How do you find the domain and range of #f(x)=2/(3/(x-6))-7#?

1 Answer
Dec 28, 2017

Answer:

Starting with a domain and range of the set of all real numbers, exclude domain values for which division by zero may occur, and exclude range values that would result from the forbidden domains, if no other domain maps to said range values. This yields:

Domain: #{x in RR | x ne 6}#

Range: #{y in RR | y ne -7}#

Explanation:

Our function is

#f(x) = 2/(3/(x - 6)) - 7#.

Since there are no radicals (where the imaginary number may reside), we could safely assume that the domain and range is the set of all real numbers...

#{x | x in RR}#, #{y | y in RR}#

... But we need to remember the wrath of division by zero.

To prevent any divisions by zero, we need to make sure that the denominator cannot equal zero. In this function, there is a fraction

#2/(3/(x - 6))#

where the denominator

#3/(x - 6)#

cannot equal zero:

#3/(x - 6) ne 0#

However, inside this denominator, there is also a denominator

#x - 6#

which also cannot equal zero:

#x - 6 ne 0#.

To figure out what the forbidden numbers are, we could solve for #x# as if it's an equal sign. In this case, add both sides by #6#:

#x - 6 + 6 ne 0 + 6#

#x ne 6#

But how about that bigger denominator?

#3/(x - 6) ne 0#

Well, we could first try to multiply by the smaller denominator...?

#3/(x - 6) * (x - 6) ne 0 * (x - 6)#

But wait! We need to make sure #(x - 6)# isn't zero! That means assuming #x - 6 ne 0# and therefore #x ne 6# as we have solved earlier. Simplfying out:

#3 ne 0#

Well, of course #3 ne 0#, so that's pretty much all the forbidden numbers. It's just #x ne 6#. So, the domain should be all real numbers except #6#:

Domain: #{x in RR | x ne 6}#

And the range? Is it all real numbers too? Take a moment to think about it. If #6# wasn't a forbidden number, what would its output be? What if we try to simplify the function?

#f(x) = 2/(3/(x - 6)) - 7#

Hmm... we could start by rewriting the fraction as plain division:

#f_"modified"(x) = 2 -: (3/(x - 6)) - 7#

Division should "flip" the numerator and denominator (due to being the multiplicative inverse):

#f_"modified"(x) = 2 * ((x - 6)/3) - 7#

We could multiply the #2# over:

#f_"modified"(x) = (2(x - 6))/3 - 7#

#f_"modified"(x) = (2x - 12)/3 - 7#

And "split" the #3#:

#f_"modified"(x) = (2x)/3 - 12/3 - 7#

Leaving us to simplify:

#f_"modified"(x) = 2/3 x - 4 - 7#

#f_"modified"(x) = 2/3 x - 11#

Nice, this is a linear function! Since its graph is a line, each range should also be associated to one and only one domain. With no divisions by #x#, we could input #6# in:

#f_"modified"(x) = 2/3 (6) - 11#

And solve:

#f_"modified"(x) = (2 * 6)/3 - 11#

#f_"modified"(x) = 12/3 - 11#

#f_"modified"(x) = 4 - 11#

#f_"modified"(6) = -7#

Ah! So that value, #-7#, could not possibly be in the original function's range, due to it being a linear function. Therefore,

Domain: #{x in RR | x ne 6}#

Range: #{y in RR | y ne -7}#