1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise
  3. Ex 6.4

Ex 6.4,3 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 15, 2021 by

Ex 6.4, 3 - Find approx value f(5.001), f(x) = x3 - 7x2 + 15

Ex 6.4,3 - Chapter 6 Class 12 Application of Derivatives - Part 2

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

    3. Ex 6.4

    Ex 6.5 →

Transcript

Ex 6.4, 3 Find the approximate value of f (5.001), where f (x) = x3 โ€“ 7x2 + 15.Let x = 5 and โˆ† x = 0.001 Given f (x) = x3 โ€“ 7x2 + 15 ๐‘“โ€™(x) = 3x2 โˆ’ 14x Now, โˆ†y = fโ€™(x) โˆ†๐‘ฅ = (3x2 โˆ’ 14x) 0.001 Also, โˆ†y = f (x + โˆ†x) โˆ’ f(x) f(x + โˆ†x) = f(x) + โˆ†y f (5.001) = x3 โˆ’ 7x2 + 15 + (3x2 โˆ’ 14x) 0.001 Putting value of x = 5 f (5.001) = 53 โˆ’ 7(5)2 + 15 + (0.001) [3"(5)2" โˆ’14(5)] = (125 โˆ’ 175 + 15) + (0.001) (5) = โˆ’35 + 0.005 = โˆ’34.995 Hence, the approximate value of f (5.001) is โˆ’34.995

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

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