How do you find the domain of #(x^2-x-12)^-4#?

1 Answer
George C.
May 18, 2015

#(x^2-x-12)^-4# can be written as #1/(x^2-x-12)^4#

This will be undefined if #(x^2-x-12) = 0#, but defined for all other values of #x# in #RR#.

#x^2-x-12 = (x-4)(x+3)# is zero when #x=4# or #x=-3#, so these are the only prohibited values of #x# and the domain of the function is:

#RR \\ {-3,4}#

that is #{x in RR: x != -3 ^^ x != 4}#

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