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 \((-9, 2]\).

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 \(-1\).

\(\blacktriangleright \) \(B(k) = T(k)+4\) 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 formula for \(B\) is the formula for \(T\) with \(4\) added.

Graph of \(z = B(k)\).

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

Second Question: Behavior

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 \(T\).

Define \(V(h)\) as

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 \([4, 6)\).

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 \(V\).

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 \(2\).