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

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Class 6

Class 7

Class 8

Class 9

Class 10

Class 11

Class 12

Courses

Interesting

Popular