Graph of \(y = m(f)\).

• The domain of \(m\) is \([-7,-4) \cup \left \{\answer {-3}\right \} \cup [-2,7)\).
• \(m\) has a global (and local) maximum of \(6\), which occurs at \(\answer {1}\).
• \(m\) has a global (and local) minimum of \(-9\), which occurs at \(\answer {-2}\).
• \(m\) has a local maximum of \(\answer {1}\), which occurs at \(-7\).

\(\answer {2}\) is both a local maximum and minimum of \(m\).

To see that \(2\) is a local maximum at \(-3\), imagine the interval \((-3.1, -2.9)\). For all of the domain elements in this interval, \(T(-3)=2\) is the maximum. This is automatically satisfied, since \(3\) is the only domain element of \(m\) inside this interval.

Note: The definition of local maximum never promised that there would be other domain numbers in the small neighborhood. It simply says that you need to supply an open interval around \(-3\) such that \(T(-3)\) is greater than or equal to all function values for all domain numbers inside the interval.

To see that \(2\) is a local minimum, imagine the interval \((-3.1, -2.9)\). For all of the domain elements in this interval, \(T(-3)=2\) is the minimum. This is automatically satisfied, since \(3\) is the only domain element of \(m\) inside this interval.

Note: The definition of local minimum never promised that there would be other domain numbers in the small neighborhood. It simply says that you need to supply an open interval around \(-3\) such that \(T(-3)\) is less than or equal to all function values for all domain numbers inside the interval.

\(\blacktriangleright \) Now, we’ll shift the range.

Graph of \(z = P(t) = m(t)+3\).

• The domain of \(P\) is \([-7,-4) \cup \{-3\} \cup [-2,7)\).
• \(P\) has a global (and local) maximum of \(6+\answer {3}=9\), which occurs at \(\answer {1}\).
• \(P\) has a global (and local) minimum of \(-9+\answer {3}=-6\), which occurs at \(\answer {-2}\).
• \(P\) has a local maximum of \(1+\answer {3}=4\), which occurs at \(\answer {-7}\).
• \(\answer {5}\) is both a local maximum and minimum of \(P\) occuring at \(-3\).

Graph of \(y = \sin (\theta )\).

• The zeros of \(\sin (\theta )\) are all integer multiples of \(\pi \).
• The maximum value is \(1\) and it occurs at: \(\left \{ \frac {(4k+1)\pi }{2} \, | \, k \in \textbf {Z} \right \} = \{ \cdots , \frac {-3\pi }{2}, \frac {\pi }{2}, \frac {5\pi }{2}, \cdots \}\)
• The minimum value is \(-1\) and it occurs at: \(\left \{ \frac {(4k+3)\pi }{2} \, | \, k \in \textbf {Z} \right \} = \{ \cdots , \frac {-5\pi }{2}, \frac {-\pi }{2}, \frac {3\pi }{2}, \cdots \}\)

Graph of \(y = \sin (\theta ) - 2\).

• No zeros.
• The maximum value is \(-1\) and it occurs at: \(\left \{ \frac {(4k+1)\pi }{2} \, | \, k \in \textbf {Z} \right \}\)
• The minimum value is \(-3\) and it occurs at: \(\left \{ \frac {(4k+3)\pi }{2} \, | \, k \in \textbf {Z} \right \}\)

Let \(B(r) = |r-3| + 2\).

We have shifted the absolute value function to the left right by -3 -2 2 3 and up down by 2 3.

The graph of the absolute value function looks like a “V”, which is also the shape of the graph of \(B\).

The graph of the absolute value function has a corner at \((0, 0)\). The corner in the graph of \(B\) is now at \((3, 2)\).

Another way of saying this is that the absolute value function, \(A(t) = |t|\) has a global minimum of \(0\) at \(0\). This minimum occurs when the inside of the vertical bars equals \(0\).

For \(B\), this happens when \(r-3=0\). \(r=\answer {3}\).