## University Calculus: Early Transcendentals (3rd Edition)

$L(x)=-x+2$ at $x=x_{0}$
$f(x)=\frac{1}{x}$, $L(x)=f(a)+f'(a)(x-a)$ $f'(x)=-\frac{1}{x^2}$ at $x_{0}=1$ the function and its derivative is close to the value 0.9 $x_{0}=1$ $f(1)=1$ $f'(1)=-1$ then $L(x)=1-1(x-1)=-x+2$ $L(x)=-x+2$ at $x=x_{0}$ Thus, the final answer is: $L(x)=-x+2$ at $x=x_{0}$