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. Here's how we can proceed:
First, let's define a new function g(x) = f(x) - x. We want to prove that g(x) = 0 for all -6 ≤ x ≤ 6.
Now, according to the Mean Value Theorem, if a function is differentiable on a closed interval [a, b] and continuous on the open interval (a, b), then there exists at least one c in (a, b) such that f'(c) = (f(b) - f(a))/(b - a).
Let's consider the interval [-6, 6]. We know that f(x) is differentiable for all real values of x and -1 ≤ f'(x) ≤ 1 for all x. Therefore, g(x) is also differentiable for all real values of x.
Now, let's consider g(-6) and g(6):
g(-6) = f(-6) - (-6) = -6 + 6 = 0
g(6) = f(6) - 6 = 6 - 6 = 0
So, we have g(-6) = g(6) = 0, which means that g(x) takes the same value at both endpoints of the interval [-6, 6].
Since g(x) is differentiable, continuous, and takes the same value at both endpoints, we can apply the Mean Value Theorem. Therefore, there exists at least one c in (-6, 6) such that g'(c) = (g(6) - g(-6))/(6 - (-6)) = 0.
But g'(x) = f'(x) - 1 (since g(x) = f(x) - x). So, at c, we have f'(c) - 1 = 0, which implies f'(c) = 1.
However, we know that -1 ≤ f'(x) ≤ 1 for all x. Therefore, f'(c) can only be equal to 1 if f'(x) is identically equal to 1 for all x in (-6, 6). In other words, the derivative of f(x) is constantly 1 within this interval.
Integrating the derivative f'(x) = 1 over (-6, 6), we get f(x) = x + C, where C is a constant.
Now, we can use the given information: f(-6) = -6 and f(6) = 6:
f(-6) = -6 = -6 + C => C = 0
f(6) = 6 = 6 + C => C = 0
So, the constant C is 0, leading to f(x) = x.
Therefore, f(x) = x holds for all -6 ≤ x ≤ 6.