Answer
$L(x)=\frac{x}{4}+\frac{1}{4}$
Work Step by Step
$f(x)={\frac{x}{x+1}}$,
$L(x)=f(a)+f'(a)(x-a)$
$f'(x)=\frac{x+1-x}{(x+1)^2}=\frac{1}{(x+1)^2}$
The value of$x_{0}$ is chosen close to 1.3; we choose a value of 1 for simplicity
$x_{0}=1$
$f(1)={\frac{1}{1+1}}=\frac{1}{2}$
$f'(8)=\frac{1}{(1+1)^2}=\frac{1}{4}$
then $L(x)=\frac{1}{2}+\frac{1}{4}(x-1)=\frac{x}{4}+\frac{1}{4}$
$L(x)=\frac{x}{4}+\frac{1}{4}$
Thus, the final answer is: $L(x)=\frac{x}{4}+\frac{1}{4}$
