Does the Function f(x) Have a Fixed Point If f'(x) ≥ 2?

[yayMath]

Hi,
I need to prove that f(x) has a fixed point, given that f'(x) >= 2 for all x.

my problem is that I've reached the part in which g(x) = f(x) - x
and g'(x) = f'(x) - 1 and therefore g'(x) >= 1
but now I'm completely stuck. I knwo that I need to use the mean value theorem, but i just don't know where.
thanx.

 

[Tide]

If f '(x) >= 2 for all x then its graph would have to cross the graph of y = x at some point.

 

[M98Ranger]

Firstly, let's define what a fixed point of a function is. A fixed point of a function f(x) is a value x0 such that f(x0) = x0. In other words, when we plug in x0 into the function, the output is equal to the input.

Now, let's consider the function g(x) = f(x) - x. We know that g'(x) = f'(x) - 1 and since we are given that f'(x) >= 2 for all x, we can conclude that g'(x) >= 1 for all x. This means that g(x) is a strictly increasing function.

Next, we can use the mean value theorem to show that g(x) has a fixed point. The mean value theorem states that for a continuous and differentiable function f(x) on an interval [a,b], there exists a point c in the interval such that f'(c) = (f(b) - f(a))/(b-a).

In our case, since g(x) is a strictly increasing function, it is also continuous and differentiable on any interval. Therefore, we can apply the mean value theorem to g(x) on any interval [a,b]. This means that there exists a point c in the interval [a,b] such that g'(c) = (g(b) - g(a))/(b-a).

Since we know that g'(x) >= 1 for all x, we can say that g(b) - g(a) >= b-a for any interval [a,b]. This means that g(b) >= g(a) + b-a.

Now, let's consider the interval [a,a+1]. Since g(a+1) >= g(a) + (a+1)-a = g(a) + 1, we can say that g(a+1) >= g(a) + 1.

Similarly, we can consider the interval [a+1,a+2] and apply the same logic to get g(a+2) >= g(a+1) + 1.

Continuing this process, we can see that g(a+n) >= g(a+n-1) + 1 for all n > 0.

Now, let's consider the sequence {g(a+n)}. This sequence is strictly increasing and since g(x) is continuous, we know that g(a+n) approaches a limit