Question

Linearize the function $f(x) = \ln(1 - x)$ around $a = 0$ and use it to approximate $\ln(1.23)$. ? $\ln(1 - x) \approx \frac{1}{1 - x}$ and $\ln(1.23) \approx \frac{1}{1.23}$ $\ln(1 - x) \approx 1 - x$ and $\ln(1.23) \approx 1.23$ $\ln(1 - x) \approx -x$ and $\ln(1.23) \approx 0.23$ $\ln(1 - x) \approx 1 + x$ and $\ln(1.23) \approx 0.77$

          Linearize the function $f(x) = \ln(1 - x)$ around $a = 0$ and use it to approximate $\ln(1.23)$. 
?
$\ln(1 - x) \approx \frac{1}{1 - x}$ and $\ln(1.23) \approx \frac{1}{1.23}$
$\ln(1 - x) \approx 1 - x$ and $\ln(1.23) \approx 1.23$
$\ln(1 - x) \approx -x$ and $\ln(1.23) \approx 0.23$
$\ln(1 - x) \approx 1 + x$ and $\ln(1.23) \approx 0.77$

Linearize the function f(x) = ln(1 - x) around a = 0 and use it to approximate ln(1.23).
?
ln(1 - x) ≈(1)/(1 - x) and ln(1.23) ≈(1)/(1.23)
ln(1 - x) ≈ 1 - x and ln(1.23) ≈ 1.23
ln(1 - x) ≈ -x and ln(1.23) ≈ 0.23
ln(1 - x) ≈ 1 + x and ln(1.23) ≈ 0.77

Added by Gregory B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Collin Parker

Step 1

The formula for the linear approximation is: f(x) ≈ f(a) + f'(a)(x - a) In this case, a = 0 and f(x) = ln(1 - x). Show more…

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Linearize the function f(x) = ln(1 - x) around a = 0 and use it to approximate ln(1.23). ln(1 - x) â‰ˆ 1 - x and ln(1.23) â‰ˆ 1.23 ln(1 - x) â‰ˆ -x and ln(1.23) â‰ˆ 0.23 ln(1 - x) â‰ˆ 1 + x and ln(1.23) â‰ˆ 0.77