Let A = {-4, -2, -1, 0, 3, 5} and f: A righta | vedclass

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(A) Given the function $f: A ightarrow R$ where $A = \{-4, -2, -1, 0, 3, 5\}$ and $f(x) = \begin{cases} 3x - 1 & 	ext{for } x > 3 \ x^2 + 1 & 	ex

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$\{-11, 5, 2, 1, 10, 14\}$

Vedclass · Answer

(A) Given the function $f: A ightarrow R$ where $A = \{-4, -2, -1, 0, 3, 5\}$ and $f(x) = \begin{cases} 3x - 1 & 	ext{for } x > 3 \ x^2 + 1 & 	ext{for } -3 \leq x \leq 3 \ 2x - 3 & 	ext{for } x We calculate the value of $f(x)$ for each element in $A$:For $x = -4$ $(x For $x = -2$ $(-3 \leq x \leq 3)$: $f(-2) = (-2)^2 + 1 = 4 + 1 = 5$.For $x = -1$ $(-3 \leq x \leq 3)$: $f(-1) = (-1)^2 + 1 = 1 + 1 = 2$.For $x = 0$ $(-3 \leq x \leq 3)$: $f(0) = 0^2 + 1 = 1$.For $x = 3$ $(-3 \leq x \leq 3)$: $f(3) = 3^2 + 1 = 9 + 1 = 10$.For $x = 5$ $(x > 3)$: $f(5) = 3(5) - 1 = 15 - 1 = 14$.Thus,the range of $f$ is $\{-11, 5, 2, 1, 10, 14\}$.

Function	Range
$A. f(x) = \|x\|$	$I. [0, \infty)$
$B. f(x) = x^2$	$II. \mathbb{R}$
$C. f(x) = x^3$	$III. [0, \infty)$
$D. f(x) = \text{sgn}(x)$	$IV. \{-1, 0, 1\}$

Let $A = \{-4, -2, -1, 0, 3, 5\}$ and $f: A \rightarrow R$ be defined by $f(x) = \begin{cases} 3x - 1 & \text{for } x > 3 \\ x^2 + 1 & \text{for } -3 \leq x \leq 3 \\ 2x - 3 & \text{for } x < -3 \end{cases}$. Then the range of $f$ is

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