Question

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

Answer 1

range is -∞<y<∞

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

`(-\infty,+\infty)`

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.