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

Slide45.JPG

  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
Ask Download

Transcript

Ex 6.4,2 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, ∆𝑥 & ∆𝑦 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

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
Jail