Misc 10 - Find points f(x) = (x-2)4 (x+1)3 has local maxima - Miscellaneous

part 2 - Misc 10 - Miscellaneous - Serial order wise - Chapter 6 Class 12 Application of Derivatives
part 3 - Misc 10 - Miscellaneous - Serial order wise - Chapter 6 Class 12 Application of Derivatives part 4 - Misc 10 - Miscellaneous - Serial order wise - Chapter 6 Class 12 Application of Derivatives

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Misc 10 Find the points at which the function f given by f (š‘„) = (š‘„āˆ’2)^4 (š‘„+1)^3 has (i) local maxima (ii) local minima (iii) point of inflexionf(š‘„)= (š‘„āˆ’2)^4 (š‘„+1)3 Finding f’(š’™) f’(š‘„) = (š‘‘ ((š‘„ āˆ’ 2)^4 (š‘„ + 1)^3 ))/š‘‘š‘„ = 怖((š‘„āˆ’2)^4 )^′ (š‘„+1)怗^3+((š‘„+1)^3 )^′ (š‘„āˆ’2)^4 = 4(š‘„āˆ’2)^3 (š‘„+1)^3+3(š‘„+1)^2 (š‘„āˆ’2)^4 = (š‘„āˆ’2)^3 (š‘„+1)^2 [4(š‘„+1)+3(š‘„āˆ’2)] = (š‘„āˆ’2)^3 (š‘„+1)^2 [4š‘„+4+3š‘„āˆ’6] = (š’™āˆ’šŸ)^šŸ‘ (š’™+šŸ)^šŸ [šŸ•š’™āˆ’šŸ] Putting f’(š’™)=šŸŽ (š‘„āˆ’2)^3 (š‘„+1)^2 (7š‘„āˆ’2)=0 Hence, š‘„=2 & š‘„=āˆ’1 & š‘„=2/7 = 0.28 (š‘„āˆ’2)^3 = 0 š‘„ – 2 = 0 š’™=šŸ (š‘„+1)^2=0 (š‘„+1)=0 š’™ = –1 7š‘„ – 2 = 0 7š‘„ = 2 š’™ = šŸ/šŸ• Thus, š‘„=āˆ’šŸ is a point of Inflexion š‘„=šŸ/šŸ• is point of maxima š‘„=šŸ is point of minima

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