Chapter 6 Class 12 Application of Derivatives
Serial order wise

Ex 6.4, 3 - Find approx value f(5.001), f(x) = x3 - 7x2 + 15

Ex 6.4,3 - Chapter 6 Class 12 Application of Derivatives - Part 2

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Question 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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.