Final answer:
The function f(x) has at least one x-intercept (a zero of the function) by the Intermediate Value Theorem, and at least one critical point (where the derivative is zero) by the Mean Value Theorem.
Step-by-step explanation:
The question asks about the number of solutions for a function f(x) that is continuous on the interval [2, 4] and differentiable on (2, 4) with given values of the function at points 2, 3, and 4. First, we consider the continuity and differentiability of the function which implies, by the Intermediate Value Theorem, that since f(x) takes on both positive and negative values within the interval, it must have crossed the x-axis at least once, guaranteeing at least one root.
Furthermore, the Mean Value Theorem tells us that because the function is differentiable on (2, 4), there exists a number c in (2, 4) such that f'(c) equals the average rate of change of the function over that interval. Since f(2) = f(4), but f(3) is different, there must be at least one point where the derivative is 0, indicating at least one relative extremum.
Therefore, without knowing the specific function, we are guaranteed at least one x-intercept (a zero of the function) by the Intermediate Value Theorem, and at least one critical point (where the derivative is zero) by the Mean Value Theorem.