Given
[tex]f(x)=\begin{cases}x^2-2\text{ x<2} \\ 4+|x-5|\text{ x}\ge2\end{cases}[/tex]
We want to find the domain, f(-1), f(0), f(2), f(4)
Solution
To find the domain
We can see that
[tex]\begin{gathered} \text{first,} \\ x<2\text{ which is (-inf,2)} \\ we\text{ also have} \\ x\ge2\text{ which is \lbrack{}2,inf)} \\ so\text{ we take the union} \\ (-\text{inf, 2)U\lbrack{}2, inf) = (-inf, inf)} \end{gathered}[/tex]
The domain is (-inf, inf)
To find f(-1)
Since -1<2, we use
[tex]\begin{gathered} f(x)=x^2-2 \\ f(-1)=(-1)^2-2=1-2=-1 \end{gathered}[/tex]
To find f(0)
Since 0<2, we use
[tex]\begin{gathered} f(x)=x^2-2 \\ f(0)=0^2-2=-2 \end{gathered}[/tex]
To find f(2)
Now, since x=2, we use
[tex]\begin{gathered} f(x)=4+|x-5| \\ f(2)=4+|2-5| \\ f(2)=4+|-3| \\ f(2)=4+3 \\ f(2)=7 \end{gathered}[/tex]
To find f(4)
Again x=4>2, we still use
[tex]\begin{gathered} f(x)=4+|x-5| \\ f(4)=4+|4-5| \\ f(4)=4+1 \\ f(4)=5 \end{gathered}[/tex]
