# Misc 13 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 14, 2019 by Teachoo

Last updated at Dec. 14, 2019 by Teachoo

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Misc 13 Find the points at which the function f given by f (π₯) = (π₯β2)^4 (π₯+1)^3 has (i) local maxima (ii) local minima (iii) point of inflexionf(π₯)= (π₯β2)^4 (π₯+1)3 Step 1: Finding fβ(π₯) fβ(π₯) = (π ((π₯β2)^4 (π₯+1)^3 ))/ππ₯ using product rule as (π’π£)^β²=π’^β² π£+π£^β² π’ = γ((π₯β2)^4 )^β² (π₯+1)γ^3+((π₯+1)^3 )^β² (π₯β2)^4 = 4(π₯β2)^3 (π₯+1)^3+3(π₯+1)^2 (π₯β2)^4 = (π₯β2)^3 (π₯+1)^2 [4(π₯+1)+3(π₯β2)] = (π₯β2)^3 (π₯+1)^2 [4π₯+4π₯+3π₯β6] = (π₯β2)^3 (π₯+1)^2 [7π₯β2] Step 2: Putting fβ(π₯)=0 (π₯β2)^3 (π₯+1)^2 (7π₯β2)=0 (π₯β2)^3 = 0 π₯ β 2 = 0 π₯=2 (π₯+1)^2=0 (π₯+1)=0 π₯ = β1 7π₯ β 2 = 0 7π₯ = 2 π₯ = 2/7 Hence, π₯=2 & π₯=β1 & π₯=2/7 = 0.28 Thus π₯=β1 is a point of inflexion π₯=2/7 is point of maxima & π₯=2 is point of minima

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.