Let us assume that

Also, let x = 400 so that x + Δx = 401

⇒ 400 + Δx = 401

∴ Δx = 1

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

We know

When x = 400, 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 = 1

⇒ Δf = (0.025)(1)

∴ Δf = 0.025

Now, we have f(401) = f(400) + Δf

⇒ f(401) = 20 + 0.025

∴ f(401) = 20.025

Thus,