Answer
$L(x)=0.5+1.5x$
Work Step by Step
Step 1. Given $f(x)=\sqrt {1+x}+sin(x)-0.5$, we have $f'(x)=\frac{1}{2\sqrt {1+x}}+cos(x)$
Step 2. Calculate function values at $x=0$: $f(0)=\sqrt {1+0}+sin(0)-0.5=0.5$ and $f'(0)=\frac{1}{2\sqrt {1+0}}+cos(0)=1.5$
Step 3. The linearization of $f(x) $at $x=0$ can be written as $L(x)=f(0)+f'(0)(x-0)=0.5+1.5x$
