Question

Q1- Find the approximate value of f (3.02) where f (x) =3x²+5x+3

Q2- find the approximate value of(31.9)^1/5

Answer 1

f(x) = 3x2 + 15x + 3.


Let x = 3 and ∆x = 0.02,

then f(3.02) = f(x + ∆x)

= 3(x + ∆x)2 + 15(x + ∆x) + 3.


As, ∆y = f(x + ∆x) – f(x)

Therefore, f(x + ∆x) = f(x) + ∆y = f(x) + f’(x).∆x


And f(3.02) = 3x2 + 15x + 3 + (6x + 15).∆x

= 3(3)2 + 15(3) + 3 + [6(3) + 15] (0.02)

= 27 + 45 + 3 + (33)(0.02)

= 75+ 0.66
= 75.66