Answer
$L(x)=1+\frac{3}{2}x$
The linearizaion of a sum of functions is the sum of the individual linearizations.
Work Step by Step
Step 1. $f(x)=\sqrt {x+1}+sin(x), f(0)=1$, $f'(x)=\frac{1}{2\sqrt {x+1}}+cos(x), f'(0)=\frac{3}{2}$
Step 2. The linearization at $x=0$: $L(x)=1+\frac{3}{2}x$
Step 3. For $g(x)=\sqrt {x+1}, g(0)=1, g'(x)=\frac{1}{2\sqrt {x+1}}, g'(0)=\frac{1}{2}, L_1(x)=1+\frac{1}{2}x$
Step 4. $h(x)=sin(x), f(0)=0, f'(x)=cos(x), f'(0)=1, L_2(x)=0+x=x$
Step 5. We have $L(x)=L_1(x)+L_2(x)$, which means that the linearization of a sum of functions is the sum of the individual linearizations.
