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

Answer:

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