Miscellaneous

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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Misc 7 Find the intervals in which the function f given by f (x) = x3 + 1/đĽ^3 , đĽ â  0 is (i) increasing (ii) decreasing. f(đĽ) = đĽ3 + 1/đĽ3 Finding fâ(đ) fâ(đĽ) = đ/đđĽ (đĽ^3+đĽ^(â3) )^. = 3đĽ2 + (â3)^(â3 â 1) = 3đĽ2 â 3đĽ^(â4) = 3đĽ^2â3/đĽ^4 = 3(đĽ^2â1/đĽ^4 ) Putting fâ(đ) = 0 3(đĽ^2â1/đĽ^4 ) = 0 (đĽ^6 â 1)/đĽ^4 = 0 đ^đâđ = 0 (đĽ^3 )^2â(1)^2=0 (đ^đâđ)(đ^đ+đ)=đ Hence, đ = 1 & â1 đĽ^3+1 = 0 đĽ3 = â1 đ = â1 Plotting points on number line So, f(đĽ) is strictly increasing on (ââ , â1) & (1 , â) & f(đĽ) strictly decreasing on (â1 , 1) But we need to find Increasing & Decreasing fâ(đĽ) = 3(đĽ^2â1/đĽ^4 ) Thus, f(đĽ) is increasing on (ââ , âđ] & [đ , â) & f(đĽ) is decreasing on [âđ , đ] For x = â1 fâ(â1) = 0 For x = 1 fâ(1) = 0