Shifting Range

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same characteristics

Outside

One way to create a new function from an existing function is to shift all of the range numbers (function values) in the function pairs.

\[ (a, b) \mapsto (a, b + k) \]

$\blacktriangleright $ How do we describe a shifting range?

We’ll begin with a graphical description and then translate to an algebraic description.

Watching the effects on the graph is helpful to understanding the algebra.

The Initial Function

Below is the piecewise defined function, $T(v)$.
$v$ is representing the domain values in $(-4,-1] \cup [1,7)$.
$T(v)$ represents the range number paired with $v$.
Therefore, $T(v)$ represents numbers in $(-9, 2]$.

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

On the interval $(-4, -1]$, the graph should be a line segment for $T(v) = \answer {2v-1}$. There should be a(n) open closed dot when $v = -4$.

There should be an open closed dot when $v = -1$.

On the interval $[1, 7)$, the graph should be another line segment for $T(v) = \answer {-v+3}$. There should be a(n) open closed dot when $v = 1$.

There should be an open closed dot when $v = 7$.

Graph of $y = T(v)$.

The domain of $T$ has two maximal intervals, $(-4,-1]$ and $[1,7)$. These correspond to two line segments on the graph. The endpoints give us four strategic points on the graph:

$(-4, -9)$, which is an open point on the graph.
$(-1, -3)$, which is a closed point on the graph.
$(1, 2)$, which is a closed point on the graph.
$(7, -4)$, which is an open point on the graph.

The function $T$ has no minimum value. It has a global maximum of $2$, which occurs at $1$. This is also a local maximum. There is another local maximum value of $-3$, which occurs at $-1$.

The function $T$ is increasing decreasing on the interval $(-4,-1]$.
The function $T$ is increasing decreasing on the interval $[1, 7)$.

Using $T$ as an initial function, let’s shift the range.

A New Function : B

Let $B(k)$ be a new function, but its definition is based on $T(v)$.

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

$\blacktriangleright $ $B(k) = T(k)+4$ with the induced domain.

First Question: Domain

In the definition of $B(k)$, we see $T(k)$, which means $k$ represents the domain of $T$, which is $(-4,-1] \cup [1,7)$.
$k$ is also representing the domain of $B$.
Therefore, the domain $B$ is also $(-4,-1] \cup [1,7)$.

The addition of $4$ is not inside the domain parentheses next to $T$. It is on the outside of the domain parentheses. $T(k)$ represents range values or values of the function, $T$, and these are increased by $\answer {4}$ to get the values of $B$.

The formula for $B$ is the formula for $T$ with $4$ added.

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

Graph of $z = B(k)$.

The shape of the graph has not changed. It just slid up.

Second Question: Behavior

$T$ has a global maximum of $2$, which occurs at $1$. Therefore, $B(k)$ has a global maximum of $\answer {6}$, which occurs at $1$. This is also a local maximum. There is another local maximum value of $-3+4=1$, which occurs at $\answer {-1}$.

The function $B$ is increasing on the interval $(-4,-1]$, just like $T$.

The function $B$ is decreasing on the interval $[-1, 7)$, just like $T$.

Shifting the Range

Define $V(h)$ as

\[ V(h) = \begin{cases} -2h-3 & [-6, -2] \\ -(h+3)(h-3) & (-2, 4] \\ \frac {7h}{4} - 8 & (4,6) \end{cases} \]

A graph always helps our thinking. Here is the graph of $y = V(h)$.

From the graph, it appears that...

The domain of $V$ is $[-6,6)$.
$V$ has a global maximum of $9$, which occurs at $\answer {-6}$ and at $\answer {0}$.
$V$ has a global minimum of $-7$, which occurs at $4$.
$V$ has a local minimum of $1$, which occurs at $-2$.
$V$ has a jump discontinuity at $-2$.
$V$ has a jump discontinuity at $4$.
$V$ is decreasing on $[-6, -2]$.
$V$ is increasing on $[-2, 0]$.
$V$ is decreasing on $\left [\answer {0}, \answer {4}\right ]$.
$V$ is increasing on $[4, 6)$.

$\blacktriangleright $ Let’s define a new function based on $V$.

Let $f(x) = V(x)-2$ with the induced domain.

The domain of $f$ is $\left [\answer {-6}, \answer {6}\right )$, same as $V$. The range or function values $f$ are all $\answer {2}$ less than the function values of $V$.

\[ f(x) = \begin{cases} -2x-5 & [-6, -2] \\ -(x+3)(x-3)-2 & (-2, 4] \\ \frac {7x}{4} - 10 & (4,6) \end{cases} \]

From the graph, it appears that...

The domain of $f$ is $[-6,6)$.
$f$ has a global maximum of $7$, one of which occurs at $-6$ and the other occurs at $\answer {0}$.
$f$ has a global minimum of $-9$, which occurs at $4$.
$V$ has a local minimum of $-1$, which occurs at $-2$.
$f$ has a jump discontinuity at $-2$.
$f$ has a jump discontinuity at $4$.
$f$ is decreasing on $[-6, -2]$.
$f$ is increasing on $[-2, 0]$.
$f$ is decreasing on $\left [\answer {0}, \answer {4}\right ]$.
$f$ is increasing on $[4, 6)$.

The places in the domain where characteristics and features occur did not changed. The maximums and minimums have all dropped by $2$. The endpoints are all still solid or hollow. Their vertical coordinates have all dropped by $2$. Their horizontal coordinates have remained the same.

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

2026-06-08 00:37:14