$\blacktriangleright$ 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 linear function, $L_o(x)$, and an inside linear 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

We would like to focus on the inside function as a linear function.

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

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

$\blacktriangleright$ How does $L$ affect $f$?

The main issue here is the range of $L$ intersecting the domain of $f$.

$\star$ Inside = Linear

In this section, our inside function will always be a linear function.

$$(Outside \circ L)(a) = Outside(L(a))$$

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

Let’s consider a quadratic function: $Q(h) = (h+1)(h-4)$

We can consider $Q(h)$ as a composition. The outside function is $Q(h) = (h+1)(h-4)$ and the inside function is the Identity function: $I(h)=h$.

The inside function doesn’t do much. It pairs every real number with itself. Then this is handed to $H$. Not much of a composition, but it will help us illustrate some ideas.

The natural domain of $I(h)=h$ is $\mathbb {R}$. We can think of moving along the real line left to right from $-\infty$ to $\infty$. Each real number we run across is put into $I(h)$, which just gives it right back. Thus, the range of $I(h)$ also runs along the real line left to right from $-\infty$ to $\infty$ at the exact same speed as the domain. The range of $I(h)$ then becomes the domain for the outside function, $Q(h)$.

$\blacktriangleright$ What happens when we have a different inside function?

Suppose $inside(h) = 5h$.

The values of $inside(h)$ still run along the real line left to right from $-\infty$ to $\infty$. What’s the difference?

The difference is that the identity function ran across its domain and range from $-\infty$ to $\infty$ at exactly the same rate - because the input and out were equal - it was the identity function.

Now the range is running through the real line left to right from $-\infty$ to $\infty$ - five times as fast as the domain. A little movement in the domain results in a lot of movement in the range and this range is the domain for $Q(h)$.

As we run across the horizontal axis, these values are going into $inside(h) = 5h$. The output values, $5h$ are the new (hidden) horizontal axis for the inputs (domain) into $Outside(h) = Q(h)$.

As you run across the $h$-axis, those values are not going into $Q$. $5$ times those values are going into $Q$.

All of the Q values are still there. The output of $Q \circ inside$ are the same values of $Q$. They are just connected to faster running domain values then before.

Slower

How do we slow down $Q$?

We feed $Q$ its domain slower.

Let $L(x) = \frac {x}{2}$. $L$ takes in real numbers and gives half their value. If we feed these into $Q$, then $Q$ will think it is only getting half the value that that we see on the horizontal axis.

Graph of $= (Q \circ L)(x) = Q(L(x))$

The graph is wider, because $Q$ thinks it is getting half the value as the $x$-axis.

The vertical heights haven’t changed. The formula $Q(L(x))$ stills say that the values of $Q$ will be the outputs - just not at the same places.

The minimum value of $Q(h) = (h+1)(h-4)$ is $-\frac {25}{4} = -6.25$, which occurs at $\frac {3}{2}$.

The minimum value of $Q \circ L$ is still $-\frac {25}{4} = 6.25$. It now occurs at $\frac {3}{2} \cdot 2 = 3$. It occurs at $3$, so that when $L$ multiplies by $\frac {1}{2}$, the number $\frac {3}{2}$ is fed into $Q$.

The vertex on $y = Q$ was $\left (\frac {3}{2}, -\frac {25}{4} \right )$. This has moved to $\left (3, -\frac {25}{4} \right )$ for $y = Q \circ L$.

The zeros for $Q$ are $-1$ and $4$. These move to $-2$ and $8$ for $Q \circ L$.