How do you find the domain and range of #f(x) = sqrt(4+x) / (1-x)#?

1 Answer
Aug 19, 2017

Answer:

Domain: #" "[-4, 1) uu (1, + oo)#

Range: #" " (-oo, + oo)#

Explanation:

The domain of the function represents all the values that #x# can take for which #f(x)# is defined.

Right from the start, you should be able to say that the domain of the function cannot include #x = 1# because that would make the function undefined.

#1 - x != 0 implies x != 1#

Moreover, notice that the function contains the square root of an expression that depends on the value of #x#. As you know, when working with real numbers, you cannot take the square root of a negative number.

This implies that you need

#4 + x >= 0 implies x >= - 4#

Therefore, you can say that the domain of the function will be

#x in [-4, 1) uu (1, +oo)#

THis tells you that the function is defined for any value of #x# that it greater than or equal to #-4# and is not equal to #1#.

The range of the function tells you all the possible values that #f(x)# can take for valid domain values.

In this case, the square root of a positive number will produce a positive number, which means that regardless what value of #x# you plug in from the domain of the function, you will have

#sqrt(4 -x ) >= 0 #

Now, for any value of #x in [-4, 1)#, you will have

#{( sqrt(4 + x) >= 0), (1 - x > 0) :} implies f(x) >= 0#

and for any value of #x in (1, +oo)#, you will have

#{( sqrt(4 + x) > 0), (1 - x < 0) :} implies f(x) < 0#

This means that the range of the function is

#(-oo, 0] uu (0, +oo) = (- oo, +oo)#

graph{sqrt(4+x)/(1-x) [-10, 10, -5, 5]}