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

Last updated at April 15, 2021 by Teachoo

Ex 6.4

Ex 6.4, 1 (i)
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Last updated at April 15, 2021 by Teachoo

Ex 6.4, 3 Find the approximate value of f (5.001), where f (x) = x3 â 7x2 + 15.Let x = 5 and â x = 0.001 Given f (x) = x3 â 7x2 + 15 đâ(x) = 3x2 â 14x Now, ây = fâ(x) âđĽ = (3x2 â 14x) 0.001 Also, ây = f (x + âx) â f(x) f(x + âx) = f(x) + ây f (5.001) = x3 â 7x2 + 15 + (3x2 â 14x) 0.001 Putting value of x = 5 f (5.001) = 53 â 7(5)2 + 15 + (0.001) [3"(5)2" â14(5)] = (125 â 175 + 15) + (0.001) (5) = â35 + 0.005 = â34.995 Hence, the approximate value of f (5.001) is â34.995