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

Last updated at April 15, 2021 by Teachoo

Last updated at April 15, 2021 by Teachoo

Transcript

Ex 6.4, 3 Find the approximate value of f (5.001), where f (x) = x3 โ 7x2 + 15.Let x = 5 and โ x = 0.001 Given f (x) = x3 โ 7x2 + 15 ๐โ(x) = 3x2 โ 14x Now, โy = fโ(x) โ๐ฅ = (3x2 โ 14x) 0.001 Also, โy = f (x + โx) โ f(x) f(x + โx) = f(x) + โy f (5.001) = x3 โ 7x2 + 15 + (3x2 โ 14x) 0.001 Putting value of x = 5 f (5.001) = 53 โ 7(5)2 + 15 + (0.001) [3"(5)2" โ14(5)] = (125 โ 175 + 15) + (0.001) (5) = โ35 + 0.005 = โ34.995 Hence, the approximate value of f (5.001) is โ34.995

Ex 6.4

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Chapter 6 Class 12 Application of Derivatives (Term 1)

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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.