Question

Using differential, find the approximate value of f(2.01), where f(x) = 4x³ + 5x² + 2.

Answer 1

Let x = 2,

Δx = 0.01 

Where f(x) = 4x³ + 5x² + 2

⇒ x + Δx = 2 + 0.01 = 2.01

By definition, approximate value of f(x) is