$$T(v) = \begin {cases} 2v-1 & \text { if } \, -4 < v \leq -1 \\ -v+3 & \text { if } \, 1 \leq v < 7 \end {cases}$$

Graph of $y = T(v)$.

$$g(x) = \begin {cases} -2x-3 & \text { if } \, -\frac {3}{2} \leq x \leq 2 \\ \frac {7}{4}x-8 & \text { if } \, 2 < x < 8 \end {cases}$$

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

This means the range of $g$ needs to be inside the domain of $T$. Since there are two pieces to each of $T$ and $g$, let’s follow the pieces separately. The chart below shows the parts of the range of $g$ coming from the individual linear pieces as well as the parts of the domain of $T$. We can see that there are numbers in the full range of $g$ that are not in the full domain of $T$.

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

We have two numbers in the domain of $g$ that $g$ maps to $0$. One from each piece.

• $g\left ( -\frac {3}{2} \right ) = -2 \left ( -\frac {3}{2} \right ) - 3 = 3 - 3 = 0$
• $g\left ( \frac {32}{7} \right ) = -\frac {7}{4} \cdot \left ( \frac {32}{7} \right ) - 8 = 8 - 8 = 0$

We cannot include these in the domain of the composition, because they would result in $T(0)$, which is not defined.

Therefore, we cannot include $-\frac {3}{2}$ nor $\frac {32}{7}$ in the composition domain we are constructing.

What other numbers from the domain of $g$ do we need to leave out?

We need to map the domain of $T$ back onto the range of $g$, find the intersection, and then work our way back into the domain of $g$, in order to restrict the domain of $g$ to just the numbers that work in the composition.

First, let’s map the domain of $T$ back to the range of $g$. We need to see what each piece is doing. So, we’ll plot the domain of $T$ on the vertical axis of the graph of $g$.

The intersection is $(-4, -1] \cup [1, 6)$.

These are the common intervals in both the domain of $T$ and range of $g$.

Now we need to connect these up to the domain of $g$. Which parts of the domain of $g$ does $g$ map to these intervals?

Let’s follow the endpoints backwards through $g$ to the domain of $g$.

Focusing on the interval endpoints (included or excluded) of this intersection, we need the preimages of $-4$, $-1$, $1$, and $6$ in the domain of $g$.

$g$ has two pieces.

$\blacktriangleright$ The interval $(-4, -1]$ is in both pieces of the range of $g$. It has a preimage from each piece.

Let’s examine each piece separately. First, $-2x - 3$.

• What is the preimage of $-4$?
• What is the preimage of $-1$?

Graphically, we begin at $-4$ on the vertical axis. It is the lowest endpoint. We follow it to the right to the graph of $g$. Once we hit the graph, we move up until we hit the horizontal axis. Graphically, it appears to be around $0.5$.

Graphically, we begin at $-1$ on the vertical axis. It is the top endpoint of the lower interval. We follow it to the left to the graph of $g$. Once we hit the graph, we move up until we hit the horizontal axis. Graphically, it appears to be around $-1$.

Let’s see if our graphical approximations are close to the actual values.

• $g(x) = -2x-3 = -4$
  • $-2x = -1$
  • $x = \frac {1}{2}$

• $g(x) = -2x-3 = -1$
  • $-2x = 2$
  • $x = -1$

We guessed fairly well from the graph. We now have the preimage of the bottom interval through one piece of the formula for $g$.

We get the interval $\left [ -1, \frac {1}{2} \right )$.

Now for the other piece of $g$.

The other piece of $g$ is $\frac {7}{4}x - 8$.

What is the preimage of $(-4, -1]$?

• What is the preimage of $-4$?
• What is the preimage of $-1$?

Graphically, we begin at $-4$ on the vertical axis. It is the lowest endpoint. We follow it to the right to the other piece of the graph of $g$. Once we hit the graph, we move up until we hit the horizontal axis. Graphically, it appears to be around $2.25$.

Graphically, we begin at $-1$ on the vertical axis. It is the top endpoint of the lower interval. We follow it to the right to the graph of $g$. Once we hit the graph, we move up until we hit the horizontal axis. Graphically, it appears to be around $4$.

Let’s see if our graphical approximations are close to the actual values.

• $g(x) = \frac {7}{4}x -8 = -4$
  • $\frac {7}{4}x = 4$
  • $x = \frac {16}{7} \approx 2.3$

• $g(x) = \frac {7}{4}x -8 = -1$
  • $\frac {7}{4}x = 7$
  • $x = 4$

We guessed fairly well from the graph. We now have the preimage of the bottom interval through the second piece of the formula for $g$.

We get the interval $\left ( \frac {16}{7}, 4 \right ]$.

Now for the other interval in the range of $g$.

$\blacktriangleright$ The interval $[1, 6)$ has a preimage on only one piece.
$1$ and $6$ have preimages only through $g(x) = -2x-3$.

• What is the preimage of $1$?
• What is the preimage of $6$?

Graphically, we begin at $1$ on the vertical axis. It is the lower endpoint of the top interval. We follow it to the right to the larger piece of the graph of $g$. Once we hit the graph, we move down until we hit the horizontal axis. Graphically, it appears to be around $5$.

Graphically, we begin at $6$ on the vertical axis. It is the upper endpoint of the top interval of our intersection (not the top of $g$, but the top of our common interval). We follow it to the right to the graph of $g$. Once we hit the graph, we move down until we hit the horizontal axis. Graphically, it appears to be around $8$.

Let’s see if our graphical approximations are close to the actual values.

• $g(x) = \frac {7}{4}x -8 = 1$
  • $\frac {7}{4}x = 9$
  • $x = \frac {36}{7} \approx 5.1$

• $g(x) = \frac {7}{4}x -8 = 6$
  • $\frac {7}{4}x = 14$
  • $x = 8$

We guessed fairly well from the graph. We now have the preimage of the top interval through the larger piece of $g$.

We get the interval $\left [ \frac {36}{7}, 8 \right )$.

The domain of $T \circ g$ is $\left ( -1, \frac {1}{2} \right ] \cup \left ( \frac {16}{7}, 4 \right ] \cup \left [ \frac {36}{7}, 8 \right )$

This is inside the full domain of $g$. It is not the full domain of $g$. We needed to leave some domain numbers out, because the full domain of $g$ would give the full range of $g$, which was bigger than the domain of $T$.

There are three pieces to the domain of our composition:

• $\left [ -1, \frac {1}{2} \right )$
• $\left ( \frac {16}{7}, 4 \right ]$
• $\left [ \frac {36}{7}, 8 \right )$

The composition will have a formula for each piece.