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

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

Ex 6.4, 2 Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2.Let x = 2 and โˆ† x = 0.01 Given f(x) = 4x2 + 5x + 2 fโ€™(x) = 8x + 5 Now, โˆ†๐‘ฆ = fโ€™(x) โˆ† x = (8x + 5) 0.01 Also, โˆ†y = f (x + โˆ†x) โˆ’ f(x) f(x + โˆ† x) = f (x) + โˆ†๐‘ฆ f (2.01) = 4x2 + 5x + 2 + (8x + 5)(0.01) Putting value of x = 2 ๐‘“ (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

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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.