For the function f(x), the real number a where f(a)=a is called the fixed point of the function f(x). Show that when f(x) is continuous and differentiable for all real numbers x, if f’(x)≠1, then f(x) has at most one fixed point.

To show that when f(x) is continuous and differentiable for all real numbers x, if f'(x) \neq 1, then f(x) has at most one fixed point, we can use the Mean Value Theorem.

Let's assume that f(x) has two fixed points, a and b, where a \neq b. This means that f(a) = a and f(b) = b.

By the Mean Value Theorem, there exists a value c between a and b such that:

f'(c) = \frac{f(b) - f(a)}{b - a}

Since f(a) = a and f(b) = b, we have:

f'(c) = \frac{b - a}{b - a} = 1

This contradicts the given condition that f'(x) \neq 1.

Therefore, our assumption that f(x) has two distinct fixed points is incorrect. Hence, the function f(x) has at most one fixed point.

\boxed{\text{Answer}}: the function f(x) has at most one fixed point.