Question

What is the range of the function y=3 Sqrt x+8

Answer 1

Answer:

range is -∞<y<∞

Step-by-step explanation:

$y=\sqrt[3]{x+8}$

Range of the given function is same as the domain of the inverse function

LEts find the inverse for the given equation

Swap the variables x and y . then solve for y

$y=\sqrt[3]{x+8}$

$x=\sqrt[3]{y+8}$

Take cube on both sides

$x^3= y+8$

Subtract 8 from both sides

$y=x^3-8$

Inverse function is a cubic function. For cubic function the domain is set of allr eal numbers. -∞<x<∞

Range of the given function is same as the domain of the inverse function

So range is -∞<y<∞

Answer 2

Answer:

$(-\infty,+\infty)$

Step-by-step explanation:

The given function is

$y=\sqrt[3]{x+8}$

The range refers to the values of y for which  x is defined.

We solve for x to obtain;

$y^3=x+8$

$x=y^3-8$

This is a polynomial function in y.

x is defined for all values of y.

Therefore the range is all real numbers.

or

$(-\infty,+\infty)$

The first choice is correct.