1] We calculate f(x) at đĽ= â1 , â2 & â3 Hence maximum value of f(đ)=đđđ at đ = â2 in the interval [â3,â1]