Let f be a function such that f(−6)=−6, f(6)=6, f is differentiable for all real values of x and −1≤f'x≤1 for all real values of x. Prove that f(x)=x for all −6≤x≤6
I tried applying the mean value theorem here but I'm not sure where to start.
To prove that f(x) = x for all -6 ≤ x ≤ 6, we can use the mean value theorem.
The mean value theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in (a, b) where the derivative of f at c is equal to the average rate of change of f over the interval [a, b].
First, let's consider the function F(x) = f(x) - x. We want to show that F(x) = 0 for all x between -6 and 6.
Since F(-6) = f(-6) - (-6) = -6 + 6 = 0, and F(6) = f(6) - 6 = 6 - 6 = 0, we have F(-6) = F(6) = 0.
Now, we need to show that F'(x) = 0 for all x between -6 and 6 to conclude that F(x) = 0.
We know that -1 ≤ f'(x) ≤ 1 for all real values of x. Since f is differentiable for all real values of x, this means that f'(x) is continuous on the closed interval [-6, 6] and differentiable on the open interval (-6, 6).
By the mean value theorem, there exists a point c in (-6, 6) such that F'(c) = (f(c) - c)' = (f(c))' - 1 = f'(c) - 1 = 0.
Since -1 ≤ f'(x) ≤ 1, and we found a point c where F'(c) = 0, this means that F(x) = f(x) - x = 0 for all x between -6 and 6.
Thus, we have proved that f(x) = x for all -6 ≤ x ≤ 6.
To prove that f(x) = x for all -6 ≤ x ≤ 6, we can indeed make use of the Mean Value Theorem. Here's how you can approach it:
1. Define a new function g(x) = f(x) - x. We want to prove that g(x) = 0 for all -6 ≤ x ≤ 6.
2. Since g(x) is differentiable for all real values of x (as f(x) is differentiable), we can apply the Mean Value Theorem to g(x) on the interval [-6, 6].
3. According to the Mean Value Theorem, there exists a value c in the open interval (-6, 6) such that g'(c) = (g(6) - g(-6))/(6 - (-6)).
4. Since g(x) = f(x) - x, we have g'(c) = (f'(c) - 1).
5. Given that -1 ≤ f'(x) ≤ 1 for all real values of x, it follows that -2 ≤ (f'(x) - 1) ≤ 0 for all real values of x.
6. Consequently, we have -2 ≤ g'(c) ≤ 0, which implies that -2 ≤ (f'(c) - 1) ≤ 0.
7. Since g'(c) = (f'(c) - 1) and -2 ≤ (f'(c) - 1) ≤ 0, it follows that -2 ≤ g'(c) ≤ 0.
8. By the Mean Value Theorem, this means that (g(6) - g(-6))/(6 - (-6)) = (f(6) - f(-6))/(6 - (-6)) = (6 - (-6))/(6 - (-6)) = 1.
9. It implies that g(6) - g(-6) = 6 - (-6).
10. Simplifying, we get g(6) - g(-6) = 12.
11. However, g(6) - g(-6) = f(6) - 6 - (f(-6) - (-6)) = 6 - 6 - (-6 - 6) = 12 - 12 = 0.
12. Therefore, we have g(6) - g(-6) = 0.
13. Combining this result with Step 10, we have 12 = 0, which is a contradiction.
14. Since we have reached a contradiction, our initial assumption that g(x) = f(x) - x is incorrect.
15. Thus, we conclude that g(x) = 0 for all -6 ≤ x ≤ 6, which means f(x) = x for all -6 ≤ x ≤ 6.