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(B) Given the function $f(x) = \begin{cases} 3x-1, & x > 1 \ x^2+1, & x \leq 1 \end{cases}$.
$1$. The domain $A$ is the set of all possib

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$(-\infty, 1)$

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(B) Given the function $f(x) = \begin{cases} 3x-1, & x > 1 \ x^2+1, & x \leq 1 \end{cases}$.
$1$. The domain $A$ is the set of all possible input values $x$. Since the function is defined for all $x > 1$ and $x \leq 1$,the domain $A = (-\infty, 1] \cup (1, \infty) = (-\infty, \infty) = R$.
$2$. To find the range $B$,we analyze the function in two parts:
For $x > 1$,$f(x) = 3x - 1$. As $x 	o 1^+$,$f(x) 	o 3(1) - 1 = 2$. Thus,$f(x) \in (2, \infty)$.
For $x \leq 1$,$f(x) = x^2 + 1$. The minimum value occurs at $x = 0$,where $f(0) = 1$. As $x 	o -\infty$,$f(x) 	o \infty$. Thus,$f(x) \in [1, \infty)$.
$3$. The range $B$ is the union of these intervals: $B = (2, \infty) \cup [1, \infty) = [1, \infty)$.
$4$. Finally,$A - B = R - [1, \infty) = (-\infty, 1)$.

If $A$ is the domain and $B$ is the range of the function $f(x) = \begin{cases} 3x-1, & x > 1 \\ x^2+1, & x \leq 1 \end{cases}$ then $A-B=$

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