Range and domain of functions and composite functions

Question

I really getting stuck at finding range of function . For example suppose there is this function $ f: {R}^+ \to R $ where $ f(x) = |x-1| $ and $ g:[-1 , \infty) \to R $ where $g(x) =e^{x} $ . Then the domain and range of $fog(x) is $

I'm being taught to fit the $ x $ in some inequality and then recreate the $ fog(x) $ to get the range where the function lies. I know how to recreate the $ fog(x) $ but I get confused between the conditions for which $fog(x)$ exists and where to recreate the $fog(x)$ so that it yields an inequality giving the range of $fog(x)$.I'll be thankful if someone explains that to me.

Answer 1

$(f\circ g)(x)=f(g(x)).$

Let $D(h)$ be a domain of the function $h$.

Thus, $$D(f\circ g)=D(g)\cap\{x|g(x)\in D(f)\}=[-1,+\infty)\cap\mathbb R=[-1,+\infty).$$

The range of $f\circ g$ it's $[0,+\infty)$ because $f\circ g$ is a continuous function, $(f\circ g)(x)\geq0,$

$(f\circ g)(0)=0$ and $\lim\limits_{x\rightarrow+\infty}(f\circ g)(x)=+\infty.$

Answer 2

Given the domain $D_1$ of $f$ and $D_2$ of $g$, to find the domain $D$ of $f\circ g$ we can solve these relations: \begin{cases} x\in D_2,\\g(x)\in D_1. \end{cases} In this problem, these relastions become \begin{cases} x\geq -1,\\e^x>0 \end{cases} implying that the domain of $f\circ g$ is $[-1,\infty)$.

To find the range of $f\circ g$, we note that $y\in Range(f\circ g)\iff y=f(g(x))\text{ for some }x\in D\iff y\in Range(f|_{Range(g|_D)})$. So $$Range(f\circ g)=Range(f|_{Range(g|_D)}).$$ In this problem, $Range(g|_D)=[e^{-1},\infty)$ and $Range(f|_{[e^{-1},\infty)})=[0,+\infty)$ so $Range(f\circ g)=[0,+\infty)$.