To complete the definition of the piecewise function [tex]\( f(x) \)[/tex] by identifying the domains of each piece, we first need to determine the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex].
Given the piecewise function:
[tex]\[
f(x) = \begin{cases}
-x, & x < a \\
-1, & a \leq x < b \\
(x-3)^2 - 1, & x \geq b
\end{cases}
\][/tex]
After solving the conditions for continuity at the transition points [tex]\( x = a \)[/tex] and [tex]\( x = b \)[/tex], we have:
[tex]\[ a = 1 \][/tex]
[tex]\[ b = 3 \][/tex]
Using these values, we can now specify the domains for each piece of the function:
1. For the first piece [tex]\( f(x) = -x \)[/tex]:
[tex]\[ x < a \implies x < 1 \][/tex]
The domain of this piece is [tex]\( x < 1 \)[/tex].
2. For the second piece [tex]\( f(x) = -1 \)[/tex]:
[tex]\[ a \leq x < b \implies 1 \leq x < 3 \][/tex]
The domain of this piece is [tex]\( 1 \leq x < 3 \)[/tex].
3. For the third piece [tex]\( f(x) = (x - 3)^2 - 1 \)[/tex]:
[tex]\[ x \geq b \implies x \geq 3 \][/tex]
The domain of this piece is [tex]\( x \geq 3 \)[/tex].
Thus, the completed definition of the piecewise function with the identified domains is:
[tex]\[
f(x) = \begin{cases}
-x, & x < 1 \\
-1, & 1 \leq x < 3 \\
(x-3)^2 - 1, & x \geq 3
\end{cases}
\][/tex]
