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

Question

Eli Assouline · Accepted 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}#

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

Range: #{y in RR | y ne -7}# Our role is #f(x) = 2/(3/(x - 6)) - 7#. We can safely assume that the domain and range are the set of all real numbers because there are no radicals, which is where the imaginary number could reside. #{x | x in RR}#, #{y | y in RR}# However, we must never forget the fury of division by zero. There is a fraction in this function, and we need to ensure that the denominator cannot equal zero in order to prevent any divisions by zero. #2/(3/(x - 6))# where the numerator #3/(x - 6)# not equal zero #3/(x - 6) ne 0# But there is another denominator inside this one as well. #x - 6# which is likewise incompatible with 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# What about that larger denominator, though? #3/(x - 6) ne 0# We could attempt to multiply by the smaller denominator first, though. #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# I see, to start, we could write the fraction as simple division: #f_"modified"(x) = 2 -: (3/(x - 6)) - 7# Because division is the multiplicative inverse, the numerator and denominator should "flip": #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# Let's get simple now: #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 resolve: #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}#

Nathan Axel · Answer

undefined To find the domain and range of $ f(x) = \frac{2}{\frac{3}{x-6}} - 7 $:

**Domain:** The function is defined for all $ x $ values except where the denominator becomes zero. Therefore, the domain is all real numbers except $ x = 6 $.

**Range:** To find the range, consider the behavior of the function as $ x $ approaches positive and negative infinity. As $ x $ approaches positive or negative infinity, the fraction $ \frac{3}{x-6} $ approaches zero. Thus, $ f(x) $ approaches $ -7 $ as $ x $ approaches infinity. Therefore, the range of $ f(x) $ is all real numbers except $ -7 $.