Can someone help rephrase this? Even with the solution I don't understand. I understand the basic concept of IVT, but I don't get how the solution is contradictory. I would have said something along the lines of: for a given x on a continuous line/curve, there can be only one f(x), otherwise it is not continuous. I don't understand this contradiction stuff.
Problem: Suppose f is continuous on [1, 5] and the only solutions of the equation f(x) = 6 are x = 1 and x = 4. Iff(2) = 8, explain why f(3) > 6.
Solution: Perform a proof by contradiction. Assume that α ∈ (2, 4) such that f(α) ≤ 6. Then by theIntermediate Value Theorem (letting a = 2, b = α, and N = 6) there exists a c such that f(c) = 6. Howeverthis violates the definition of f(x) since c , 1 and c , 4. Therefore the initial assumption that α is such thatf(α) < 6 is false. Therefore f(α) > 6. Since 3 is a viable choice for α, f(3) > 6.
Problem: Suppose f is continuous on [1, 5] and the only solutions of the equation f(x) = 6 are x = 1 and x = 4. Iff(2) = 8, explain why f(3) > 6.
Solution: Perform a proof by contradiction. Assume that α ∈ (2, 4) such that f(α) ≤ 6. Then by theIntermediate Value Theorem (letting a = 2, b = α, and N = 6) there exists a c such that f(c) = 6. Howeverthis violates the definition of f(x) since c , 1 and c , 4. Therefore the initial assumption that α is such thatf(α) < 6 is false. Therefore f(α) > 6. Since 3 is a viable choice for α, f(3) > 6.