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.

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

0.1 Domain of $T \circ g$

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

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

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

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

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

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 formulas

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

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

$\blacktriangleright $ On $\left (\frac {16}{7}, 4\right ]$, we have $T(v) = \answer {2v - 1}$.

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

$\blacktriangleright $ On $\left [\frac {36}{7}, 8\right )$, we have $T(v) = \answer {-v + 3}$.

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

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

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

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

2025-01-07 00:32:11