Check sibling questions

Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 20

Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 21
Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 22 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 23

Maths Crash Course - Live lectures + all videos + Real time Doubt solving!


Transcript

Ex 6.5, 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 (π‘₯) = π‘₯/2 + 2/π‘₯ , π‘₯ > 0g (π‘₯) = π‘₯/2 + 2/π‘₯, π‘₯ > 0 Finding g’(𝒙) g’(π‘₯)=𝑑/𝑑π‘₯ (π‘₯/2+2/π‘₯) =𝑑/𝑑π‘₯ (π‘₯/2)+𝑑/𝑑π‘₯ (2π‘₯^(βˆ’1) ) =1/2βˆ’2π‘₯^(βˆ’2) =1/2βˆ’2/π‘₯^2 Putting g’(𝒙)=𝟎 1/2βˆ’2/π‘₯^2 =0 (π‘₯^2βˆ’ 4)/(2π‘₯^2 )=0 π‘₯^2βˆ’4=0 Γ—2π‘₯^2 (π‘₯βˆ’2)(π‘₯+2)=0 So, π‘₯=2 & π‘₯=βˆ’2 Since π‘₯>0 is given, we consider only π‘₯=2 Finding g’’(𝒙) g’(π‘₯)=1/2βˆ’2/π‘₯^2 g’’(π‘₯)=𝑑/𝑑π‘₯ (1/2βˆ’2/π‘₯^2 ) = 0 – 2 . (βˆ’2) π‘₯^(βˆ’2βˆ’1) = 4π‘₯^(βˆ’3) = 4/π‘₯^3 Putting value π‘₯=2 in g’’(π‘₯) g’’(π‘₯)=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(π‘₯) is g(π‘₯)=π‘₯/2+2/π‘₯ Putting π‘₯=2 g(2)= 2/2+2/2 = 1 + 1 = 2

Ask a doubt (live)
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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.