Question

I challenge the world to solve this:

Problem: Let f(x) be a continuous function defined on the closed interval [0, 1] such that f(0) = f(1) = 0. Prove that there exists a point c in the interval (0, 1) such that f(c) = f(c + 1/2).

Answer 1

Explanation:

Here is one possible way to solve it:

First, we define a new function g(x) = f(x) - f(x + 1/2) for x in [0, 1/2]. Note that g is also continuous on [0, 1/2], since f is continuous on [0, 1].

Next, we observe that g(0) = f(0) - f(1/2) = -f(1/2) and g(1/2) = f(1/2) - f(1) = f(1/2). Therefore, g(0) and g(1/2) have opposite signs, unless f(1/2) = 0, in which case we are done.

Then, we apply the intermediate value theorem to g on the interval [0, 1/2]. Since g is continuous and g(0) and g(1/2) have opposite signs, there exists a point c in (0, 1/2) such that g© = 0.

Finally, we conclude that g© = 0 implies that f© = f(c + 1/2), and c is the point we are looking for.

Answer 2

Shown below

Explanation:

Proof:

Since f(x) is continuous on the closed interval [0, 1], it follows from the Intermediate Value Theorem that for any two values y and z, where y < z, there exists a point c in the open interval (y, z) such that f(c) = y.

Let y = f(0) = 0 and z = f(1) = 0. Then, by the Intermediate Value Theorem, there exists a point c in the open interval (0, 1) such that f(c) = 0.

Now, let g(x) = f(x) - f(c). Then, g(x) is also a continuous function on the closed interval [0, 1].

Moreover, g(0) = g(1) = 0.

Again, by the Intermediate Value Theorem, there exists a point c' in the open interval (0, 1) such that g(c') = 0.

Therefore, f(c') = f(c).

But c' is in the open interval (0, 1), so c' is also in the interval
`\sf \left(0, (1)/(2)\right).`

Therefore, there exists a point c in the interval (0, 1) such that:

`\sf f(c) = f\left(c + (1)/(2)\right).`