Ex 6.3, 3 (vi) - Chapter 6 Class 12 Application of Derivatives

Last updated at June 12, 2023 by Teachoo

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Book a free demo

Transcript

Ex 6.3, 3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (vi) g (𝑥) = 𝑥/2 + 2/𝑥 , 𝑥 > 0g (𝑥) = 𝑥/2 + 2/𝑥, 𝑥 > 0 Finding g’(𝒙) g’(𝑥)=𝑑/𝑑𝑥 (𝑥/2+2/𝑥) =𝑑/𝑑𝑥 (𝑥/2)+𝑑/𝑑𝑥 (2𝑥^(−1) ) =1/2−2𝑥^(−2) =1/2−2/𝑥^2 Putting g’(𝒙)=𝟎 1/2−2/𝑥^2 =0 (𝑥^2− 4)/(2𝑥^2 )=0 𝑥^2−4=0 ×2𝑥^2 (𝑥−2)(𝑥+2)=0 So, 𝑥=2 & 𝑥=−2 Since 𝑥>0 is given, we consider only 𝑥=2 Finding g’’(𝒙) g’(𝑥)=1/2−2/𝑥^2 g’’(𝑥)=𝑑/𝑑𝑥 (1/2−2/𝑥^2 ) = 0 – 2 . (−2) 𝑥^(−2−1) = 4𝑥^(−3) = 4/𝑥^3 Putting value 𝑥=2 in g’’(𝑥) g’’(𝑥)=4/(2)^3 = 4/8= 1/2 Since g’’ (x) > 0, for x = 2 g(x) is minimum at x = 2 Minimum value of g(𝑥) is g(𝑥)=𝑥/2+2/𝑥 Putting 𝑥=2 g(2)= 2/2+2/2 = 1 + 1 = 2

Next: Ex 6.3, 3 (vii) Important → Ask a doubt

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Class 6

Class 7

Class 8

Class 9

Class 10

Class 11

Class 12

Courses

Interesting

Popular