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Chapter 1: Problem 2

Use the differential formulas in this chapter to solve these problems. Approximate \(\sqrt{25.02}\)

Short Answer

Expert verified
The approximate value of \(\sqrt{25.02}\) is 5.002.

Step by step solution

01

Identify the knowns

The function is \(f(x) = \sqrt{x}\), its derivative is \(f'(x) = \frac{1}{2\sqrt{x}}\), and the approximate point is 25. The number we are approximating to is 25.02.
02

Find \(f'(25)\)

Substitute 25 into the derivative: \(f'(25) = \frac{1}{2\sqrt{25}} = \frac{1}{10}\).
03

Apply the linear approximation formula

The linear approximation formula is \(L(x) = f(a) + f'(a)(x - a)\). Substituting our values into the formula, we get \(L(x) = \sqrt{25} + \frac{1}{10}(x - 25)\).
04

Approximate \(\sqrt{25.02}\)

Substitute 25.02 for x in the equation from Step 3: \(L(25.02) = 5 + \frac{1}{10}(25.02 - 25) = 5.002\).

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