If \(f(x)=x^2-9\) and \(g(x)=x+3\), what is the simplified form of \(\left(\frac{f}{g}\right)(x)\), where \(x\neq-3\)?

Correct answer: (x-3)

The direct answer is option A: x−3, with the original restriction x≠−3. The quotient of functions is \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x^2-9}{x+3}\). Factor the numerator as a difference of squares: x^2−9=(x−3)(x+3). Thus \(\frac{(x-3)(x+3)}{x+3}=x-3\), but only when x+3≠0, so x≠−3 remains essential. Option A is correct. Option B, x+3, would result from an incorrect factor cancellation or sign error. Option C, x^2−3, does not come from factoring and simplifying the quotient. Option D, x−9, confuses the constant 9 with the factor x−3; it is not algebraically equivalent. A subtle point is that the simplified rule x−3 cannot be used to say the original quotient is defined at x=−3: the original denominator is zero there. Always carry the excluded value after cancellation. The memory cue is difference of squares: \(x^2-a^2=(x-a)(x+a)\).