Question

given the function below, fill in the table of values, use the table of values to graph the function, and then identify the function’s domain and range.
$y = \sqrt3{x + 1} + 4$

$x$	$x + 1$	$\sqrt3{x + 1}$	$y = \sqrt3{x}$
$-2$
$-1$
$0$
$7$

Explanation:

Step1: Calculate $x+1$ for each $x$

For $x=-9$: $-9 + 1 = -8$
For $x=-2$: $-2 + 1 = -1$
For $x=-1$: $-1 + 1 = 0$
For $x=0$: $0 + 1 = 1$
For $x=7$: $7 + 1 = 8$

Step2: Calculate $\sqrt[3]{x+1}$

For $x+1=-8$: $\sqrt[3]{-8} = -2$
For $x+1=-1$: $\sqrt[3]{-1} = -1$
For $x+1=0$: $\sqrt[3]{0} = 0$
For $x+1=1$: $\sqrt[3]{1} = 1$
For $x+1=8$: $\sqrt[3]{8} = 2$

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

For $\sqrt[3]{x+1}=-2$: $-2 + 4 = 2$
For $\sqrt[3]{x+1}=-1$: $-1 + 4 = 3$
For $\sqrt[3]{x+1}=0$: $0 + 4 = 4$
For $\sqrt[3]{x+1}=1$: $1 + 4 = 5$
For $\sqrt[3]{x+1}=2$: $2 + 4 = 6$

Step4: Find domain and range

Cube roots accept all real numbers, so the domain is all real $x$. The cube root outputs all real numbers, so adding 4 keeps the range all real $y$.

Answer:

Completed Table:

$x$	$x+1$	$\sqrt[3]{x+1}$	$y=\sqrt[3]{x+1}+4$
$-2$	$-1$	$-1$	$3$
$-1$	$0$	$0$	$4$
$0$	$1$	$1$	$5$
$7$	$8$	$2$	$6$

Domain and Range:

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