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
