Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Approximations (using Differentiation)
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Approximations (using Differentiation)
Last updated at May 29, 2023 by Teachoo
Question 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