We have already seen a couple of versions of composition.

\(\blacktriangleright \) Pointwise composition was seen via individual numbers:

The number, \(a\), in the domain of \(G\) was connected to its range partner, \(G(a)\). \(G\) was evaluated at \(a\) and the function value, \(G(a)\), was then viewed as a member of the domain of \(F\). As a member of the domain of \(F\), \(F\) can be evaluated at \(G(a)\) to get \(F(G(a))\).

\(\blacktriangleright \) Linear composition between two linear functions produced a whole new function - a linear function. Instead of thinking of domain numbers individually, this composition was viewed as an operation on linear functions.

There is an outside function, \(L_o(x)\), and an inside function \(L_i(x)\). The composition operation, \(\circ \), is applied and a new linear function is created.

Our symbol for this function is \((L_o \circ L_i)\) or \(L_o \circ L_i\). The parentheses are used to clear up communication.

In our investigations, we have discovered that \((L_o \circ L_i)\) “is” \(L_o\), just shifted, stretched, and reflected horizontally.

We would like to extend this idea of a function operation beyond linear functions.

Composition

This time, we would like to focus on the outside function as a linear function.

Let \(f(x)\) be any function.
Let \(L(y)\) be any linear function.

Form the composition \(L(f(z))\).

\(\blacktriangleright \) How does \(L\) affect \(f\)?

The main issue here is the range of \(f\) intersecting the domain of \(L\). However, since the natural domain of a linear function is all real numbers, there shouldn’t be a problem.

\(\star \) Outside = Linear

In this section, our \(Outside\) function will always be a linear function.

where \(L(x) = a \, x + b\), with \(a\) and \(b\) real numbers and \(a \ne 0\).

Let’s consider a quadratic function: \(Q(h) = (h+1)(h-4)\) as the \(inside\) function.

From left to right, the range values, or function values, for \(Q\) begin very big and positive. These values decrease to \(0\) and continue negative until they reach a value of \(-6.25\). Then, they increase to \(0\) again and continue to very big and positive values.

Now we will transform these function values with a linear function.

Let \(L(t) = -\frac {1}{4} t + 3\) with domain \(\mathbb {R}\).

• \(L\) will take a function value from \(Q\) and compress it by a factor of \(\frac {1}{4}\). Our parabola will be squished vertically a bit.
• Then \(L\) will negate the values. This will vertically reflect the parabola over the horizontal axis.
• Then \(L\) will add \(3\) to all of the values. This will shift the parabola up by \(3\).

The vertical measurements have all been processed linearly, which means the shape doesn’t change. It is still a parabola and all of its features are relatively in the same place.

The minimum of \(Q\) is \(-6.25\) and this occurrs at \(1.5\). We are flipping vertically, so the maximum of the composition is still at \(1.5\). There were no horizontal transformations. The maximum is \(L(-6.25) = 4.5625\).

The zeros of \(L \circ Q\) occur when \(L(t) = 0\), which is at \(t = 12\). Therefore, the zeros of \(L \circ Q\) occur when \(Q(h) = 12\).

\(Q(h) = (h+1)(h-4)\)

\begin{align*} Q(h) = (h+1)(h-4) & = 12 \\ h^2 - 3h - 4 & = 12 \\ h^2 - 3h - 16 & = 0 \\ \end{align*}

This does not factor easily. We’ll use the quadratic formula.

We have two zeros: \(\frac {3 + \sqrt {73}}{2} \approx 5.77\) and \(\frac {3 - \sqrt {73}}{2} \approx -2.77\), which agrees with our graph.