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Misc 2 - Show that f(x) = log x / x has maximum at x = e

Misc 2 - Chapter 6 Class 12 Application of Derivatives - Part 2
Misc 2 - Chapter 6 Class 12 Application of Derivatives - Part 3 Misc 2 - Chapter 6 Class 12 Application of Derivatives - Part 4 Misc 2 - Chapter 6 Class 12 Application of Derivatives - Part 5

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Transcript

Misc 2 Show that the function given by f(x) = log⁡𝑥/𝑥 is maximum at x = e.Let f(𝑥) = log⁡𝑥/𝑥 Finding f’(𝒙) f’(𝑥) = 𝑑/𝑑𝑥 (log⁡𝑥/𝑥) f’(𝑥) = (𝑑(log⁡𝑥 )/𝑑𝑥 " " . 𝑥 − 𝑑(𝑥)/𝑑𝑥 " . log " 𝑥)/𝑥2 f’(𝑥) = (1/𝑥 × 𝑥 − log⁡𝑥)/𝑥2 f’(𝑥) = (1 − log⁡𝑥)/𝑥2 Using quotient rule as (𝑢/𝑣)^′ = (𝑢^′ 𝑣−𝑣^′ 𝑢)/𝑣^2 Putting f’(𝒙) = 0 (1 − log⁡𝑥)/𝑥2=0 1 – log 𝑥 = 0 log 𝑥 = 1 𝒙 = e Finding f’’(𝒙) f’(𝑥) = (1 − log⁡𝑥)/𝑥2 Diff w.r.t. 𝑥 f’’(𝑥) = 𝑑/𝑑𝑥 ((1 − log⁡𝑥)/𝑥2) log 𝑥=1 is only possible when 𝒙=𝒆 Using quotient rule as (𝑢/𝑣)^′=(𝑢^′ 𝑣 − 𝑣^′ 𝑢)/(𝑣2 .) f’’(𝑥) = (𝑑(1 − log⁡𝑥 )/𝑑𝑥 . 𝑥2− 𝑑(𝑥2)/𝑑𝑥 . (1 − log⁡𝑥 ))/(𝑥^2 )^2 = ((0 − 1/𝑥) . 𝑥2 − 2𝑥(1 − log⁡𝑥 ))/𝑥4 = ((−1)/𝑥 × 𝑥2 − 2𝑥(1 − log⁡𝑥 ))/𝑥^4 = (−𝑥 − 2𝑥(1 − log⁡𝑥 ))/𝑥^4 = (−𝑥[1 + 2(1 − log⁡𝑥 )])/𝑥4 = (−𝑥[3 − 2 log⁡𝑥 ])/𝑥4 ∴ f’’(𝑥) = (−(3 − 2 log⁡𝑥 ))/𝑥3 Putting 𝒙 = e f’’(𝑒) = (−(3 − 2 log⁡𝑒 ))/𝑒3 = (−(3 − 2))/𝑒3 = (−1)/𝑒3 = –(1/𝑒3) < 0 (∵1/𝑒𝑥>0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 ∈𝑹) Since f’’(𝑥) < 0 at 𝑥 = e . ∴ 𝑥 = e is point of maxima Hence, f(𝑥) is maximum at 𝒙 = e.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.