Question

I need to find the domain and range and fill out the t chart

Answer 1

Given:

$y=\sqrt[3]{x}+1$

Sol:.

$\begin{gathered} y=\sqrt[3]{x}+1 \\ x=-8 \\ y=\sqrt[3]{-8}+1 \\ y=-2+1 \\ y=-1 \end{gathered}$

At x=-1

$\begin{gathered} y=\sqrt[3]{x}+1 \\ y=\sqrt[3]{-1}+1 \\ y=-1+1 \\ y=0 \end{gathered}$

At x=0

$\begin{gathered} y=\sqrt[3]{x}+1 \\ y=\sqrt[3]{0}+1 \\ y=1 \end{gathered}$

At x=1

$\begin{gathered} y=\sqrt[3]{x}+1 \\ y=\sqrt[3]{1}+1 \\ y=2 \end{gathered}$

At x=8

$\begin{gathered} y=\sqrt[3]{x}+1 \\ y=\sqrt[3]{8}+1 \\ y=2+1 \\ y=3 \end{gathered}$

Graph of function is:

Domain of function :

$x\in\mathfrak{\Re }\colon x\ge0$

All possible real number use input for the function so the all real number is domain

Range of function:

$y\in\mathfrak{\Re }\colon y\ge1$