find the x-values (if any) at which f is not continuous. which of the discontinuities are removable? f(x)=(x-3)/(x^2-9)           

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Factor the denominator into (x+3) and (x-3).  Any value of x that makes the denominator = 0 is not allowed so these values are x = 3 and x = -3.  Since (x-3) is in the numerator also it could be removed.