Misc 4 - Find intervals f(x) = x3 + 1/x3 x = 0 is increasing - Miscellaneous

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

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Misc 4 Find the intervals in which the function f given by f (x) = x3 + 1/š‘„^3 , š‘„ ≠ 0 is (i) increasing (ii) decreasing. f(š‘„) = š‘„3 + 1/š‘„3 Finding f’(š’™) f’(š‘„) = š‘‘/š‘‘š‘„ (š‘„^3+š‘„^(āˆ’3) )^. = 3š‘„2 + (āˆ’3)^(āˆ’3 āˆ’ 1) = 3š‘„2 – 3š‘„^(āˆ’4) = 3š‘„^2āˆ’3/š‘„^4 = 3(š‘„^2āˆ’1/š‘„^4 ) Putting f’(š’™) = 0 3(š‘„^2āˆ’1/š‘„^4 ) = 0 (š‘„^6 āˆ’ 1)/š‘„^4 = 0 š’™^šŸ”āˆ’šŸ = 0 (š‘„^3 )^2āˆ’(1)^2=0 (š’™^šŸ‘āˆ’šŸ)(š’™^šŸ‘+šŸ)=šŸŽ Hence, š’™ = 1 & –1 Plotting points on number line So, f(š‘„) is strictly increasing on (āˆ’āˆž , āˆ’1) & (1 , āˆž) & f(š‘„) strictly decreasing on (āˆ’1 , 1) But we need to find Increasing & Decreasing f’(š‘„) = 3(š‘„^2āˆ’1/š‘„^4 ) Thus, f(š‘„) is increasing on (āˆ’āˆž , āˆ’šŸ] & [šŸ , āˆž) & f(š‘„) is decreasing on [āˆ’šŸ , šŸ]

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