Example 3

Find an approximate value for (0.99)⁵.

Let

f(x) = x⁵, c = 1.

Then

f(c) = 1⁵ = 1, f'(c) = 5c⁴ = 5.

We put

0.99 = c + Δx, Δx = -0.01.

Then the approximate value is

To get an error estimate we see that f"(u) = 20u³, so |f"(u)| ≤ 20 for u between 0.99 and 1. Then M = 20, and

or

0.949 ≤ (0.99)⁵ ≤ 0.951.

Theorem 1 is closely related to the Increment Theorem in Section 2.2. The relation between them can be seen when we write them next to each other.