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Misc 7 - Find intervals f(x) = x3 + 1/x3 x = 0 is increasing

Misc 7 - Chapter 6 Class 12 Application of Derivatives - Part 2
Misc 7 - Chapter 6 Class 12 Application of Derivatives - Part 3
Misc 7 - Chapter 6 Class 12 Application of Derivatives - Part 4

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Misc 7 Find the intervals in which the function f given by f (x) = x3 + 1/π‘₯^3 , π‘₯ β‰  0 is (i) increasing (ii) decreasing. f(π‘₯) = π‘₯3 + 1/π‘₯3 Finding f’(𝒙) f’(π‘₯) = 𝑑/𝑑π‘₯ (π‘₯^3+π‘₯^(βˆ’3) )^. = 3π‘₯2 + (βˆ’3)^(βˆ’3 βˆ’ 1) = 3π‘₯2 – 3π‘₯^(βˆ’4) = 3π‘₯^2βˆ’3/π‘₯^4 = 3(π‘₯^2βˆ’1/π‘₯^4 ) Putting f’(𝒙) = 0 3(π‘₯^2βˆ’1/π‘₯^4 ) = 0 (π‘₯^6 βˆ’ 1)/π‘₯^4 = 0 𝒙^πŸ”βˆ’πŸ = 0 (π‘₯^3 )^2βˆ’(1)^2=0 (𝒙^πŸ‘βˆ’πŸ)(𝒙^πŸ‘+𝟏)=𝟎 Hence, 𝒙 = 1 & –1 π‘₯^3+1 = 0 π‘₯3 = βˆ’1 𝒙 = βˆ’1 Plotting points on number line So, f(π‘₯) is strictly increasing on (βˆ’βˆž , βˆ’1) & (1 , ∞) & f(π‘₯) strictly decreasing on (βˆ’1 , 1) But we need to find Increasing & Decreasing f’(π‘₯) = 3(π‘₯^2βˆ’1/π‘₯^4 ) Thus, f(π‘₯) is increasing on (βˆ’βˆž , βˆ’πŸ] & [𝟏 , ∞) & f(π‘₯) is decreasing on [βˆ’πŸ , 𝟏] For x = βˆ’1 f’(βˆ’1) = 0 For x = 1 f’(1) = 0

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.