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Misc 4 Find the intervals in which the function f given by f (x) = x3 + 1/๐‘ฅ^3 , ๐‘ฅ โ‰  0 is (i) increasing (ii) decreasing. f(๐‘ฅ) = ๐‘ฅ3 + 1/๐‘ฅ3 Finding fโ€™(๐’™) fโ€™(๐‘ฅ) = ๐‘‘/๐‘‘๐‘ฅ (๐‘ฅ^3+๐‘ฅ^(โˆ’3) )^. = 3๐‘ฅ2 + (โˆ’3)^(โˆ’3 โˆ’ 1) = 3๐‘ฅ2 โ€“ 3๐‘ฅ^(โˆ’4) = 3๐‘ฅ^2โˆ’3/๐‘ฅ^4 = 3(๐‘ฅ^2โˆ’1/๐‘ฅ^4 ) Putting fโ€™(๐’™) = 0 3(๐‘ฅ^2โˆ’1/๐‘ฅ^4 ) = 0 (๐‘ฅ^6 โˆ’ 1)/๐‘ฅ^4 = 0 ๐’™^๐Ÿ”โˆ’๐Ÿ = 0 (๐‘ฅ^3 )^2โˆ’(1)^2=0 (๐’™^๐Ÿ‘โˆ’๐Ÿ)(๐’™^๐Ÿ‘+๐Ÿ)=๐ŸŽ Hence, ๐’™ = 1 & โ€“1 Plotting points on number line So, f(๐‘ฅ) is strictly increasing on (โˆ’โˆž , โˆ’1) & (1 , โˆž) & f(๐‘ฅ) strictly decreasing on (โˆ’1 , 1) But we need to find Increasing & Decreasing fโ€™(๐‘ฅ) = 3(๐‘ฅ^2โˆ’1/๐‘ฅ^4 ) Thus, f(๐‘ฅ) is increasing on (โˆ’โˆž , โˆ’๐Ÿ] & [๐Ÿ , โˆž) & f(๐‘ฅ) is decreasing on [โˆ’๐Ÿ , ๐Ÿ]

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About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo