Linear Approximations — Question 3

Section 4-11 · Calculus pages 879–880

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Question 3

Use a linear approximation to estimate the value of ln⁡(1.05)\ln(1.05).

  • Identify a function f(x)f(x), choose a point aa, and write the linear approximation L(x)L(x).

  • Use L(x)L(x) to estimate ln⁡(1.05)\ln(1.05).

  • Compare the estimate to the actual value.

Question 3 - Solution

Let f(x)=ln⁡(x)f(x) = \ln(x), and choose a=1a = 1 because ln⁡(1)=0\ln(1) = 0 is easy to compute.

(a) Linear approximation formula: L(x)=f(a)+f′(a)(x−a)L(x) = f(a) + f'(a)(x - a)

Compute: f(x)=ln⁡(x),f′(x)=1x,f′(1)=1f(x) = \ln(x), \quad f'(x) = \frac{1}{x}, \quad f'(1) = 1

So: L(x)=ln⁡(1)+1⋅(x−1)=0+(x−1)=x−1L(x) = \ln(1) + 1 \cdot (x - 1) = 0 + (x - 1) = x - 1

(b) Estimate: L(1.05)=1.05−1=0.05L(1.05) = 1.05 - 1 = 0.05

(c) Actual value: ln⁡(1.05)≈0.04879(using calculator)\ln(1.05) \approx 0.04879 \quad \text{(using calculator)}

Conclusion: The linear approximation gives: ln⁡(1.05)≈0.05\boxed{\ln(1.05) \approx 0.05} which is a very good estimate.