What is the domain of #f(x)=log_3# #(9-x^2)/(x-3)#?

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Answer 1 · 2017-09-17T15:29:51

Domain: #{x| x < -3, x in RR}#

We have:

#f(x) = log_3 -(x^2 - 9)/(x - 3)#

#f(x) = log_3 -((x +3)(x -3))/(x -3)#

#f(x) = log_3 -(x + 3)#

It follows that

#f(x) = log_3 (-x - 3)#

We need the part in parentheses to be greater than #0#.

#-x - 3 > 0#

#-x > 3#

#x < - 3#

The graph of the function confirms.

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There is a vertical asymptote at #x= -3#.

Hopefully this helps!