6.3 The Shape of Graphs

We know if $f$ is differentiable at $x = a$ then the graph of $f$ is locally linear at $x = a$ and $f^{'} (a)$ is the slope of the tangent line at the point $(a, f (a))$ . In this section, we explore how local behavior near a point can be extrapolated to global behavior over an interval. First, we review Definition 1.7 from Section 1.2:

Definition

Let $f$ be a function defined on an interval $I$ . Then $f$ is said to be:

increasing on $I$ if, whenever $a < b$ , then $f (a) < f (b)$ . (i.e., as inputs increase, outputs increase.)
NOTE: The graph of an increasing function rises as one moves from left to right.
decreasing on $I$ if, whenever $a < b$ , then $f (a) > f (b)$ . (i.e., as inputs increase, outputs decrease.)
NOTE: The graph of a decreasing function falls as one moves from left to right.
constant on $I$ if $f (a) = f (b)$ for all $a$ , $b$ in $I$ . (i.e., outputs don't change with inputs.)
NOTE: The graph of a function that is constant over an interval is a horizontal line.

Suppose a function satisfies $f prime ⁢ open parenthesis x close parenthesis greater than 0$ for all $x$ in an open interval¹ $I$ . Then we know that not only is the graph of $f$ locally linear on $I$ , but the slopes of all of the tangent lines are positive. This means that all of the tangent lines are increasing so it stands to reason that the function $f$ is likewise increasing on $I$ . In other words, if a function is locally increasing on $I$ , then it is globally increasing on $I$ as well.

We can apply the same reasoning above to situations where $f prime ⁢ open parenthesis x close parenthesis less than 0$ for all $x$ in $I$ , which implies $f$ is decreasing on $I$ or $f prime ⁢ open parenthesis x close parenthesis equals 0$ on $I$ , which implies $f$ is constant on $I$ . In Calculus, you'll learn this fact is a consequence of the Mean Value Theorem.² In this text, we'll just accept the following theorem is true and hope we've done enough hand-waving to deem it reasonable.

Theorem

Suppose $f$ is differentiable on an open interval $I$ :

If $f prime ⁢ open parenthesis x close parenthesis greater than 0$ for all $x$ in $I$ , then $f$ is increasing on $I$ .
If $f prime ⁢ open parenthesis x close parenthesis less than 0$ for all $x$ in $I$ , then $f$ is decreasing on $I$ .
If $f prime ⁢ open parenthesis x close parenthesis equals 0$ for all $x$ in $I$ , then $f$ is constant on $I$ .

Theorem 6.4 may be visualized as follows:

$f prime ⁢ open parenthesis x close parenthesis greater than 0$ for all $x$ in $I$ :
Figure 6.50
Figure 6.51
$f prime ⁢ open parenthesis x close parenthesis less than 0$ for all $x$ in $I$ :
Figure 6.52
Figure 6.53
$f prime ⁢ open parenthesis x close parenthesis equals 0$ for all $x$ in $I$ :
Figure 6.54

We can use Theorem 6.4 to help us determine the (open) intervals over which a function $f$ is increasing, decreasing, and constant by making a sign diagram for the derivative $f^{'}$ .

In order to avoid us having to go through the (somewhat lengthy) process of finding $f^{'} (x)$ using Definition 6.8, we'll just use some properties of derivatives from Calculus behind the scenes and present you with both a function and its derivative. It's time for an example.

Example 1

Let $f (x) = x^{3} - 3 x^{2} - 9 x + 5$ . Use the fact that $f^{'} (x) = 3 x^{2} - 6 x - 9$ to find the open intervals over which $f$ is increasing, decreasing, and constant. Check your answer graphically.

To make use of Theorem 6.4, we make a sign diagram for $f^{'} (x)$ . Since $f^{'}$ is a polynomial, $f^{'}$ is continuous so the per the Intermediate Value Theorem, Theorem 2.14, $f^{'}$ will only change sign on either side of zeros. Hence, our first step is to solve $f prime ⁢ open parenthesis x close parenthesis equals 0$ .

We are given $f^{'} (x) = 3 x^{2} - 6 x - 9$ . Solving $f^{'} (x) = 3 x^{2} - 6 x - 9 = 0$ gives $3 (x^{2} - 2 x - 3) = 0$ or $3 (x - 3) (x + 1) = 0$ . We get two solutions: $x = - 1$ and $x equals 3$ which divides the $x$ -axis into three regions: $x less than minus 1$ , $minus 1 less than x less than 3$ and $x greater than 3$ .

Next we select a test value in each of these three regions to determine the sign of $f^{'} (x)$ . For the interval $x less than minus 1$ , we select $x = - 3$ : $f^{'} (- 3) = 3 {(- 3)}^{2} - 6 (- 3) - 9 = (+)$ . For $minus 1 less than x less than 3$ , we select $x equals 0$ : $f^{'} (0) = 3 {(0)}^{2} - 6 (0) - 9 = (-)$ . Finally, for $x greater than 3$ , we select $x equals 4$ : $f^{'} (4) = 3 {(4)}^{2} - 6 (4) - 9 = (+)$ .

Below on the left is a sign diagram for $f^{'} (x)$ and on the right is what this means for the graph of $y = f (x)$ :

Figure 6.55

Figure 6.56

We find $f$ is increasing on $(- \infty, - 1)$ and again on $open parenthesis 3, ∞ close parenthesis$ while $f$ is decreasing on $open parenthesis minus 1, 3 close parenthesis$ . At the points $x = - 1$ and $x equals 3$ , we have $f prime ⁢ open parenthesis x close parenthesis equals 0$ so the graph of $f$ is locally flat there.

Since $f$ changes from increasing just to the left of $x = - 1$ to decreasing just to the right of $x = - 1$ , it stands to reason that $f$ has a local maximum at $x = - 1$ . This is indeed the case and we find that the local maximum value is $f (- 1) = {(- 1)}^{3} - 3 {(- 1)}^{2} - 9 (- 1) + 5 = 10$ .

Similarly, since $f$ changes from decreasing just to the left of $x equals 3$ to increasing just to the right of $x equals 3$ , $f$ has a local minimum at $x equals 3$ . The local minimum value is $f (3) = {(3)}^{3} - 3 {(3)}^{2} - 9 (3) + 5 = - 22$ .

A quick check using desmos confirms our results.

Figure 6.57

We generalize our observations about local extrema in the following result.

Theorem

The (First)³ Derivative Test for Local Extema: Let $f$ be continuous on an open interval $I$ containing a critical number $c$ .⁴ If $f$ is differentiable on $I$ , except possibly at $c$ , then

If $f^{'} (x)$ changes from $(+)$ for $x < c$ to $(-)$ for $x > c$ , $f$ has a local maximum at $x = c$ .
If $f^{'} (x)$ changes from $(-)$ for $x < c$ to $(+)$ for $x > c$ , $f$ has a local minimum at $x = c$ .
If $f^{'} (x)$ doesn't change sign going from $x < c$ to $x > c$ , $f$ does not have a local extremum at $x = c$ .

Example 2

Let $f (x) = x^{4 / 3} - 4 x^{1 / 3}$ . Use the fact that $f^{'} (x) = \frac{4}{3} x^{1 / 3} - \frac{4}{3} x^{- 2 / 3}$ to help you find:

the open intervals over which $f$ is increasing, decreasing, and constant.
the local extrema.

In order to make a sign diagram for $f^{'} (x)$ , we rewrite $f^{'} (x)$ as a single fraction:
$f^{'} (x) = \frac{4}{3} x^{1 / 3} - \frac{4}{3} x^{- 2 / 3} = \frac{4 x^{1 / 3}}{3} - \frac{4}{3 x^{2 / 3}} = \frac{4 x^{1 / 3}}{3} \cdot \frac{x^{2 / 3}}{x^{2 / 3}} - \frac{4}{3 x^{2 / 3}} = \frac{4 x}{3 x^{2 / 3}} - \frac{4}{3 x^{2 / 3}} = \frac{4 x - 4}{3 x^{2 / 3}} .$
Unlike the derivative in Example 6.3.1, $f^{'} (x) = \frac{4 x - 4}{3 x^{2 / 3}}$ is undefined when $3 x^{2 / 3} = 0$ , that is, when $x equals 0$ , so we need to record this on our sign diagram with the customary `‽.'
Next, we solve $f^{'} (x) = \frac{4 x - 4}{3 x^{2 / 3}} = 0$ to get $4 x - 4 = 0$ or $x equals 1$ . The usual machinations produces the sign diagram for $f^{'} (x)$ below on the left. We interpret what this means for $f$ below on the right.
Figure 6.58
Figure 6.59
We get $f$ is decreasing for $x less than 0$ as well as from $0 less than x less than 1$ . Since $0$ is in the domain of $f$ , we splice the two intervals together so $f$ is decreasing from $open parenthesis minus ∞, 1 close parenthesis$ . We see $f$ is increasing from $open parenthesis 1, ∞ close parenthesis$ .
We note that $f$ satisfies the conditions of Theorem 6.5 since $f$ is continuous everywhere and $f^{'}$ exists for all $x not equal to 0$ . Since $f$ changes from decreasing just to the left of $x equals 1$ to increasing just to the right of $x equals 1$ , the graph of $f$ has a local minimum at $x equals 1$ . The local minimum value is $f (1) = {(1)}^{4 / 3} - 4 {(1)}^{1 / 3} = - 3$ .
What is happening at $x equals 0$ ? Since $f^{'} (x)$ doesn't change sign on either side of $0$ , the graph of $f$ doesn't have a local extremum there. The sign diagram indicates $f$ is decreasing through that point. A quick check using desmos reveals `unsual steepness' at $x equals 0$ , a phenomenon which is called a vertical tangent. This means the function locally resembles a vertical line.⁵
Figure 6.60

Concavity and the Second Derivative

In section Section 4.2, we introduced the notion of concavity. In that section, we described curves as being concave up over an interval if it resembles a portion of a ` $⌣$ ' shape and concave down over an interval if resembles part of a ` $⌢$ ' shape. Now that we've had some exposure to Calculus, we can more precisely define these notions.

If we take the time to study a generic concave up curve, the ` $⌣$ ' shape can be divided into a decreasing and increasing arc:

slopes are increasing towards $0$

slopes are increasing away from $0$

In both of these cases, the slopes of the tangent line are increasing.

Likewise, we can dissect a generic ` $⌢$ ' shape curve into an increasing and decreasing arc:

slopes are decreasing towards $0$

slopes are decreasing away from $0$

Here, the slopes of the tangent line are decreasing.

We know from Theorem 6.4 that the derivative of a function can tell us where that function is increasing and decreasing. Since the function which gives us the slopes of tangent lines is the derivative, $f^{'} (x)$ , we could use the derivative of $f^{'} (x)$ to determine where the slopes of the tangent lines were increasing and decreasing. This leads us to define the second derivative, $f^{''} (x)$ as the derivative of $f^{'} (x)$ .

We present the following theorem without proof, but hopefully sufficiently motivated.

Theorem

Suppose $f$ is twice differentiable on an open interval $I$ :

If $f double prime ⁢ open parenthesis x close parenthesis greater than 0$ for all $x$ in $I$ , then slopes are increasing and $f$ is concave up on $I$ .
If $f double prime ⁢ open parenthesis x close parenthesis less than 0$ for all $x$ in $I$ , then slopes are decreasing and $f$ is concave down on $I$ .

Example 3

Let $f (x) = x^{3} - 3 x^{2} - 9 x + 5$ . Use the fact that $f^{''} (x) = 6 x - 6$ to find the intervals over which the graph of $f$ is concave up and concave down.

To analyze the concavity of the graph of $f$ , we need to make a sign diagram for $f^{''} (x)$ .

Solving $f^{''} (x) = 6 x - 6 = 0$ gives $x equals 1$ . We find $f^{''} (0) = 6 (0) - 6 = (-)$ and $f^{''} (2) = 6 (2) - 6 = (+)$ .

We have our sign diagram below on the left and our interpretation below on the right.

Figure 6.65

Figure 6.66

We find $f$ is concave down on $open parenthesis minus ∞, 1 close parenthesis$ and concave up on $open parenthesis 1, ∞ close parenthesis$ .

At $x equals 1$ , the concavity changes. We find $f (1) = {(1)}^{3} - 3 {(1)}^{2} - 9 (1) + 5 = - 6$ and we call the point $open parenthesis 1, minus 6 close parenthesis$ an inflection point. In this case since the concavity changes from concave down to concave up, the point $open parenthesis 1, minus 6 close parenthesis$ is the point on the graph of $y = f (x)$ were the slopes stop decreasing and start to increase.

A quick check using desmos confirms our results.

Figure 6.67

Note that we can use concavity to help us distinguish local extrema.

For the function above, both $f^{'} (- 1) = 0$ and $f prime ⁢ open parenthesis 3 close parenthesis equals 0$ . Note that $f double prime ⁢ open parenthesis minus 1 close parenthesis less than 0$ which means $f$ is concave down there. This forces $f$ to have a local maximum at $open parenthesis minus 1, 10 close parenthesis$ . Likewise, $f double prime ⁢ open parenthesis 3 close parenthesis greater than 0$ which means $f$ is concave up there. This forces $f$ to have a local minimum at $open parenthesis 3, minus 22 close parenthesis$ . We generalize this observation below.

Theorem

The Second⁶ Derivative Test for Local Extrema: Suppose $f$ is differentiable on an open interval $I$ containing $c$ and $f prime ⁢ open parenthesis c close parenthesis equals 0$ :

If $f double prime ⁢ open parenthesis c close parenthesis greater than 0$ then $f$ has a local minimum at $x = c$ .
If $f double prime ⁢ open parenthesis c close parenthesis less than 0$ then $f$ has a local maximum at $x = c$ .
If $f double prime ⁢ open parenthesis c close parenthesis equals 0$ then the test is inconclusive. $f$ may or may not have a local extremum at $x = c$ .
(In this case, we would appeal to the first derivative test.)

Example 4

Let $f (x) = x^{4 / 3} - 4 x^{1 / 3}$ . Use the fact that $f^{''} (x) = \frac{4}{9} x^{- 2 / 3} + \frac{8}{9} x^{- 5 / 3}$ to help you find:

the open intervals over which the graph of $f$ is concave up and concave down.
the inflection points in the graph.

As in Example 6.3.2, our first step is to rewrite $f^{''} (x) = \frac{4}{9} x^{- 2 / 3} + \frac{8}{9} x^{- 5 / 3}$ as a single fraction:
$f^{''} (x) = \frac{4}{9} x^{- 2 / 3} + \frac{8}{9} x^{- 5 / 3} = \frac{4}{9 x^{2 / 3}} + \frac{8}{9 x^{5 / 3}} = \frac{4}{9 x^{2 / 3}} \cdot \frac{x^{3 / 3}}{x^{3 / 3}} + \frac{8}{9 x^{5 / 3}} = \frac{4 x}{9 x^{5 / 3}} + \frac{8}{9 x^{5 / 3}} = \frac{4 x + 8}{9 x^{5 / 3}}$
We see $f^{''} (x) = \frac{4 x + 8}{9 x^{5 / 3}}$ is undefined when $9 x^{5 / 3} = 0$ , that is, when $x equals 0$ .
Solving $f^{''} (x) = \frac{4 x + 8}{9 x^{5 / 3}} = 0$ gives $4 x + 8 = 0$ so $x = - 2$ .
Going through the usual routine, we obtain our sign diagram for $f^{''} (x)$ is below.
Figure 6.68
Figure 6.69
We see $f$ is concave up on $(- \infty, - 2)$ and again from $open parenthesis 0, ∞ close parenthesis$ . $f$ is concave down on $open parenthesis minus 2, 0 close parenthesis$ .
Since $f$ changes concavity at both $x = - 2$ and $x equals 0$ , we have inflection point at both of these values.
We find: $f (- 2) = {(- 2)}^{4 / 3} - 4 {(- 2)}^{1 / 3} = 2 {(2)}^{1 / 3} + 4 {(2)}^{1 / 3} = 6 {(2)}^{1 / 3}$ . So $(- 2, 6 {(2)}^{1 / 3})$ is one inflection point. When $x equals 0$ , $f (0) = {(0)}^{4 / 3} - 4 {(0)}^{1 / 3} = 0$ , so $open parenthesis 0, 0 close parenthesis$ is the other inflection point.
Checking with desmos, it's not apparent that the graph of $y = f (x)$ is concave up for $x less than minus 2$ . We invite the reader to graph $f$ and zoom out to see that characteristic of the graph.
Figure 6.70

Our last example offers a twist on these sorts of curve-sketching problems.

Example 5

Below is the graph of the derivative of a function. Assume as $x \to \pm \infty$ , $f prime ⁢ open parenthesis x close parenthesis → minus ∞$ .

Figure 6.71

The graph of $y = f^{'} (x)$

Use the graph of $y = f^{'} (x)$ to determine the open intervals where $f$ is increasing and decreasing.
Find the $x$ -coordinates of the local extrema.
Use the graph of $y = f^{'} (x)$ make a sign diagram for $y = f^{''} (x)$ .
List the open intervals over which the graph of $f$ is concave up and concave down.
Find the $x$ -coordinates of the inflection points.
Sketch a possible graph of $y = f (x)$ .

Recall from algebra, the solutions to $f prime ⁢ open parenthesis x close parenthesis less than 0$ are the $x$ -values where the graph of $y = f^{'} (x)$ is below the $x$ -axis. This happens on the intervals $open parenthesis minus ∞, 0 close parenthesis$ and $open parenthesis 4, ∞ close parenthesis$ , so this means $f$ is decreasing here.
Likewise, the solutions to $f prime ⁢ open parenthesis x close parenthesis greater than 0$ are the $x$ -values where $y = f^{'} (x)$ is above the $x$ -axis. This happens on the interval $open parenthesis 0, 4 close parenthesis$ , so $f$ is increasing here.
Since $f$ goes from decreasing to the left of $x equals 0$ to increasing to the right of $x equals 0$ , $f$ has a local minimum at $x equals 0$ . Since $f$ goes from increasing to the left of $x equals 4$ to decreasing to the right of $x equals 4$ , $f$ has a local maximum at $x equals 4$ .
Since $f^{''} (x)$ is the derivative of $f^{'} (x)$ , we know $f double prime ⁢ open parenthesis x close parenthesis greater than 0$ on $open parenthesis minus ∞, 2 close parenthesis$ since $f^{'} (x)$ is increasing there. We see $f double prime ⁢ open parenthesis 2 close parenthesis equals 0$ since $f^{'} (x)$ is locally flat at $open parenthesis 2, 4 close parenthesis$ . Lastly, we see $f double prime ⁢ open parenthesis x close parenthesis less than 0$ on $open parenthesis 2, ∞ close parenthesis$ since $f^{'} (x)$ is decreasing there. We put all this together in a sign diagram below.
Figure 6.72
Figure 6.73
We have $f$ is concave up on $open parenthesis minus ∞, 2 close parenthesis$ and concave down on $open parenthesis 2, ∞ close parenthesis$ .
Since $f$ changes concavity at $x equals 2$ , there is an inflection point there.
A plausible graph of $y = f (x)$ is below. We cannot determine any $y$ -coordinates (why not?)
Figure 6.74
A possible graph of $y = f (x)$

Exercises

In Exercises -, use the given function $f$ and its (first) derivative $f^{'}$ to help you find:

the open intervals over which $f$ is increasing, decreasing, and constant.
the local extrema.

Check your answers using a graphing utility.

$f (x) = 2 x^{3} - 3 x^{2} - 12 x + 1$ , $f^{'} (x) = 6 x^{2} - 6 x - 12$
$f (x) = \frac{10 x}{x^{2} + 1}$ , $f^{'} (x) = \frac{10 - 10 x^{2}}{{(x^{2} + 1)}^{2}}$
$f (x) = x \sqrt[3]{x - 2}$ , $f^{'} (x) = \frac{4 x - 6}{3 {(x - 2)}^{\frac{2}{3}}}$

In Exercises -, use the given function $f$ and its second derivative $f^{''}$ to help you find:

the open intervals over which the graph of $f$ is concave up and concave down.
the inflection points in the graph.

Check your answers using a graphing utility.

$f (x) = 2 x^{3} - 3 x^{2} - 12 x + 1$ , $f^{''} (x) = 12 x - 6$
$f (x) = \frac{10 x}{x^{2} + 1}$ , $f^{''} (x) = \frac{20 x^{3} - 60 x}{{(x^{2} + 1)}^{3}}$
$f (x) = x \sqrt[3]{x - 2}$ , $f^{''} (x) = \frac{4 (x - 3)}{9 {(x - 2)}^{\frac{5}{3}}}$
If $a not equal to 0$ , we showed in Exercise in Section 6.2 that if $f (x) = a x^{2} + b x + c$ , then $f^{'} (x) = 2 a x + b$ . Solving $f prime ⁢ open parenthesis x close parenthesis equals 0$ produced $x = - \frac{b}{2 a}$ , the $x$ -coordinate of the vertex of the parabola $y = f (x)$ . This Exercise shows this is part of a pattern.
1. If $a not equal to 0$ , show the $x$ -coordinate of the $x$ -intercept of the graph of $y = a x + b$ is $x = - \frac{b}{a} = - \frac{b}{1 a}$ .
2. If $a not equal to 0$ , for $f (x) = a x^{3} + b x^{2} + c x + d$ it turns out that $f^{''} (x) = 6 a x + 2 b$ . Show $x = - \frac{b}{3 a}$ is the $x$ -coordinate of the inflection point of the graph of $y = f (x)$ .
In Exercise in Section 5.2, we observed that average cost appeared to be minimized when average cost was approximately equal to marginal cost. In this Exercise, we use Calculus and the tools from this section to show this.
Recall if $C (x)$ is the cost to produce $x$ items, the average cost is defined as $\bar{C} (x) = \frac{C (x)}{x}$ , $x greater than 0$ , is the cost per item.
1. It turns out that ${\bar{C}}^{'} (x) = \frac{x C^{'} (x) - C (x)}{x^{2}}$ . Show $C bar prime ⁢ open parenthesis x close parenthesis equals 0$ when $C^{'} (x) = \bar{C} (x)$ .
2. It turns out that ${\bar{C}}^{''} (x) = \frac{x^{2} C^{''} (x) - 2 x C^{'} (x) + 2 C (x)}{x^{3}}$ .
  Show we can rewrite this as: ${\bar{C}}^{''} (x) = \frac{x C^{''} (x) - 2 C^{'} (x) + 2 \bar{C} (x)}{x^{2}}$ .
3. Show that when $C^{'} (x) = \bar{C} (x)$ , then ${\bar{C}}^{''} (x) = \frac{C^{''} (x)}{x}$ .
4. It is usually assumed in most economic settings that for cost functions,⁷ $C double prime ⁢ open parenthesis x close parenthesis greater than 0$ . Use this and your results from parts and to prove that a minimum is produced when $C^{'} (x) = \bar{C} (x)$ .
  NOTE: In Exercise in Section 6.2, we saw how $C^{'} (x)$ can be used to approximate the marginal cost, $M C (x)$ , so we have established that in order to minimize average cost, we should look where the average cost matches the marginal cost.
The complete graph of $y = f (x)$ is shown below. Use the graph to answer the following questions.
NOTE: Assume $open parenthesis 2.029, 1.82 close parenthesis$ is a local maximum and that $open parenthesis 1.077, 0.948 close parenthesis$ is an inflection point.
Figure 6.75
1. Determine the $x$ -values where:
  $f ⁢ open parenthesis x close parenthesis equals 0$ :
  $f prime ⁢ open parenthesis x close parenthesis equals 0$ :
2. List the open intervals over which:
  $f ⁢ open parenthesis x close parenthesis greater than 0$ :
  $f ⁢ open parenthesis x close parenthesis less than 0$ :
  $f prime ⁢ open parenthesis x close parenthesis greater than 0$ :
  $f prime ⁢ open parenthesis x close parenthesis less than 0$ :
  $f double prime ⁢ open parenthesis x close parenthesis greater than 0$ :
  $f double prime ⁢ open parenthesis x close parenthesis less than 0$ :
Below is the graph of $y = f^{'} (x)$ for a continuous function $f$ .
Using Example 6.3.5 as a guide, sketch a probable graph of $y = f (x)$ .
Figure 6.76
The graph below was taken from https://ohiohospitals.org/covid19data on January 28th, 2021:
Figure 6.77
With help from your classmate, highlight and label one segment on the graph which (roughly) represents the following scenarios.
For brevity, we'll use `patients' to mean `inpatient COVID positive patients' and `the rate of change' to mean `the rate of change of inpatient COVID positive patients with respect to time.'
1. The number of patients is decreasing and the rate of change is decreasing. (Label this `a.')
2. The number of patients is decreasing and the rate of change is increasing. (Label this `b.')
3. The number of patients is increasing and the rate of change is increasing. (Label this `c.')
4. The number of patients is increasing and the rate of change is decreasing. (Label this `d.')
5. Discuss with your classmates what the phrase `flatten the curve' could mean in terms of first and second derivatives. (See below for an illustration.)
  Figure 6.78

Answers

increasing: $(- \infty, - 1)$ , $open parenthesis 2, ∞ close parenthesis$ ; decreasing: $open parenthesis minus 1, 2 close parenthesis$ ; local max: $open parenthesis minus 1, 8 close parenthesis$ ; local min: $open parenthesis 2, minus 19 close parenthesis$ .
increasing: $open parenthesis minus 1, 1 close parenthesis$ ; decreasing: $(- \infty, - 1)$ , $open parenthesis 1, ∞ close parenthesis$ ; local max: $open parenthesis 1, 5 close parenthesis$ ; local min: $(- 1, - 5)$ .
increasing: $(\frac{3}{2}, \infty)$ ; decreasing: $(- \infty, \frac{3}{2})$ ; local (absolute) min: $(\frac{3}{2}, - \frac{3}{2 \sqrt[3]{2}})$
concave up: $(\frac{1}{2}, \infty)$ ; concave down: $(- \infty, \frac{1}{2})$ ; inflection point: $(\frac{1}{2}, \frac{11}{2})$
concave up: $(- \sqrt{3}, 0)$ , $open parenthesis the square root of 3, ∞ close parenthesis$ ; concave down: $(- \infty, - \sqrt{3})$ , $open parenthesis 0, the square root of 3 close parenthesis$ ; inflection points: $(- \sqrt{3}, - \frac{5 \sqrt{3}}{2})$ , $open parenthesis 0, 0 close parenthesis$ , $(\sqrt{3}, \frac{5 \sqrt{3}}{2})$
concave up: $open parenthesis minus ∞, 2 close parenthesis$ , $open parenthesis 3, ∞ close parenthesis$ ; concave down: $open parenthesis 2, 3 close parenthesis$ ; inflection points: $open parenthesis 2, 0 close parenthesis$ , $open parenthesis 3, 3 close parenthesis$ .
1. To find the $x$ -intercept, we set $a x + b = 0$ and get $x = - \frac{b}{a}$ provided $a not equal to 0$ .
2. Solving $f^{''} (x) = 6 a x + 2 b = 0$ , we get $x = - \frac{2 b}{6 a} = - \frac{b}{3 a}$ , provided $a not equal to 0$ .
  The graph of $f^{''} (x) = 6 a x + 2 b$ is a line so we know on one side of $x = - \frac{b}{3 a}$ , $f double prime ⁢ open parenthesis x close parenthesis greater than 0$ and on the other side, $f double prime ⁢ open parenthesis x close parenthesis less than 0$ .
  Hence, the graph of the original function $y = f (x)$ changes concavity at $x = - \frac{b}{3 a}$ .
1. To solve ${\bar{C}}^{'} (x) = \frac{x C^{'} (x) - C (x)}{x^{2}} = 0$ , we set the numerator, $x C^{'} (x) - C (x) = 0$ . We get $x C^{'} (x) = C (x)$ so $C^{'} (x) = \frac{C (x)}{x} = \bar{C} (x)$ .
2. Divide both numerator and denominator of ${\bar{C}}^{''} (x) = \frac{x^{2} C^{''} (x) - 2 x C^{'} (x) + 2 C (x)}{x^{3}}$ by $x$ :
  ${\bar{C}}^{''} (x) = \frac{x C^{''} (x) - 2 C^{'} (x) + 2 \frac{C (x)}{x}}{x^{2}}$ and substitute $\frac{C (x)}{x} = \bar{C} (x)$ .
3. If $C^{'} (x) = \bar{C} (x)$ , then:
  ${\bar{C}}^{''} (x) = \frac{x C^{''} (x) - 2 C^{'} (x) + 2 \bar{C} (x)}{x^{2}} = \frac{x C^{''} (x) - 2 \bar{C} (x) + 2 \bar{C} (x)}{x^{2}} = \frac{x C^{''} (x)}{x^{2}} = \frac{C^{''} (x)}{x}$ .
4. When $C^{'} (x) = \bar{C} (x)$ , we have that $C bar prime ⁢ open parenthesis x close parenthesis equals 0$ and $C bar double prime ⁢ open parenthesis x close parenthesis greater than 0$ . By the Second Derivative Test for Local Extrema, Theorem 6.7, the average cost $\bar{C} (x)$ has a minimum when $C^{'} (x) = \bar{C} (x)$
1. $f ⁢ open parenthesis x close parenthesis equals 0$ : $x equals 0, π$
  $f prime ⁢ open parenthesis x close parenthesis equals 0$ : $x equals 2.029$ ( $f_{+}^{'} (0) = 0$ ., too.)
2. $f ⁢ open parenthesis x close parenthesis greater than 0$ : $open parenthesis 0, π close parenthesis$
  $f ⁢ open parenthesis x close parenthesis less than 0$ : none
  $f prime ⁢ open parenthesis x close parenthesis greater than 0$ : $open parenthesis 0, 2.029 close parenthesis$
  $f prime ⁢ open parenthesis x close parenthesis less than 0$ : $open parenthesis 2.029, π close parenthesis$
  $f double prime ⁢ open parenthesis x close parenthesis greater than 0$ : $open parenthesis 0, 1.077 close parenthesis$
  $f double prime ⁢ open parenthesis x close parenthesis less than 0$ : $open parenthesis 1.077, π close parenthesis$ .
Answers vary. below is a sketch of the key features that should be included:
Figure 6.79

We've defined derivatives as two-sided limits, so an open interval here guarantees enough `room' on either side of any given number to take such a limit. ↩
which Carl thinks is the actual `Fundamental Theorem of Calculus' since it relates local and global behavior … ↩
Why `First'? Stay tuned … ↩
Recall this means $f prime ⁢ open parenthesis c close parenthesis equals 0$ or $f^{'} (c)$ does not exist. ↩
See Example 4.1.2 for another such example and discussion. ↩
Now you know why we titled Theorem 6.5 the `First' Derivative Test for Local Extrema. ↩
Can you think of reasons why? ↩

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.