Piecewise functions are defined by using pieces of other functions. Since we are now familiar with linear and quadratic functions, let’s build the piecewise function, $V(h)$, from their pieces.

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

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

Domain

From the definition of $V(h)$, we can see the domain is $[-6, -2] \cup (-2, 4] \cup (4,6) = [-6, 6)$.

Zeros

We can see from the graph that usually the component pieces of $V$ would total four intercepts and zeros. One each from the linear pieces and two from the quadratic piece. However, the domain restrictions only admit two of them. The greater zero from the quadratic piece and the zero from the right linear function. These are both represented by intercepts on the graph.

The left line would have an intercept, except the domain does not allow it. The middle parabola has a left intercept, except the domain does allow it.

From the graph, we might estimate our two zeros to be around $3$ and $4.5$. What are the exact values?

$\blacktriangleright$ The middle piece.

$\blacktriangleright$ The right piece.

Increasing and Decreasing

• $-2h-3$ is a linear function. Its constant rate of change is $-2$. This tells us it is a decreasing function. $V$ is decreasing on $[-6, -2]$.
• $-(h+3)(h-3)$ is a quadratic on $(-2, 4]$. Here, $iRoC_V(h) = -2h$. The critical number occurs when $iRoC_V(h) = -2h = 0$, which happens when $h=0$. This agrees with the midpoint of the zeros, which are $-3$ and $3$. Therefore, $0$ is where the function switches behavior.
  • On $(-2, 0)$, $iRoC_V(h) = -2h > 0$, therefore, $V$ is increasing.
  • On $(0,4)$, $iRoC_V(h) = -2h < 0$, therefore, $V$ is decreasing.

  This agrees with the graph.

Technically: We also need to account for the endpoints. $V$ is increasing on $[-2, 4]$, since the dot at $-2$ is lower than the left side of the included parabola.

Technically: $V$ is increasing on $[4, 6)$, since the dot at $4$ is lower than the left side of the included line segment.

The intervals where $V$ is monotonic are

• $V$ is decreasing on $[-6,-2]$.
• $V$ is increasing on $[-2,0]$.
• $V$ is decreasing on $[0,4]$.
• $V$ is increasing on $[4,6)$.

Extreme Values

We have three pieces to our function and the graph will help our algebra. We’ll gather information about the pieces indiviudally and then combine them for $V$.

• $-2h-3$ is a decreasing linear function on $[-6, -2]$. Its maximum would occur at the left endpoint, which is included. Therefore, the maximum is $-2(-6)-3 = 9$. Its minimum would be at the right endpoint, which is included. Therefore, the minimum is $-2(-2)-3 = 1$.
• $-(h+3)(h-3)$ is a quadratic on $(-2, 4]$. Its graph is a parabola opening down and we have seen that $V$ increases and then decreases.

  Therefore, its maximum value corresponds to the vertex, $(0, 9)$, which is included. We get a maximum value of $9$. The candidates for minimum value would occur at the endpoints, except only the right endpoint is included. $-(4-3)(4+3) = -7$ is a possible minimum value.

  We could have expanded the factored form: $-(h+3)(h-3) = -h^2 + 9$. This is now in vertex form, $-(h-0)^2 + 9$. From this expression we can see that the vertex is $(0, 9)$, and the negative leading coefficient tells us that we have a maximum of $9$ occuring at $0$.

  We could have observed the expanded form as a standard form, $-h^2 + 0 h + 9$. Then the quadratic formula would give us $\frac {-b}{2a} = \frac {-0}{-2} = 0$ as the location of the vertex and the place where the behavior changes.

  And, we can look at the instantaneous rate of change. The middle piece of $V$ is a quadratic, therefore, the vertex form $-(h-0)^2 + 9$ gives us $iRoC(h) = -2(h-0) = -2h$. This equals $0$ at $0$, which means $0$ is a critical number and a possible location of changing behavior.

  Lastly,
  
• $\frac {7h}{4} - 8$ is an increasing linear function on $[4, 6)$. Its maximum would be encoded in the right endpoint, which is excluded. Its minimum would be encoded in the left endpoint, which is also excluded. This piece is contributing no extreme values to $V$.

Our analysis of the pieces leaves us with a global maximum value of $9$ for $V$. $9$ occurs twice. Once at $-6$ and then again at $0$.

Our analysis of the pieces leaves us with a global minimum value of $-7$ for $V$. $-7$ occurs once at $4$.

Each of these global extrema are also local extrema. In addition we have $1$, which is a local minimum occuring at $-2$.

Critical Numbers

$\blacktriangleright$ We know that the instantaneous rate of change of the middle quadratic is $iRoC_V(h) = -2h$

$\blacktriangleright$ We know that the instantaneous rate of change of the left linear is $iRoC_V(h) = -2$

$\blacktriangleright$ We know that the instantaneous rate of change of the right linear is $iRoC_V(h) = \frac {7}{4}$

The only place where the $iRoC_V(h) = 0$ is at $0$, therefore, $0$ is a critical number.

We also include domain numbers where the $iRoC_V$ does not exist. That would be domain numbers where there is no corresponding tangent line. That would include $-2$ and $4$.

The set of critical numbers is $\{ -2, 0, 4 \}$.

This doesn’t include all of the places where we were investigating for extreme values. We also looked at $-6$. There is a tangent line at $(-6, 9)$, which means $-6$ is not a critical number. Instead, $-6$ is an endpoint of a maximal interval of the domain. That automatically makes $-6$ a candidate for the location of an extreme value of $V$.

Critical numbers and endpoints: $\{ -6, -2, 0, 4 \}$.

These are exactly the places we have been investigating for possible maximums and minimums.

Range

Since each piece of $V$ is continuous, the ranges of the three pieces of $V$ are $[1, 9]$, $[-7, 9]$, and $\left ( -1, \frac {2}{5} \right )$.

The range of $V$ is their union.

$$[1, 9] \cup [-7, 9] \cup \left ( -1, \frac {2}{5} \right ) = [-7, 9]$$

Continuity

Linear and quadratic functions are continuous functions. $V(h)$ is continuous on $[-6, -2]$, $(-2, 4]$, and $(4, 6)$. These correspond to the intervals in the definition of $V(h)$, because there were no isolated points.

Therefore, we only need to check the endpoints of maximal intervals. Here, those would be $\{ -6, -2, 4, 6 \}$.

This will help our algebraic explanations.

Extrema

The extreme values of a function are the maximum and minimum values. There are two types of each.

The global maximum and mimnium values are the overall greatest and least values of the function. There can be at most one global maximum value (there might not be an actual global maximum value). There can be at most one global minimum value (there might not be an actual global minimum value). These two values can occur at many domain numbers.

Local maximum and minimum values are greatest and least values that occur in a “neighborhood” of a domain number. There can be many local extrema values that occur at many domain numbers.

• Graphically, global maximum values are visually encoded as the highest point on the graph.
• Graphically, global minimum values are visually encoded as the lowest point on the graph.
• Graphically, local maximum values are visually encoded as the tops of hill on the graph.
• Graphically, global minimum values are visually encoded as the bottoms of valleys on the graph.

• Graphically, local extrema values are visually encoded as endpoints.
• Graphically, local extrema values are visually encoded as singleton points.

We can sum up this ideas as follow: