What is the domain and range of #f(x) = 1/(2x+4)#?

1 Answer
Aug 16, 2015

Answer:

Domain: #(-oo, -2) uu (-2, + oo)#
Range: #(-oo, 0) uu (0, + oo)#

Explanation:

First, notice that you can rewrite your function as

#f(x) = 1/(2 * (x + 2))#

This function is defined for any value of #x in RR# except the value that would make the denominator equal to zero.

More specifically, you need to exclude from the domain of the function the value of #x# that would make

#x + 2 = 0 implies x = -2#

Therefore, the domain of the function will be #RR - {-2}#, or #(-oo, -2) uu (-2, + oo)#.

Notice that since you're dealing with a fraction that has a constant numerator, the function has no way of ever being equal to zero.

#f(x) !=0", "(AA)x in RR - {-2}#

The range of the function will thus be #RR - {0}#, or #(-oo, 0) uu (0, + oo)#.

graph{1/(2x + 4) [-6.243, 6.243, -3.12, 3.123]}