Question

Fixed Points A number \(a\) is called a fixed point of a function \(f\) if \(f\left( a \right) = a\). Prove that if \(f'\left( x \right) \ne 1\) for all real numbers \(x\), then \(f\) has at most one fixed point.

Short answer

It is proved that if \(f'\left( x \right) \ne 1\) for all real numbers \(x\), then \(f\) has at most one fixed point.

Step-by-step solution

The mean value theorem

Let \(f\) be a function that is continuous on the closed interval \(\left( {a,b} \right)\) and differentiable on the open interval \(\left( {a,b} \right)\). Then there is a number \(c \in \left( {a,b} \right)\) such as that \(f'\left( c \right) = \frac{{f\left( b \right) - f\left( b \right)}}{{b - a}}\).

Proof the statement

Let \(f'\left( x \right) \ne 1\) for all real numbers \(x\). So, \(f'\left( x \right)\) is defined for all real numbers and differentiable on open interval on \(\mathbb{R}\). If \(f\left( x \right)\) is differentiable on some open interval then it also continuous on closed interval as well.

Let there are two fixed points \(f\left( a \right) = a\) and \(f\left( b \right) = b\).

So, by the Mean Value Theorem, there exists a point \(c \in \left( {a,b} \right)\) such that

\(\begin{aligned}{c}f'\left( c \right) &= \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\\f'\left( c \right) &= \frac{{b - a}}{{b - a}}\\f'\left( c \right) &= 1\end{aligned}\)

So, when \(f\left( x \right)\) has more than one fixed point then \(f'\left( c \right) = 1\).

So, if \(f'\left( x \right) \ne 1\) then \(f\left( x \right)\) has at most one fixed point.