Question 2 - Approximations (using Differentiation) - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by

Ex 6.4, 2 - Find approximate value f(2.01), f(x) = 4x2+5x+2

Ex 6.4,2 - Chapter 6 Class 12 Application of Derivatives - Part 2

Go Ad-free

Transcript

Question 2 Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2.Let x = 2 and ∆ x = 0.01 Given f(x) = 4x2 + 5x + 2 f’(x) = 8x + 5 Now, ∆𝑦 = f’(x) ∆ x = (8x + 5) 0.01 Also, ∆y = f (x + ∆x) − f(x) f(x + ∆ x) = f (x) + ∆𝑦 f (2.01) = 4x2 + 5x + 2 + (8x + 5)(0.01) Putting value of x = 2 𝑓 (2.01)=4 (2)^2+ 5(2)+2+(0.01)[8(2)+5] = (16+10+2)+(21) (0.01) = 28+0.21 =28.21 Hence, the approximate value of f (2.01) is 28.21

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