Question

the function’s domain and range.
$y = -\frac{1}{2}sqrt3{x - 4}$
column by column
$ - 4$ $sqrt3{x - 4}$ $y = -\frac{1}{2}sqrt3{x}$
answer
attempt 1 out of 2

Explanation:

Step1: Pick x-values for calculation

Choose $x=-4, 4, 12, -13, 31$ (values that make $x-4$ a perfect cube for simplicity)

Step2: Calculate $\sqrt[3]{x-4}$ for each x

• For $x=-4$: $\sqrt[3]{-4-4}=\sqrt[3]{-8}=-2$
• For $x=4$: $\sqrt[3]{4-4}=\sqrt[3]{0}=0$
• For $x=12$: $\sqrt[3]{12-4}=\sqrt[3]{8}=2$
• For $x=-13$: $\sqrt[3]{-13-4}=\sqrt[3]{-27}=-3$
• For $x=31$: $\sqrt[3]{31-4}=\sqrt[3]{27}=3$

Step3: Calculate $y=-\frac{1}{2}\sqrt[3]{x-4}$

• For $x=-4$: $y=-\frac{1}{2}(-2)=1$
• For $x=4$: $y=-\frac{1}{2}(0)=0$
• For $x=12$: $y=-\frac{1}{2}(2)=-1$
• For $x=-13$: $y=-\frac{1}{2}(-3)=\frac{3}{2}$
• For $x=31$: $y=-\frac{1}{2}(3)=-\frac{3}{2}$

Step4: Find domain of the function

Cube roots accept all real numbers, so $x-4$ can be any real number. Thus, $x \in \mathbb{R}$.

Step5: Find range of the function

Since $\sqrt[3]{x-4}$ outputs all real numbers, scaling by $-\frac{1}{2}$ still gives all real numbers. Thus, $y \in \mathbb{R}$.

Answer:

Completed Table:

$x$	$\sqrt[3]{x-4}$	$y=-\frac{1}{2}\sqrt[3]{x-4}$
$4$	$0$	$0$
$12$	$2$	$-1$
$-13$	$-3$	$\frac{3}{2}$
$31$	$3$	$-\frac{3}{2}$

Domain: All real numbers, or $(-\infty, \infty)$

Range: All real numbers, or $(-\infty, \infty)$