Given f(x) = 4x² + 5x + 2

Let x = 2 so that x + Δx = 2.01

⇒ 2 + Δx = 2.01

∴ Δx = 0.01

On differentiating f(x) with respect to x, we get

We know and derivative of a constant is 0.

When x = 2, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.01

⇒ Δf = (21)(0.01)

∴ Δf = 0.21

Now, we have f(2.01) = f(2) + Δf

⇒ f(2.01) = 4(2)² + 5(2) + 2 + 0.21

⇒ f(2.01) = 16 + 10 + 2 + 0.21

∴ f(2.01) = 28.21

Thus, f(2.01) = 28.21