Stating domain and range from functions?

[Mellic]

i just cannot seem to wrap my head in this.
im supposed to take a function like
y = 4x-6.2
and state its domain and range which from what my notes tell me is supposed to look like
D = {X|XER}
R = {Y|YER}
but i cant seem to find a way to start or solve it , anyone got an ideal? thanks for any help.

 

[JeffM]

i just cannot seem to wrap my head in this.
im supposed to take a function like
y = 4x-6.2
and state its domain and range which from what my notes tell me is supposed to look like
D = {X|XER}
R = {Y|YER}
but i cant seem to find a way to start or solve it , anyone got an ideal? thanks for any help.

First, the implicit assumption is that, unless otherwise specified, we're talking about real numbers.

Are there any real values of x for which the function is not a real number? Those values are not in the domain; all other values are.

Are there any real values that y cannot have? Those values are not in the range; all other values are.

Example 1: $\displaystyle y = \sqrt{x} - 1.$

If x is negative, the square root of x is not defined in the real numbers. The domain is all non-negative numbers. Because the square root of x is not less than zero, y cannot be less than - 1. The range is all real numbers greater than or equal to -1.

Example 2: $\displaystyle y = \left | \dfrac{1}{x^2 + 5x + 6} \right | + 2.$

Here x cannot equal x = - 2 or - 3, which would make the denominator zero and render the function undefined in the real numbers. So the domain is all real numbers except -2 and -3. Obviously, y cannot equal 2, nor can it be less than 2. So the range is all real numbers greater than 2.

 

[JeffM]

Oh, I really should give a third example.

$\displaystyle 0 < x < 1 \implies y = \dfrac{1}{x}.$

in this case, the function is not defined outside (0, 1). So the domain is all real numbers in the interval
(0, 1). Furthermore, y cannot be less than or equal to 1 so the range is all real numbers greater than 1.