Slide1.JPG

Slide2.JPG
Slide3.JPG
Slide4.JPG
Slide5.JPG

Go Ad-free

Transcript

Misc 1 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 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) 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 Since f’’(𝑥) < 0 at 𝑥 = e . ∴ 𝑥 = e is point of maxima Hence, f(𝑥) is maximum at 𝒙 = e.

Davneet Singh's photo - Co-founder, Teachoo

Made by

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