Mathematics - Functions Question with Solution | TestHub

$$f(x)=|x-1|=\{\begin{array}{ll}1-x& 0<x<1\\ x-1& x\ge 1\end{array}$$

$$g(x)=e^{x}x\ge -1$$

$$(f\circ g)(x)=\{\begin{array}{ll}1-g(x)& 0<g(x)<1i.e.,-1\le x<0\\ g(x)-1& g(x)\ge 1i.e.,0\le x\end{array}=\{\begin{array}{ll}1-e^x& -1\le x<0\\ e^x-1& x\ge 0\end{array}$$

$$\begin{gathered} 1-e^x\&-1\le x<0 \\ {e}^x-1\&x\ge 0 \end{gathered}$$

Therefore, the domain is $[-1,\infty )$.

$f\circ g$ is decreasing in $[-1,0]$ and increasing in $(0,\infty )$.

$f\circ g(-1)=1-\frac{1}{e}$ and $f\circ g(0)=0$.

As $x\to \infty$, $f\circ g(x)=\infty$. Therefore, the range is $[0,\infty )$.