Question

If $f\left(x\right)=\{\begin{array}{cc}x-1,& -1\le x\le 0\\ x^2,& 0<x\le 1\end{array},$ $g\left(x\right)=\sin x$ and $h\left(x\right)=f\left(\left|g\right(x\left)\right|\right)+\left|f\right(g\left(x\right)\left)\right|$ then find h(0).

Answer 1

$h\left(0\right)=f\left(|g\right(0\left)|\right)+|f\left(g\right(0\left)\right)|$, substitute $x=$ in $h\left(x\right)$

Now $g\left(0\right)=\sin 0=0$

$h\left(0\right)=f\left(|g\right(0\left)|\right)+|f\left(g\right(0\left)\right)|=f\left(|0|\right)+|f\left(0\right)|=f\left(0\right)+|f\left(0\right)|=(0-1)+|(0-1)|=-1+1=0$