A function \(f(x)\) is continuous in the interval [0,2]. It is known that…

Answer: There exists a point y in (0,1) such that f(y) = f(y+1).

Reasoning:

Define the continuous function h(x) = f(x) - f(x+1) on the closed interval [0,1].

• Compute h(0) = f(0) - f(1) = -1 - 1 = -2.

• Compute h(1) = f(1) - f(2) = 1 - (-1) = 2.

Since h is continuous on [0,1] and h(0) and h(1) have opposite signs, the Intermediate Value Theorem guarantees some y in (0,1) with h(y) = 0. Therefore f(y) = f(y+1).

Why the other statements need not be true (brief counterexamples):

1. Statement claiming f(y) = f(2-y) for every y in (0,1) is too strong. For instance, define a continuous piecewise function by f(x) = -1 + 2x + 0.5 x(1-x) on [0,1] and f(x) = 3 - 2x on [1,2]. Then f(0.5) = 0.125 while f(1.5) = 0, so the equality fails.

2. Statement that the maximum on (0,2) is 1 is not guaranteed. Example: take f(x) = -1 + 2x on [0,1], f(x) = 1 + 2(x-1) on [1,1.5] (so f(1.5) = 2), and f(x) = 2 - 6(x-1.5) on [1.5,2] (so f(2) = -1). This continuous function attains value 2 in (0,2).

3. Statement asserting existence of y with f(y) = -f(2-y) need not hold. Using the first piecewise example above, a zero of f in (0,1) occurs near x ≈ 0.4385 but f(2 - 0.4385) ≈ -0.123, so equality does not hold at that point; thus such a y is not guaranteed by the given conditions.

Conclusion: The only statement that must be true from the given information is the existence of y in (0,1) with f(y) = f(y+1), established by applying the Intermediate Value Theorem to h(x) = f(x) - f(x+1).