Final answer:
For the function f(x) = x² + ax + b to be differentiable at x = 3, it must be continuous and its derivative must exist at that point. Since f(x) is a polynomial, it is naturally continuous everywhere, and its derivative, f'(x) = 2x + a, also a polynomial, is continuous everywhere, including at x = 3. Thus, f(x) is differentiable at x = 3 for any real values of a and b.
Step-by-step explanation:
To determine the values of a and b that make the function f(x) = x2 + ax + b differentiable at x = 3, we must ensure that both the function is continuous at x = 3 and its derivative is continuous at x = 3. The function will be continuous if it's defined at that point and the limit as x approaches 3 from both sides is equal to the function's value at x = 3.
First, the function is a polynomial, and polynomials are continuous everywhere, so f will be continuous at x = 3 for any value of a and b.
Secondly, for the derivative to be continuous at x = 3, we need to take the derivative of f(x) and ensure it exists at x = 3. The derivative of f(x) = x2 + ax + b is f'(x) = 2x + a. For f'(x) to exist at x = 3, we don't need to find a specific value because the derivative of a polynomial is also a polynomial, and hence it is continuous everywhere including at x =3.
Therefore, f(x) is differentiable at x = 3 for any real numbers a and b because its derivative exists and it's continuous at that point.