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 $r-3=0$. $r=\answer {3}$.