Linear Composition

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linear linear linear

The composition of two functions, $F$ and $G$, is a new function symbolized by $F \circ G$. A hollow circle hovers between the two function names.

Composition

Given the doamin number, $a$, in the domain of $G$, such that $G(a)$ is in the domain of $F$, $(F \circ G)(a)$ is defined to be $F(G(a))$.

If $Dom_G$ and $Dom_F$ represent the domains of $G$ and $F$ respectively, the induced or implied domain of $F \circ G$ is

\[ Dom_{F \circ G} = \{ g \in Dom_G \, | \, G(g) \in Dom_F \} \]

The order matters. Generally speaking, $F \circ G \ne G \circ F$. Composition is not commutative.

Linear Composition

In this section, we will explore the composition of linear functions.

Let $A(x) = a_1 \cdot x + a_0$ be a linear function with $\mathbb {R}$ as its domain.
Let $B(w) = b_1 \cdot w + b_0$ be a linear function with $\mathbb {R}$ as its domain.

Then $A \circ B$ is a new function. What is its formula?

To keep everything separate, let’s use $v$ as the variable for the formula of $A \circ B$.

If we write $A \circ B(v)$, then people will miss read this to say $B(v)$. So, let’s put parentheses around $A \circ B$: $(A \circ B)(v)$.

\begin{align*} (A \circ B)(v) & = A(B(v)) \\ & = A(b_1 \cdot v + b_0) \\ & = a_1 \cdot (b_1 \cdot v + b_0) + a_0 \\ & = a_1 \cdot b_1 \cdot v + a_1 \cdot b_0 + a_0 \\ & = (a_1 \cdot b_1) v + (a_1 \cdot b_0 + a_0) \end{align*}

$\blacktriangleright $ The composition of two linear functions is again a linear function.

The rate-of-change of the composition is the sum difference productquotient of the rates-of-change of the original linear functions.

Linear Composition

Let $G(y) = 4 y + 3$ be a linear function with $\mathbb {R}$ as its domain.
Let $H(t) = 3 t -5$ be a linear function with $\mathbb {R}$ as its domain.

Define two new functions, $U(x)$ and $V(w)$, as compositions by

$U(x) = (G \circ H)(x) = G(H(x)) = 4 \left (\answer {3 x - 5}\right ) + 3 = 12 x -17$
$V(w) = (H \circ G)(w) =H(G(w)) = 3 \left (\answer {4 w + 3}\right ) - 5 = 12 w + 4$

Two different functions. Both are linear functions. The constant rate-of-change is $12$ for both.

Their graphs would be parallel lines.

Linear Composition

Let $k(m) = 2 m + 1$ be a linear function with $\mathbb {R}$ as its domain.
Let $h(n)$ be a linear function with $\mathbb {R}$ as its domain.

Let $r(v) = (k \circ h)(v) = k(h(v)) = 3 v - 2$

$\blacktriangleright $ Determine a formula for $h(n)$.

explanation

$k$ is linear and the composition, $r$, is linear. A good guess is that $h$ is linear.

In that case, the rates-of-change would multiply and $h$ would start off looking like $h(n) = \frac {3}{2} n + n_0$, for some $n_0$.

So, far we have

\[ r(v) = k(h(v)) = 2 \left (\frac {3}{2} v + n_0\right ) + 1 = 3v + 2 n_0 + 1\]

This is the composition $(k \circ h)(v)$, which we know equals $3 v - 2$. Therefore, we must have

\[ 3 v + 2 n_0 + 1 = 3 v - 2 \]

We need $2 n_0 + 1 = -2$. Therefore, $n_0 = \answer {-\frac {3}{2}}$.

We have $h(n) = \frac {3}{2} n - \frac {3}{2}$

Linear Composition (reverse order)

Let $k(m) = 2 m + 1$ be a linear function with $R$ as its domain.
Let $h(n)$ be a linear function with $R$ as its domain.

Let $r(v) = (h \circ k)(v) = h(k(v)) = 3 v - 2$

$\blacktriangleright $ Determine a formula for $h(n)$.

explanation

$k$ is linear and the composition, $r$, is linear. A good guess is that $h$ is linear.

In that case the rates-of-change would mulitply and $h$ would start off looking like $h(n) = \frac {3}{2} n + n_0$, for some $n_0$.

So, far we have

\[ r(v) = h(k(v)) = \frac {3}{2} \left ( \answer {2 v + 1} \right ) + n_0 = 3 v + \frac {3}{2} + n_0\]

This is the composition $(h \circ k)(v)$, which we also know equals $3 v - 2$. Therefore, we must have

\[ 3 v + \frac {3}{2} + n_0 = 3 v - 2 \]

We need $\frac {3}{2} + n_0 = -2$. Therefore, $n_0 = \answer {-\frac {7}{2}}$.

We have $h(n) = \frac {3}{2} n - \frac {7}{2}$

Restricted Domains

Let’s restrict the domains of our linear functions.

\[ T(v) = -v+3 \, \text { with domain } \, [1,7) \]

Graph of $y = T(v)$.

\[ g(x) = \frac {7}{4}x-8 \, \text { with domain } \, (2,8) \]

Let’s create the composition $T \circ g = T(g)$.

This means the range of $g$ needs to be inside the domain of $T$. But as the chart below shows, there are numbers in the range of $g$ that are not in the domain of $T$.

For instance, $0$ is in the range of $g$, but not in the domain of $T$.

Therefore, we need to remove the numbers in the domain of $g$, whose function value is $0$.

There are more numbers in the domain of $g$, where the function value of $g$ is not in the domain of $T$. We need to remove these numbers from the domain of $T \circ g$.

What part of the range of $g$ can we use?

Since $g$ is an increasing linear function, the range of $g$ is the interval $\left ( -\frac {9}{2}, 6 \right )$.

The domain of $T$ is $[1,7)$.

How do these fit together?

The full range of $g$ does not fit inside the domain of $T$.

It looks like the interval $\left ( -\frac {9}{2}, 1 \right )$ is in the range of $g$, but not in the domain in of $T$.

We need to remove the number from the domain of $g$ whose function values are inside $\left ( -\frac {9}{2}, 1 \right )$.

Chaining $T$ and $g$

The whole range of $g$ is $\left ( -\frac {9}{2}, 6 \right )$.
The whole domain of $T$ is $[1,7)$.

This whole range of $g$ cannot be used as domain values of $T$. That means we cannot use the whole domain of $g$ as the domain of $T \circ g$.

It looks like we can use the interval $\left [ \answer {1}, \answer {6} \right )$ from the range of $g$.

Therefore we need to restrict the domain of $g$, so that the range is just $[1, 6)$. We need to find the preimages of $6$ and $1$ in the domain of $g$.

$\blacktriangleright $ The preimage of $6$.

We need to solve $\frac {7}{4}x-8 = 6$.

\begin{align*} \frac {7}{4}x-8 &= 6 \\ \frac {7}{4}x &= \answer {14} \\ x &= \frac {4}{7} \cdot 14 = \answer {8} \end{align*}

The preimage of $6$ is $8$.

If we take a closer look at the range of $g$, we see that $6$ is an endpoint that is not included.

Therefore, we need to exclude $8$ as an endpoint of the restricted domain we are building.

$\blacktriangleright $ The preimage of $1$.

\begin{align*} \frac {7}{4}x-8 &= 1 \\ \frac {7}{4}x &= 9 \\ x &= \frac {4}{7} \cdot 9 = \frac {36}{7} \approx 5.14 \end{align*}

$1$ is included in the range of $g$ and also included in the domain of $T$, so we want to include it in the domain of $T \circ g$.

We cannot use the whole domain of $g$. We can only use $\left [ \answer {\frac {36}{7}}, \answer {8} \right )$.

With this restricted domain, the range of $g$ will be $[1, 6)$, which is the domain of $T$.

We now have a domain for our composition: $Dom_{T \circ g} = \left [\frac {36}{7}, 8\right )$

Our domain of $T \circ g$ is $\left [\frac {36}{7}, 8\right )$.

The image of this interval under $g$ is $[1,6)$, which sits inside the domain of $T$.

Now, we need the image of $[1,6)$ under $T$. This will give us the range of $T \circ g$.

$T(v) -v + 3$

$T(1) = -1+3 = 2$, included
$T(6) = -6+3 = -3$, excluded

The range of our composition is the image interval $(-3,2]$.

What do we have?

$T \circ g$

We have a composition: $T \circ g$. We know this is a linear function, because it is the composition of two linear functions.

Our composition has a formula, let’s use $c$ for its variable.

$(T \circ g)(c) = -\left (\frac {7}{4}c - 8\right ) + 3 = -\frac {7}{4}c + 11$

\[ -\frac {7}{4} \cdot \frac {36}{7} + 11 = 2 \]

\[ -\frac {7}{4} \cdot 8 + 11 = -3 \]

It’s domain is $\left [ \answer {\frac {36}{7}}, \answer {8} \right )$.

It’s range is $\left ( \answer {-3}, \answer {2} \right ]$.

Graph of $y = T(g(c))$.

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more examples can be found by following this link
More Examples of Piecewise Composition

2026-05-30 17:27:14