Give the function of $f:\mathbb{R} \to \mathbb{R}$ and $f(x) = x^2 - 4x$, determine: $(f\circ f)(x)$, and evaluate $(f\circ f)^{-1}(\{0\})$

Question

Given the function $f: \mathbb{R} \to \mathbb{R}$ and $f(x) = x^2 - 4x$, determine: $(f\circ f)(x)$, and evaluate $(f\circ f)^{-1}(\{0\})$.

I'm perplexed with this question because when I interchange the $x$ and $y$ parameters, it gives me a function that I can not isolate the $x$ in one side of the equation. Can anyone give me a suggestion or answer for this?

Answer 1

We can picture the composition of the function $f$ with itself as a diagram: \begin{array}{ccccc} \color{green}{\mathbb{R}} &\xrightarrow{f} &\color{teal}{\mathbb{R}} &\xrightarrow{f} &\color{blue}{\mathbb{R}} \\ \color{green}{x} &\mapsto &\color{teal}{y} &\mapsto &\color{blue}{z} \end{array} which says that for any $\color{green}{x}$, we name $\color{teal}{y} = f(\color{green}{x})$ and $\color{blue}{z} = f(\color{teal}{y})$. If the function is for transportation then the composition is the express route: $\color{blue}{z} = (f \circ f)(\color{green}{x})$ \begin{array}{ccc} \color{green}{\mathbb{R}} &\xrightarrow{\hskip1.25em f \,\circ\, f \hskip1.25em} &\color{blue}{\mathbb{R}} \\ \color{green}{x} &\mapsto &\color{blue}{z} \end{array}

Putting them together is straightforward: \begin{align} \color{blue}{z} &= \color{teal}{y}^2 - 4\color{teal}{y} \\ &= \color{teal}{(}\color{green}{x}^2 - 4\color{teal}{)}^2 - 4\color{teal}{(}\color{green}{x}^2 - 4\color{teal}{)} \\ &= \color{green}{x}^4 - 12\color{green}{x}^2 + 32 \end{align} so $(f \circ f)(x) = x^4 - 12x^2 + 32$ is a quartic polynomial, as expected since $\deg(f) = 2$ and $2 \cdot 2 = 4$. This means that for any $z$, solving $(f \circ f)(x) = z$ for $x$ will yield at most $4$ solutions.

We're concerned with the task of finding preimages of the particular value $z = 0$. But this isn't just any old quartic equation (difficult), we can use the structure of $f \circ f$ as a composition to our advantage.

It's a two-step process: First find all $y$ such that $f(y) = 0$. This is a quadratic equation and it turns out there are two them, say $y_1$ and $y_2$. Then find all $x$ such that either $f(x) = y_1$ or $f(x) = y_2$. Each of these quadratic equations will yield two solutions, so there are four preimages $x$ in total.

\begin{align} y^2 - 4y &= 0 \\ y \, (y - 4) &= 0 \\ y = 0 \quad&\text{ or }\quad y = 4 \end{align}

Now, for $y = 0$, the equation $x^2 - 4x = 0$ is identical, so we find our first two solutions: $x = 0$ or $x = 4$.

For the other two solutions, we have to solve with $y = 4$:

\begin{align} x^2 - 4x &= 4 \\ x^2 - 4x + 4 &= 4 + 4 \\ (x - 2)^2 &= 8 \\ x = 2 - \sqrt8 \quad&\text{ or }\quad x = 2 + \sqrt8 \end{align}

Putting them all together, in increasing order,

$$ (f \circ f)^{-1}(\{0\}) = \bigl\{ 2 - 2\sqrt2,\, 0,\, 4,\, 2 + 2\sqrt2 \bigr\} $$

Here's a sketch of $y = f(x)$ (quadratic) and $z = (f \circ f)(x)$ (quartic). Notice that the four roots of $f \circ f$ on the $x$-axis correspond in pairs to inputs $x$ such that the outputs ar $y$-values where either $f(x) = 0$ or $f(x) = 4$.