Range of composite function method

Question

Let $f(x)$ and $g(x) $be functions I've made a method to find range of $f(g(x))$

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$$ f(x)= \begin{cases} 2+x,& 0\leq x<2\\ x^2,& 2\leq x<4 \end{cases} $$

$$ g(x)= \begin{cases} 1+x,& 0\leq x<1\\ 3-x,& 2\leq x<5 \end{cases} $$

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Now to find the range what I did was put $g(x)$ in $f(x)$ and find intersection of function and x Like putting

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1. 2+(1+x)......0<=(1+x)<2 and 0<=x=<1

2. Taking the intersection I got this function will be satisfactory for 0=<x<1. Apparently the answer isnt coming sme and I'm not able to figure out what is wrong in my method . Is there any simpler method

Answer 1

Step 1: examine the range of $g$

• if $x\in[0,1)$ then $g(x)\in A=$ ?

$A=[1,2)$

• if $x\in[2,5)$ then $g(x)\in B=$ ?

$B=[-2,1)$

Step 2: intersect the range of $g$ with the domain of $f$

• $C=(A\cup B)\cap[0,4)=$ ?

$C=[-2,2)\cap[0,4)=[0,2)$

Step 3: the final range is $R=f(C)=$ ?

$R=[2,4)$

Bonus: it is not asked but the domain of $f\circ g$ is $D=g^{-1}(C)=$ ?

this part is a bit more difficult $C=[0,2)=[0,1)\cup[1,2)$
the first output interval $[0,1)$ is covered by $B$ so we search antecedents $x$ in $[2,5)$, this is the subset $(2,3]$
the second output interval $[1,2)$ is exactly $A$ so antecedents $x$ are the whole $[0,1)$
therefore $D=[0,1)\cup(2,3]$