The range of the function f(x) = begin{cases} | vedclass

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(C) For $x > 3$,$f(x) = 4x - 1$. Since $x > 3$,$4x > 12$,so $4x - 1 > 11$. Thus,the range for this part is $(11, \infty)$.For $-2 \leq x \leq 3$,$f(x)

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$R - (7, 11]$

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(C) For $x > 3$,$f(x) = 4x - 1$. Since $x > 3$,$4x > 12$,so $4x - 1 > 11$. Thus,the range for this part is $(11, \infty)$.For $-2 \leq x \leq 3$,$f(x) = x^2 - 2$. The minimum value is at $x = 0$,$f(0) = -2$. The maximum value is at $x = 3$,$f(3) = 7$. Thus,the range for this part is $[-2, 7]$.For $x Combining these intervals: $(-\infty, -2) \cup [-2, 7] \cup (11, \infty) = (-\infty, 7] \cup (11, \infty)$.This can be written as $R - (7, 11]$.

The range of the function $f(x) = \begin{cases} 4x - 1, & x > 3 \\ x^2 - 2, & -2 \leq x \leq 3 \\ 3x + 4, & x < -2 \end{cases}$ is:

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