If $f ( x ) = \left\{ \begin{array} { c c }
- 2 x + 1 & - 3 \leq x

Question

If $f ( x ) = \left\{ \begin{array} { c c } 
- 2 x + 1 & - 3 \leq x < - 1 \
4 & - 1 < x < 5, \
x - 2 & 5 < x < 7
\end{array} \right.$ then the domain of f is

A) $[ - 3,7 ]$
B) $[ - 3,7 )$
C) $( - 3 , - 1 ) \cup ( - 1,7 )$
D) $[ - 3 , - 1 ) \cup ( - 1,7 ]$
E) $[ - 3 , - 1 ) \cup ( - 1,5 ) \cup ( 5,7 )$

Quizplus · Accepted Answer

The function $f(x)$ is defined piecewise with three different expressions for different intervals of $x$. The first piece is defined for $-3 \leq x < -1$, the second piece for $-1 < x < 5$, and the third piece for $5 < x < 7$. Notice that $x = -1$ and $x = 5$ are not included in the domain, as the function does not define values exactly at these points. Therefore, the domain of $f$ is $[ - 3 , - 1 ) \cup ( - 1,5 ) \cup ( 5,7 )$, which matches choice E.

If f(x)={−2x+1−3≤x<−14−1<x<5,x−25<x<7f ( x ) = \left\{ \begin{array} { c c } - 2 x + 1 & - 3 \leq x < - 1 \\4 & - 1 < x < 5, \\x - 2 & 5 < x < 7\end{array} \right.f(x)=⎩⎨⎧​−2x+14x−2​−3≤x<−1−1<x<5,5<x<7​

If $f ( x ) = \left\{ \begin{array} { c c } - 2 x + 1 & - 3 \leq x < - 1 \\4 & - 1 < x < 5, \\x - 2 & 5 < x < 7\end{array} \right.$