Misc 13 - Find points f(x) = (x-2)4 (x+1)3 has local maxima - Miscellaneous

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Misc 13 Find the points at which the function f given by f (π‘₯) = (π‘₯βˆ’2)^4 (π‘₯+1)^3 has (i) local maxima (ii) local minima (iii) point of inflexionf(π‘₯)= (π‘₯βˆ’2)^4 (π‘₯+1)3 Step 1: Finding f’(π‘₯) f’(π‘₯) = (𝑑 ((π‘₯βˆ’2)^4 (π‘₯+1)^3 ))/𝑑π‘₯ using product rule as (𝑒𝑣)^β€²=𝑒^β€² 𝑣+𝑣^β€² 𝑒 = γ€–((π‘₯βˆ’2)^4 )^β€² (π‘₯+1)γ€—^3+((π‘₯+1)^3 )^β€² (π‘₯βˆ’2)^4 = 4(π‘₯βˆ’2)^3 (π‘₯+1)^3+3(π‘₯+1)^2 (π‘₯βˆ’2)^4 = (π‘₯βˆ’2)^3 (π‘₯+1)^2 [4(π‘₯+1)+3(π‘₯βˆ’2)] = (π‘₯βˆ’2)^3 (π‘₯+1)^2 [4π‘₯+4π‘₯+3π‘₯βˆ’6] = (π‘₯βˆ’2)^3 (π‘₯+1)^2 [7π‘₯βˆ’2] Step 2: Putting f’(π‘₯)=0 (π‘₯βˆ’2)^3 (π‘₯+1)^2 (7π‘₯βˆ’2)=0 (π‘₯βˆ’2)^3 = 0 π‘₯ – 2 = 0 π‘₯=2 (π‘₯+1)^2=0 (π‘₯+1)=0 π‘₯ = –1 7π‘₯ – 2 = 0 7π‘₯ = 2 π‘₯ = 2/7 Hence, π‘₯=2 & π‘₯=βˆ’1 & π‘₯=2/7 = 0.28 Thus π‘₯=βˆ’1 is a point of inflexion π‘₯=2/7 is point of maxima & π‘₯=2 is point of minima

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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.