Find all possible functions \(f\) such that \(f(x) = ax + b\) and \(f\left(f\left(f(x)\right)\right) = bx - a^2\) with \(a \neq 0\), \(b \neq 0\), \(a\), \(b \in \mathbb{R}\).

Solution

First, using the appropriate substitutions, determine the function \(f\left(f\left(f(x)\right)\right)\) given that \(f(x) = ax + b\):

Find all possible functions \(f\) such that \(f(x) = ax + b\) and \(f\left(f\left(f(x)\right)\right) = bx - a^2\) with \(a \neq 0\), \(b \neq 0\), \(a\), \(b \in \mathbb{R}\).

Solution

We have \({\color{NavyBlue}f\left(f\left(f(x)\right)\right) = a^{3}x + a^{2}b + ab + b}\) and \({\color{Mulberry}f\left(f\left(f(x)\right)\right) = bx - a^{2}}\), \(a\), \(b \neq 0\), \(a\), \(b \in \mathbb{R}\), so

Equating coefficients, we have

Substituting \((1)\) into \((2)\) and simplifying,

This polynomial equation can be solved by factoring.

Find all possible functions \(f\) such that \(f(x) = ax + b\) and \(f\left(f\left(f(x)\right)\right) = bx - a^2\) with \(a \neq 0\), \(b \neq 0\), \(a\), \(b \in \mathbb{R}\).

Solution

Using the factor theorem, if \(P(a) = a^{3} + a^{2} + a + 1\) and \(P(-1) = 0\), then \((a + 1)\) is a factor of \(P(a)\). So,

Expanding \((a + 1)(a^2 + 1)\) gives \(a^{3} + a^{2} + a + 1\), so

Alternatively, we can factor by grouping.

Find all possible functions \(f\) such that \(f(x) = ax + b\) and \(f\left(f\left(f(x)\right)\right) = bx - a^2\) with \(a \neq 0\), \(b \neq 0\), \(a\), \(b \in \mathbb{R}\).

Solution

The solutions to the equation

are \(a = 0\) or \(a = -1\). Note that \(a^2 + 1 \neq 0\) for \(a \in \mathbb{R}\).

However, \(a \neq 0\), so \(a = -1\).

Substituting \(a = -1\) into \((1)\), we get

Therefore, the only possible function \(f\), which satisfies the conditions, is \(f(x) = -x - 1\).

• The composition of functions applies one function to the result of another function. For two functions \(f\) and \(g\), applying \(f\) to the output of \(g\) results in the composite function defined by \(\left(f \circ g\right)(x) = f\left(g(x)\right)\).
• To determine \(\left(f \circ g\right)(x)\), substitute the expression for \(g(x)\) into \(f(x)\), for every occurrence of \(x\) in the expression for \(f(x)\).
• To determine \(\left(f \circ g\right)(a)\), where \(a \in \mathbb{R}\), substitute the value of \(g(a)\) into \(f(x)\) for \(x\), or substitute \(x = a\) into the expression for \(\left(f \circ g\right)(x)\).
• The domain of \(\left(f \circ g\right)(x)\) is the set of all values, \(x\), in the domain of \(g\) that produce an output, \(g(x)\), that lies in the domain of \(f\).
• Different approaches can be used to graph composite functions depending on the functions involved.
• The decomposition of a function involves identifying two or more (often simpler) functions such that the composition of these functions results in the given function.
• The composition of functions is generally not commutative. For most functions \(f\) and \(g\),\[\left(f \circ g\right)(x) \neq \left(g \circ f\right)(x)\]
• For any function \(f\) that has an inverse function \(f^{-1}\),\[f\left(f^{-1}(x)\right) = f^{-1}\left(f(x)\right) = x\] for all values of \(x\) in the domain of \(f\) and \(f^{-1}\).