Using diffe

Let us assume that f(x) = x⁵

Also, let x = 2 so that x + Δx = 1.999

⇒ 2 + Δx = 1.999

∴ Δx = –0.001

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

We know

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 = –0.001

⇒ Δf = (80)(–0.001)

∴ Δf = –0.08

Now, we have f(1.999) = f(2) + Δf

⇒ f(1.999) = 2⁵ – 0.08

⇒ f(1.999) = 32 – 0.08

∴ f(1.999) = 31.92

Thus, (1.999)⁵ ≈ 31.92