Piecewise Functions

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piecewise linear

Let’s extend our composition of linear functions to piecewise linear functions.

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

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.

The intersection of the range of $g$ and the domain of $T$ is

\[ (-4, -2] \cup [1,6]) \]

This is the only part of the range of $g$ that we can use in the composition. We need to find the domain valus of $g$ that give these function values.

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?

preimages in $g$

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

\begin{align*} \frac {7}{4}x -8 &= -4 \\ \frac {7}{4}x &= 4 \\ x &= \frac {16}{7} \approx 2.3 \end{align*}

\begin{align*} \frac {7}{4}x -8 &= -1 \\ \frac {7}{4}x &= 7 \\ x &= 4 \end{align*}

\begin{align*} \frac {7}{4}x -8 &= 1 \\ \frac {7}{4}x &= 9 \\ x &= \frac {36}{7} \approx 5.1 \end{align*}

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

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

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 formula of $T \circ g$

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

Formula for $T \circ g$

$g$ only has one formula: $\frac {7}{4}x - 8$.
$T$ has two pieces.

We need to figure out which piece goes with which formula for $T$.

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

$\blacktriangleright $ On $\left ( \frac {16}{7}, 4 \right ]$

On $\left ( \frac {16}{7}, 4 \right ]$, $g$ has the image $(-4, -1]$.

On $(-4, -1]$, $T$ uses the formula $T(v) = \answer {2v - 1}$.

$(T \circ g)(c) = 2\left ( \frac {7}{4}c - 8 \right ) - 1 = \frac {7}{2}c - 17$

$\blacktriangleright $ On $\left [ \frac {36}{7}, 8 \right )$,

On $\left [ \frac {36}{7}, 8 \right )$, $g$ has the image $[1, 6)$.

On $[1, 6)$, $T$ uses the formula $T(v) = \answer {-v + 3}$.

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

\[ (T \circ g)(c) = T(g(c)) = \begin{cases} \frac {7}{2}c - 17 & \text { on } \left (\frac {16}{7}, 4\right ] \\ -\frac {7}{4}c + 11 & \text { on } \left [\frac {36}{7}, 8\right ) \end{cases} \]

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

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

2026-05-30 19:51:41