Question

Let f be a continuous function on [0,1] with f(0)=f(1). (a) By considering g(x)=f(x)−f(x+ 1/2 ) for x∈[0, 1/2 ] show that there exist a,b∈[0,1] such that ∣a−b∣= 1/2 and f(a)=f(b)

Answer 1

To show that there exist a and b in the interval [0,1] such that |a-b| = 1/2 and f(a) = f(b), we can consider the function g(x) = f(x) - f(x + 1/2) for x in the interval [0,1/2]. By applying the intermediate value theorem, we can find two values a and b in [0,1/2] such that g(a) = g(b) = 0. For these values, f(a) = f(b) and |a - b| = 1/2.

Step-by-step explanation:

To show that there exist a and b in the interval [0,1] such that |a-b| = 1/2 and f(a) = f(b), we can consider the function g(x) = f(x) - f(x + 1/2) for x in the interval [0,1/2].

1. First, notice that g(0) = f(0) - f(1/2) and g(1/2) = f(1/2) - f(1).
2. Using the given information that f(0) = f(1), we can substitute these values into the expressions for g(0) and g(1/2) to get g(0) = g(1/2).
3. By applying the intermediate value theorem, set g(x) = 0 and find two values a and b in [0,1/2] such that g(a) = g(b) = 0.
4. For these values, we have f(a) - f(a + 1/2) = f(b) - f(b + 1/2) = 0, which implies f(a) = f(b).
5. Since |a - b| = 1/2, we have shown that there exist a and b in [0,1] such that |a-b| = 1/2 and f(a) = f(b).

Answer 2

To show that there exist a, b ∈ [0,1] such that |a - b| = 1/2 and f(a) = f(b), we can consider the function g(x) = f(x) - f(x + 1/2) for x ∈ [0, 1/2]. By the Intermediate Value Theorem, there exists some c ∈ (0, 1/2) such that g(c) = 0, which implies f(c) = f(c + 1/2).

Step-by-step explanation:

To show that there exist a, b ∈ [0,1] such that |a - b| = 1/2 and f(a) = f(b), we can consider the function g(x) = f(x) - f(x + 1/2) for x ∈ [0, 1/2].

Since f(0) = f(1), we have g(0) = f(0) - f(1/2) = f(0) - f(1) = g(1/2). Therefore, g(x) is continuous on [0, 1/2] and satisfies g(0) = g(1/2).

By the Intermediate Value Theorem, since g(x) is continuous and g(0) = g(1/2), there exists some c ∈ (0, 1/2) such that g(c) = 0. This means f(c) - f(c + 1/2) = 0, which implies f(c) = f(c + 1/2).