Ex 6.2,18 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 14, 2021 by Teachoo

Ex 6.2, 18 - Prove that f(x) = x3 - 3x2 + 3x - 100 is increasing

Ex 6.2,18 - Chapter 6 Class 12 Application of Derivatives - Part 2

Transcript

Ex 6.2, 18 Prove that the function given by f (𝑥) = 𝑥3 – 3𝑥2 + 3𝑥 – 100 is increasing in R. We need to show f(𝑥) is strictly increasing on R i.e. we need to show f’(𝒙) > 0 Finding f’(𝒙) f’(𝑥)= 3x2 – 6x + 3 – 0 = 3(𝑥2−2𝑥+1) = 3((𝑥)2+(1)2−2(𝑥)(1)) = 3(𝑥−1)2 Since Square of any number is always positive (𝑥−1)2 > 0 3(𝑥−1)2>0 f’(𝒙) > 0 Hence, f’(𝑥) > 0 for all values of 𝑥 ∴ f(𝑥) is strictly increasing on R

Ex 6.2

Ex 6.2,18 You are here

Ex 6.3 →

Chapter 6 Class 12 Application of Derivatives (Term 1)

Serial order wise

Ex 6.1
Ex 6.2
Ex 6.3
Ex 6.4
Ex 6.5
Examples
Miscellaneous
Case Based Questions (MCQ)

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.