Question 2 Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2.Let x = 2 and ∆ x = 0.01 Given f(x) = 4x2 + 5x + 2 f’(x) = 8x + 5 Now, ∆𝑦 = f’(x) ∆ x = (8x + 5) 0.01 Also, ∆y = f (x + ∆x) − f(x) f(x + ∆ x) = f (x) + ∆𝑦 f (2.01) = 4x2 + 5x + 2 + (8x + 5)(0.01) Putting value of x = 2 𝑓 (2.01)=4 (2)^2+ 5(2)+2+(0.01)[8(2)+5] = (16+10+2)+(21) (0.01) = 28+0.21 =28.21 Hence, the approximate value of f (2.01) is 28.21

Next: Question 3 Important → Ask a doubt

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.

Class 6

Class 7

Class 8

Class 9

Class 10

Class 11

Class 12

Courses

Interesting

Popular