$\blacktriangleright$ Pointwise composition was seen via individual numbers:

$$(F \circ G)(a) = F(G(a))$$

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.

$$(L_o \circ L_i)(x) = L_o(L_i(x))$$

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.

$$(L \circ Inside)(z) = L(Inside(z))$$

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 $Q(h) = 12$.

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

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

$$h = \frac {-(-3) \pm \sqrt {(-3)^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} = \frac {3 \pm \sqrt {73}}{2}$$

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.