If the domain of the function f(x) = ⁻¹ ( x^2 - ax + b x^2 - 25 ) is [- 3 2 , ) - 5 , then the value of a + b is

Step-by-step solution

For the domain to extend to infinity without a vertical asymptote at x = 5 , the numerator must have (x - 5) as a factor. Substituting x = 5 into x^2 - ax + b = 0 yields 25 - 5a + b = 0 . Let the roots of the numerator be 5 and k . Then x^2 - ax + b = (x - 5)(x - k) , which gives a = k + 5 and b = 5k . The function simplifies to f(x) = ⁻¹ ( x - k x + 5 ) for x 5 . For the inverse cosine to be defined, we must have -1 x - k x + 5 1 . Solving x - k x + 5 1 -(k + 5) x + 5 0 . Since the domain extends to infinity, we m