How do you find the domain and range and determine whether the relation is a function given :x=2y^2-3?

1 Answer
Sep 20, 2017

See below.

Explanation:

For y as a function of x.

x = 2y^2 -3

Rearrange;

Add 3 to both sides:

2y^2 = x + 3

Divide by 2:

y^2 = 1/2x + 3/2 => y^2 = (x + 3)/2

Taking roots of both sides:

y = sqrt((x + 3)/2)

For positive root:

For the domain we must make sure the numerator is >= 0, since we are only interested in real numbers.

The numerator is >=0 for x >= -3

So the domain is { x in RR | x >= -3 }

The range for the positive root is:

As
x -> +oo
y -> +oo

So the range is:

{y in RR | y >= 0 }

For negative root the domain is the same for real numbers. The range is in the reverse direction:

{ y in RR | y <= 0 }

The determination of a function.

For positive root only a function is defined.
For negative root only a function is defined.

For +-sqrt((3 + x )/2 )

This is not a function. A function is only defined if an element in a domain is mapped onto one and one only element in a range. In the case of both roots this would map one element in the domain to two elements in the range. This is sometimes called a one to many relationship.

See graph:enter image source here