Find the range of the function f(x) defined b | vedclass

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(B) To find the range,we analyze the function in three intervals:
$1$. For $x < -1$,$f(x) = 2x - 3$. As $x 	o -1^-$,$f(x) 	o 2(-1) - 3 = -5$. Si

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$(-\infty, -5) \cup [0, 1] \cup (5, \infty)$

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(B) To find the range,we analyze the function in three intervals: $1$. For $x < -1$,$f(x) = 2x - 3$. As $x o -1^-$,$f(x) o 2(-1) - 3 = -5$. Since $x < -1$,$f(x) < -5$. So,the range for this part is $(-\infty, -5)$. $2$. For $-1 \leq x \leq 1$,$f(x) = 1 - x^2$. The minimum value is at $x = -1$ or $x = 1$,which is $1 - (1)^2 = 0$. The maximum value is at $x = 0$,which is $1 - 0 = 1$. So,the range for this part is $[0, 1]$. $3$. For $x > 1$,$f(x) = 3x^2 + 2$. As $x o 1^+$,$f(x) o 3(1)^2 + 2 = 5$. Since $x > 1$,$f(x) > 5$. So,the range for this part is $(5, \infty)$. Combining these,the total range is $(-\infty, -5) \cup [0, 1] \cup (5, \infty)$.

$x$	$-2$	$-1.5$	$-1$	$-0.5$	$0.25$	$0.5$	$1$	$1.5$	$2$
$y = \frac{1}{x}$	....	....	....	....	....	....	....	....	....

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

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