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Find the Approximate Value of F (2.01), Where F (X) = 4x2 + 5x + 2 - Mathematics

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प्रश्न

Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2

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उत्तर

Let x = 2 and Δx = 0.01. Then, we have:

f(2.01) = f(+ Δx) = 4(x + Δx)2 + 5(x + Δx) + 2

Now, Δy = f(x + Δx) − f(x)

∴ f(x + Δx) = f(x) + Δy

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पाठ 6: Application of Derivatives - Exercise 6.4 [पृष्ठ २१६]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 6 Application of Derivatives
Exercise 6.4 | Q 2 | पृष्ठ २१६

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