Ex 6.4, 3 Find the approximate value of f (5.001), where f (x) = x3 – 7x2 + 15.Let x = 5 and ∆ x = 0.001 Given f (x) = x3 – 7x2 + 15 𝑓’(x) = 3x2 − 14x Now, ∆y = f’(x) ∆𝑥 = (3x2 − 14x) 0.001 Also, ∆y = f (x + ∆x) − f(x) f(x + ∆x) = f(x) + ∆y f (5.001) = x3 − 7x2 + 15 + (3x2 − 14x) 0.001 Putting value of x = 5 f (5.001) = 53 − 7(5)2 + 15 + (0.001) [3"(5)2" −14(5)] = (125 − 175 + 15) + (0.001) (5) = −35 + 0.005 = −34.995 Hence, the approximate value of f (5.001) is −34.995

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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