Question #N993

If $f(x) = \frac{x^2-1}{x+1}$ for $x \ne -1$, what is the value of $f(-3)$?
A. -4
B. -2
C. 2
D. 4

Correct Answer is: D

Substituting -3 for x in the function, we get: $f(-3) = \frac{(-3)^2-1}{-3+1}$. Simplifying the expression, we get $f(-3) = \frac{9-1}{-2} = \frac{8}{-2} = -4$. However, since the function is undefined for x = -1, this value is not a valid solution. Therefore, we need to factor the expression, simplify, and then substitute x = -3. Factoring the expression, we get: $f(x) = \frac{(x+1)(x-1)}{x+1}$. Since $x \ne -1$, we can cancel out the common factor of (x+1), leaving us with $f(x) = x-1$. Substituting -3 for x in this simplified expression, we get $f(-3) = -3 - 1 = -4$.