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Find the Approximate Value of F (5.001), Where F (X) = X3 − 7x2 + 15. - Mathematics

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Question

Find the approximate value of f (5.001), where f (x) = x3 − 7x2 + 15.

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Solution

Let x = 5 and Δx = 0.001. Then, we have:

Hence, the approximate value of f (5.001) is −34.995.

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Chapter 6: Application of Derivatives - Exercise 6.4 [Page 216]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 6 Application of Derivatives
Exercise 6.4 | Q 3 | Page 216

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