1980 AHSME Problems/Problem 14
If the function is defined by satisfies for all real numbers except , then is
As , we can plug that into and simplify to get . However, we have a restriction on x such that if we have an undefined function. We can use this to our advantage. Plugging that value for x into yields , as the left hand side simplifies and the right hand side is simply the value we have chosen. This means that .
Alternatively, after simplifying the function to , multiply both sides by and divide by to yield . This can be factored to . This means that both and either one of or are equal to 0. yields and the other two yield . The clear solution is