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

Graph of $y = T(v)$.

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

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 $T$.

We need to identify the intersection of the range of $g$ and the domain of $T$.

We need to map the domain of $T$ back onto the range of $g$ 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.

Focusing on the common endpoints (hollow or solid), we need to find the preimages of $-4$, $-1$, $1$, and $6$ in $g$.

Domain of $T \circ g$

Which numbers in the domain of $g$ have $-4$, $-1$, $1$, and $6$ as function values?

These are the domain numbers of $g$, that have function values in $(-4,1] \cup [1,7)$, which is the intersection of the range of $g$ and domain of $T$.

Now for the formulas

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

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