Proving Intermediate Value Theorem by contradiction - not understanding

Strat

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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.
 
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.
The function is continuous between 2 and 4 and f(2)=8 and f(4)=6. Now suppose f(3) were less than 6 [it can't be six since x=1 and x=4 are the only places it can be six]. Since the function is continuous it must take on all values between 8 and f(3) in the interval (2,4) [note the open interval]. By our assumption, 6 is between 8 and f(3). But this can't be since the only place f(x)=6 is x=1 and x=4. By contradiction, f(3) can't be less that 6.

I hope that helps.
 
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.
The point of the IVT is that, for a function continuous on [a, b], if you know the values of f(a) and f(b), then you know that f(x) must take on every value between f(a) and f(b). This is often used to narrow down zeroes. For instance, if you know that a function g(x) is continuous everywhere, and that g(a) < 0 and g(b) > 0, then g(c) = 0 for some x = c, because 0 is between positive and negative.

The IVT can also be used to rope off areas. This is what your exercise is doing.

Problem: Suppose f is continuous on [1, 5] and the only solutions of the equation f(x) = 6 are x = 1 and x = 4. If f(2) = 8, explain why f(3) > 6.
The "solutions" to f(x) = 6 are the zeroes of the function g(x) = f(x) - 6. So this is saying that g(x) is continuous on [1, 5], and that g(1) = g(4) = 0. Also, it is saying that x = 1 and x = 4 are the only zeroes on this interval. So g(x) is either positive (above the x-axis) or negative (below the x-axis) everywhere in-between.

Then you are given that g(2) = 8. This means that g(x) is positive on the entire interval. And if g(x) > 0, then f(x) - 6 > 0, so f(x) > 6. In particular, f(3) must be greater than six.

Hope that helps! ;)
 
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.
No, that "for a given x there can be only one f(x)" follows from the definition of "function". It is true whether the function is continuous or not.

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. If f(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.
 
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