Ex 6.3, 5 (iii) - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by

  Slide10.JPG

Slide11.JPG
Slide12.JPG

Go Ad-free

Transcript

Ex 6.3, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (iii) f(𝑥) = 4𝑥 – 1/2 𝑥2 , 𝑥 ∈ [−2, 9/2] Putting f’(𝒙)=𝟎 4 – 𝑥=0 𝑥=4 ∴ 𝑥=4 is only critical point Since given interval 𝑥 ∈ [−2 , 9/2] Hence , calculating f(𝑥) at 𝑥 = – 2 , 4 , 9/2 Absolute Maximum value of f(x) is 8 at 𝒙 = 4 & Absolute Minimum value of f(x) is –10 at 𝒙 = –2

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