Chapter 3: Derivatives - Practice Exercises - Page 181: 120

$L(x)=2.5x-0.1$

Step 1. Given $f(x)=\frac{2}{1-x}+\sqrt {1+x}-3.1$, we have $f'(x)=\frac{2}{(1-x)^2}+\frac{1}{2\sqrt {1+x}}$ Step 2. Calculate function values at $x=0$: $f(0)=\frac{2}{1-0}+\sqrt {1+0}-3.1=-0.1$ and $f'(0)=\frac{2}{(1-0)^2}+\frac{1}{2\sqrt {1+0}}=2.5$ Step 3. The linearization of $f(x) $at $x=0$ can be written as $L(x)=f(0)+f'(0)(x-0)=-0.1+2.5x=2.5x-0.1$