Question

use the linear approximation of $f(x)=ln(x)$ at $x = 1$ to estimate $ln(1.31)$.
$ln(1.31)approx2.31$
$ln(1.31)approx0.31$
$ln(1.31)approx1.31$
$ln(1.31)approx1.27$
$ln(1.31)approx0.27$

Explanation:

Step1: Recall linear - approximation formula

The linear - approximation formula is $L(x)=f(a)+f^{\prime}(a)(x - a)$, where $a$ is the point of approximation, $x$ is the point at which we want to estimate the function value. Given $f(x)=\ln(x)$ and $a = 1$, first find $f(1)$ and $f^{\prime}(x)$.
The derivative of $f(x)=\ln(x)$ is $f^{\prime}(x)=\frac{1}{x}$. Then $f(1)=\ln(1)=0$ and $f^{\prime}(1)=\frac{1}{1}=1$.

Step2: Write the linear - approximation function

Substitute $a = 1$, $f(1)=0$, and $f^{\prime}(1)=1$ into the linear - approximation formula $L(x)=f(a)+f^{\prime}(a)(x - a)$. We get $L(x)=0 + 1\times(x - 1)=x - 1$.

Step3: Estimate $\ln(1.31)$

We want to estimate $\ln(1.31)$, so we set $x = 1.31$ in the linear - approximation function $L(x)$. Then $L(1.31)=1.31-1=0.31$.

Answer:

$\ln(1.31)\approx0.31$