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## Section3.4Using Derivatives to Describe Families of Functions

###### Motivating Questions
• Given a family of functions that depends on one or more parameters, how does the shape of the graph of a typical function in the family depend on the value of the parameters?

• How can we construct first and second derivative sign charts of functions that depend on one or more parameters while allowing those parameters to remain arbitrary constants?

Mathematicians are often interested in making general observations, say by describing patterns that hold in a large number of cases. Think about the Pythagorean Theorem: it doesn't tell us something about a single right triangle, but rather a fact about every right triangle. In the next part of our studies, we use calculus to make general observations about families of functions that depend on one or more parameters. People who use applied mathematics, such as engineers and economists, often encounter the same types of functions where only small changes to certain constants occur. These constants are called parameters. Figure3.38The graph of $y=f(t)\text{,}$ where $f(t) = a \sin(b(t-c)) + d\text{,}$ based on parameters $a\text{,}$ $b\text{,}$ $c\text{,}$ and $d\text{.}$

You are already familiar with certain families of functions. For example, $f(t) = a \sin(b(t-c)) + d$ is a stretched and shifted version of the sine function with amplitude $a\text{,}$ period $\frac{2\pi}{b}\text{,}$ phase shift $c\text{,}$ and vertical shift $d\text{.}$ We know that $a$ affects the size of the oscillation, $b$ the rapidity of oscillation, and $c$ where the oscillation starts, as shown above in Figure3.38, while $d$ affects the vertical positioning of the graph.

Here is another example: every function of the form $y = mx + b$ is a line with slope $m$ and $y$-intercept $(0,b)\text{.}$ The value of $m$ affects the line's steepness, and the value of $b$ situates the line vertically on the coordinate axes. These two parameters describe all possible non-vertical lines.

For other less familiar families of functions, we can use calculus to discover where key behavior occurs: e.g. where members of the family are increasing or decreasing, concave up or concave down, where relative extrema occur, and more, all in terms of the parameters involved. To get started, we revisit a common collection of functions to see how calculus confirms things we already know.

###### Example3.39

Let $a\text{,}$ $h\text{,}$ and $k$ be arbitrary real numbers with $a \ne 0\text{,}$ and let $f$ be the function given by the rule $f(x) = a(x-h)^2 + k\text{.}$

1. What familiar type of function is $f\text{?}$ What information do you know about $f$ just by looking at its form? (Think about the roles of $a\text{,}$ $h\text{,}$ and $k\text{.}$)

2. Next we use some calculus to develop familiar ideas from a different perspective. To start, treat $a\text{,}$ $h\text{,}$ and $k$ as constants and compute $f'(x)\text{.}$

3. Find all critical numbers of $f\text{.}$ (These will depend on at least one of $a\text{,}$ $h\text{,}$ and $k\text{.}$)

4. Assume that $a \lt 0\text{.}$ Construct a first derivative sign chart for $f\text{.}$

5. Based on the information you've found above, classify the critical values of $f$ as maxima or minima.

Hint
1. $f$ is written in vertex form for this type of function; what does that tell you?

2. Try using the chain rule to differentiate $f$ as written (instead of expanding $f$ before taking the derivative).

3. If $c$ is a critical number of $f\text{,}$ what must be true about $f'(c)\text{?}$ Remember, there are two possibilities in general.

4. What does the graph of $y=f(x)$ look like when $a\lt 0\text{?}$

5. Use what you know prior about the shape of $y=f(x)$ to support the answer you get with calculus and the first derivative sign chart you made in (d).

1. $f$ is quadratic.

2. $f'(x)=2a(x-h)\text{.}$

3. $h$ is the only critical number of $f\text{.}$

4. For $x\lt h\text{,}$ we have $f'(x)\gt0$ and $f(x)$ increasing. For $x\gt h\text{,}$ we have $f'(x)\lt0$ and $f(x)$ decreasing.

5. $k$ is a maximum of $f\text{.}$

Solution
1. Since $f$ is a second-degree polynomial in $x\text{,}$ $f$ is a quadratic function, and the graph of $y=f(x)$ is a parabola. The vertex of this parabola is at the point $(h,k)\text{,}$ and the sign of $a$ determines whether the parabola opens upward (when $a\gt0$) or downward (when $a\lt0$).

2. Because $h$ and $k$ are constants, their derivatives are both $0$ and this simplifies the application of the sum rule in taking the derivative. Furthermore, the product rule is unnecessary because $a$ is a constant as well, so we instead make use of the constant multiple rule. Doing so while also applying the chain rule, we find that

\begin{equation*} f'(x)=2a(x-h)\text{.} \end{equation*}
3. Since $f'$ is linear with slope $2a\neq0\text{,}$ it is continuous (hence, defined everywhere) and has exactly one root (where $f'(x)=0$). Thus the function $f$ will have exactly one critical number. Indeed, we see that $f'(x)=0$ only at the point $x=h\text{.}$ Therefore, the only critical number of $f$ is $h\text{.}$ We note that this agrees with our previous assertion that the vertex of $y=f(x)$ is at the point $(h,k)\text{.}$

4. If $a\lt0\text{,}$ then our algebra knowledge tells us that the graph of $y=f(x)$ is a parabola that opens down, so we should find that $f$ is increasing for $x\lt h$ and decreasing for $x\gt h\text{.}$ Testing this intuition, we note that

• When $x\lt h\text{,}$ then $x-h\lt0$ and hence $2a(x-h)\gt0$ since $a$ also is known to be negative. Therefore, on the interval $(-\infty,h)$ we have $f'(x)\gt0$ and $f$ increasing.
• When $x\gt h\text{,}$ then $x-h\gt0$ and hence $2a(x-h)\lt0$ since $a$ is known to be negative. Therefore, on the interval $(h,\infty)$ we have $f'(x)\lt0$ and $f$ decreasing.
5. Applying the first derivative test, we see that at the point $x=h\text{,}$ the value of $f'(x)$ switches from positive to negative, and thus $k\text{,}$ the critical value corresponding to the critical number $h\text{,}$ is a maximum of $f\text{.}$ This agrees with our existing knowledge of parabolas, as we had established that this parabola opens downward.

### SubsectionDescribing Families of Functions in Terms of Parameters

Our goal is to describe the key characteristics of the overall behavior of each member of a family of functions in terms of its parameters. By finding the first and second derivatives and constructing sign charts (each of which may depend on one or more of the parameters), we can often make broad conclusions about how each member of the family will appear.

###### Example3.40

Consider the two-parameter family of functions given by $g(x) = axe^{-bx}\text{,}$ where $a$ and $b$ are positive real numbers. Fully describe the behavior of a typical member of the family in terms of $a$ and $b\text{,}$ including: the location of all critical numbers; where $g$ is increasing, decreasing, concave up, and concave down; and the long-term behavior of $g\text{.}$

Hint

Start by finding the first and second derivatives of $g\text{.}$ Be careful with your use of the product rule and chain rule.

$g$ has one critical number, at $x=\frac1b\text{.}$

$g$ is increasing on $\big(-\infty,\frac1b\big)\text{,}$ decreasing on $\big(\frac1b,\infty\big)\text{,}$ concave down on $\big(-\infty,\frac2b\big)\text{,}$ and concave up on $\big(\frac2b,\infty\big)\text{.}$

$\lim_{x\to-\infty}g(x)=-\infty$ and $\lim_{x\to\infty}g(x)=0\text{.}$

Solution

We begin by computing $g'(x)\text{.}$ By the product rule,

\begin{equation*} g'(x) = \frac{d}{dx}[ax]e^{-bx} +ax \frac{d}{dx}\left[e^{-bx}\right] \text{.} \end{equation*}

By applying the chain rule and constant multiple rule, we find that

\begin{equation*} g'(x) = ae^{-bx}+axe^{-bx}(-b)\text{.} \end{equation*}

To find the critical numbers of $g\text{,}$ we note that $g'$ is defined everywhere and proceed to solve the equation $g'(x) = 0\text{.}$ By factoring $g'(x)\text{,}$ we find

\begin{equation*} 0 = ae^{-bx}(1-bx)\text{.} \end{equation*}

Since we are given that $a \ne 0$ and we know that $e^{-bx} \ne 0$ for all values of $x\text{,}$ the only way this equation can hold is when $1-bx = 0\text{.}$ Solving for $x\text{,}$ we find $x = \frac{1}{b}\text{,}$ and this is therefore the only critical number of $g\text{.}$

Because the factor $ae^{-bx}$ is always positive, the sign of $g'$ depends on the linear factor $(1-bx)\text{,}$ which is positive for $x \lt \frac{1}{b}$ and negative for $x \gt \frac{1}{b}\text{.}$ Hence we can not only conclude that $g$ is always increasing for $x \lt \frac{1}{b}$ and decreasing for $x \gt \frac{1}{b}\text{,}$ but also that $g$ has a global maximum at $\big(\frac{1}{b}, g\big(\frac{1}{b}\big)\big)$ and no local minimum. The first derivative sign chart for $g$ is shown below in Figure3.41. Figure3.41The first derivative sign chart for $g(x) = axe^{-bx}\text{.}$

We turn next to analyzing the concavity of $g\text{.}$ With $g'(x) = ae^{-bx}-abxe^{-bx}\text{,}$ we differentiate to find that

\begin{equation*} g''(x) = ae^{-bx}(-b)-ab\left(e^{-bx}+xe^{-bx}(-b)\right)\text{.} \end{equation*}

Combining like terms and factoring, we now have

\begin{equation*} g''(x) = ab^2xe^{-bx} - 2abe^{-bx} = abe^{-bx}(bx - 2)\text{.} \end{equation*}

We observe that $abe^{-bx}$ is always positive, and thus the sign of $g''$ depends on the sign of $(bx-2)\text{,}$ which is zero when $x = \frac{2}{b}\text{.}$ Since $b$ is positive, the value of $(bx-2)$ is negative for $x \lt \frac{2}{b}$ and positive for $x \gt \frac{2}{b}\text{.}$ The sign chart for $g''$ is shown below in Figure3.42. Thus, $g$ is concave down for all $x \lt \frac{2}{b}$ and concave up for all $x \gt \frac{2}{b}\text{.}$ Figure3.42The second derivative sign chart for $g(x) = axe^{-bx}\text{.}$

Finally, we analyze the long-term behavior of $g$ by considering two limits. First, we note that

\begin{equation*} \lim_{x \to \infty} g(x) = \lim_{x \to \infty} axe^{-bx} = \lim_{x \to \infty} \frac{ax}{e^{bx}}\text{.} \end{equation*}

This limit has indeterminate form $\frac{\infty}{\infty}\text{,}$ so with our current tools, we can not evaluate the limit algebraically. We can, however, reason that $ax$ is linear while $e^{bx}$ is exponential, and conclude that because exponential growth is so much more rapid than linear growth, we expect the limiting value to be zero. A more rigorous answer will be possible following Section3.6, but for now we can also use graphical evidence10A good tool to use here is desmos.com. to support the claim that $\lim_{x\to\infty}g(x)=0\text{.}$ We do so by examining the graphs of several different functions in this family, noting that each has a horizontal asymptote of $y=0$ as $x\to\infty\text{.}$ In the other direction,

\begin{equation*} \lim_{x \to -\infty} g(x) = \lim_{x \to -\infty} axe^{-bx} = -\infty\text{,} \end{equation*}

because $ax \to -\infty$ and $e^{-bx} \to \infty$ as $x \to -\infty\text{.}$ Hence, as we move left on its graph, $g$ decreases without bound, while as we move to the right, $g(x) \to 0\text{.}$

All of this information now helps us produce the graph of a typical member of this family of functions without using a graphing utility (and without choosing particular values for $a$ and $b$), as shown in Figure3.43 below. Figure3.43The graph of $g(x) = axe^{-bx}\text{.}$

Note that the value of $b$ controls the horizontal location of the global maximum and the inflection point, as neither depends on $a\text{.}$ The value of $a$ affects the vertical stretch of the graph. For example, the global maximum occurs at the point $\big(\frac{1}{b}, g\big(\frac{1}{b}\big)\big) = \big(\frac{1}{b}, \frac{a}{b}e^{-1}\big)\text{,}$ so the larger the value of $a\text{,}$ the greater the value of the global maximum.

The work we've completed in Example3.40 can often be replicated for other families of functions that depend on parameters. Normally we are most interested in determining all critical numbers, a first derivative sign chart, a second derivative sign chart, and the limit of the function as $x \to \infty\text{.}$ Throughout, we prefer to work with the parameters as arbitrary constants. In addition, we can experiment with some particular values of the parameters present to reduce the algebraic complexity of our work. What follows are several key examples where we see that the values of the parameters substantially affect the behavior of individual functions within a given family.

###### Example3.44

Consider the family of functions defined by $p(x) = x^3 - ax\text{,}$ where $a \ne 0$ is an arbitrary constant.

1. Find $p'(x)$ and determine the critical numbers of $p\text{.}$ How many critical numbers does $p$ have?

2. Construct a first derivative sign chart for $p\text{.}$ What can you say about the overall behavior of $p$ if the constant $a$ is positive? Why? What if the constant $a$ is negative? In each case, describe the relative extrema of $p\text{.}$

3. Find $p''(x)$ and construct a second derivative sign chart for $p\text{.}$ What does this tell you about the concavity of $p\text{?}$ What role does $a$ play in determining the concavity of $p\text{?}$

4. Without using a graphing utility, sketch and label typical graphs of $p(x)$ for the cases where $a\gt 0$ and $a \lt 0\text{.}$ Label all inflection points and local extrema.

5. Finally, use a graphing utility to test your observations above by entering and plotting the function $p(x) = x^3 - ax$ for at least four different values of $a\text{.}$ Write several sentences to describe your overall conclusions about how the behavior of $p$ depends on $a\text{.}$

Hint
1. When solving $p'(x) = 0\text{,}$ think about two possible cases: when $a\gt 0$ and when $a \lt 0\text{.}$

2. Remember that any quadratic function can be zero at most two times. How does the graph of $y = 3x^2 - a$ look?

3. Don't forget that $\frac{d}{dx}[a] = 0\text{.}$

4. Think about how the graph of a typical cubic polynomial behaves.

1. $p$ has two critical numbers $\left(x = \pm \sqrt{\frac{a}{3}}\right)$ whenever $a \gt 0$ and no critical numbers when $a \lt 0\text{.}$

2. When $a \lt 0\text{,}$ $p$ is always increasing and has no relative extreme values. When $a\gt 0\text{,}$ $p$ has a relative maximum at $x = -\sqrt{\frac{a}{3}}$ and a relative minimum at $x = +\sqrt{\frac{a}{3}}\text{.}$

3. $p$ is concave down for $x \lt 0$ and $p$ is concave up for $x\gt 0\text{,}$ making $x = 0$ an inflection point.

4. On the left, $a\lt0\text{;}$ on the right, $a\gt0\text{.}$ 5. See desmos.com for additional graph samples. As $a$ decreases towards $-\infty\text{,}$ the graph of $y=p(x)$ looks more and more linear. As $a$ increases towards $\infty\text{,}$ the two bumps on the graph get more pronounced.

Solution
1. We first note that $p'(x) = 3x^2 - a\text{,}$ so to find critical numbers we set $p'(x) = 0$ and solve for $x\text{.}$ This leads to the equation $3x^2 - a = 0\text{,}$ which implies

\begin{equation*} x^2 = \frac{a}{3}\text{.} \end{equation*}

If $a \gt 0\text{,}$ then the solutions to this equation are $x = \pm \sqrt{\frac{a}{3}}\text{;}$ if $a \lt 0\text{,}$ then the equation has no solution. Hence, $p$ has two critical numbers $\left(x = \pm \sqrt{\frac{a}{3}}\right)$ whenever $a \gt 0$ and no critical numbers when $a \lt 0\text{.}$

2. For the case when $a \lt 0\text{,}$ we observe that $p'(x) = 3x^2 - a$ is positive for every value of $x\text{,}$ and thus $p$ is always increasing and has no relative extreme values. (There are no critical numbers to place on the first derivative sign chart, and $p'$ is always positive.) For the case when $a\gt 0\text{,}$ we observe that $p'(x) = 3x^2 - a$ is a concave up parabola with zeros at $x = -\sqrt{\frac{a}{3}}$ and $x = +\sqrt{\frac{a}{3}}\text{.}$ It follows that for $x \lt -\sqrt{\frac{a}{3}}\text{,}$ $p'(x)\gt 0$ (so $p$ is increasing); for $-\sqrt{\frac{a}{3}} \lt x \lt \sqrt{\frac{a}{3}}\text{,}$ $p'(x)\gt 0$ (so $p$ is decreasing); and for $x\gt \sqrt{\frac{a}{3}}\text{,}$ $p'(x)\gt 0$ (so $p$ is again increasing). In this situation, we see that $p$ has a relative maximum at $x = -\sqrt{\frac{a}{3}}$ and a relative minimum at $x = +\sqrt{\frac{a}{3}}\text{.}$

3. Since $p'(x) = 3x^2 - a$ and $a$ is constant, it follows that $p''(x) = 6x\text{.}$ Note that $p''(x) = 0$ when $x = 0$ and that $p''(x) \lt 0$ for $x \lt 0$ and $p''(x)\gt 0$ for $x\gt 0\text{.}$ Hence $p$ is concave down for $x \lt 0$ and $p$ is concave up for $x\gt 0\text{,}$ making $x = 0$ an inflection point.

4. Below, we show the two possible situations. At left, for the case when $a \lt 0$ and $p$ is always increasing with an inflection point at $x = 0\text{,}$ and at right for when $a\gt 0$ and $p$ has a relative maximum at $x = -\sqrt{\frac{a}{3}}$ and a relative minimum at $x = +\sqrt{\frac{a}{3}}\text{,}$ again with an inflection point at $x = 0\text{.}$ Note, too, that $p$ has its $x$-intercepts at $x = \pm \sqrt{a}\text{.}$ 5. See desmos.com for additional graph samples. We note that as $a$ gets increasingly negative, the graph of $y=p(x)$ gets steeper near the origin, tending more and more toward a linear shape (which it never achieves). Conversely, as $a$ gets increasingly positive, the two bumps on the graph get more and more pronounced.

###### Example3.45

Consider the two-parameter family of functions of the form $h(x) = a(1-e^{-bx})\text{,}$ where $a$ and $b$ are positive real numbers.

1. Find the first derivative and the critical numbers of $h\text{.}$ Use these to construct a first derivative sign chart and determine for which values of $x$ the function $h$ is increasing and decreasing.

2. Find the second derivative and build a second derivative sign chart. For which values of $x$ is a function in this family concave up? Concave down?

3. What is the value of $\lim_{x \to \infty} a(1-e^{-bx})\text{?}$ What about $\lim_{x \to -\infty} a(1-e^{-bx})\text{?}$

4. How does changing the value of $b$ affect the shape of the curve?

5. Without using a graphing utility, sketch the graph of a typical member of this family. Write several sentences to describe the overall behavior of a typical function $h$ and how this behavior depends on $a$ and $b\text{.}$

Hint
1. Expand to write $h(x) = a - ae^{-bx}$ before differentiating.

2. Remember that $e^{-bx}$ is always positive (hence, never zero), regardless of the value of $x\text{.}$

3. Recall that $e^{-x} \to 0$ as $x \to \infty$ and $e^{-x} \to \infty$ as $x \to -\infty\text{.}$

4. Consider how $b$ affects the value of $h'(x)\text{.}$

5. Use your work in (a)-(d).

1. $h$ is an always increasing function.

2. $h$ is always concave down.

3. $\lim_{x \to \infty} a(1-e^{-bx}) = a\text{,}$ and $\lim_{x \to -\infty} a(1-e^{-bx}) = -\infty\text{.}$

4. If $b$ is large and $x$ is close to zero, $h'(x)$ is relatively large near $x = 0\text{,}$ and the curve's slope will quickly approach zero as $x$ increases. If $b$ is small, the graph is less steep near $x = 0$ and its slope goes to zero less quickly as $x$ increases.

5. Solution
1. Since $h(x) = a - ae^{-bx}\text{,}$ we have by the constant multiple and chain rules that

\begin{equation*} h'(x) = -ae^{-bx}(-b) = abe^{-bx}\text{.} \end{equation*}

Since $a$ and $b$ are positive constants and $e^{-bx} \gt 0$ for all $x\text{,}$ we see that $h'(x)$ is never zero (nor undefined), and indeed $h'(x) \gt 0$ for all $x\text{.}$ Hence $h$ is an always increasing function.

2. Because $h'(x) = abe^{-bx}\text{,}$ we have that $h''(x) = abe^{-bx}(-b) = -ab^2e^{-bx}\text{.}$ As with $h'\text{,}$ we recognize that $a\text{,}$ $b^2\text{,}$ and $e^{-bx}$ are always positive, and thus $h''(x) = -ab^2e^{-bx} \lt 0$ for all values of $x\text{,}$ making $h$ always concave down.

3. As $x \to \infty\text{,}$ $e^{-bx} \to 0\text{.}$ Thus,

\begin{equation*} \lim_{x \to \infty} a(1-e^{-bx}) = \lim_{x \to \infty} (a - ae^{-bx}) = a - 0 = a\text{.} \end{equation*}

This shows that $h$ has a horizontal asymptote at $y = a$ as we move rightward on its graph. As $x \to -\infty\text{,}$ $e^{-bx} \to \infty\text{.}$ Thus,

\begin{equation*} \lim_{x \to -\infty} a(1-e^{-bx}) = \lim_{x \to -\infty} (a - ae^{-bx}) = -\infty\text{.} \end{equation*}
4. Noting that $h'(x) = abe^{-bx}\text{,}$ we see that if we consider different values of $b\text{,}$ the slope of the graph changes. If $b$ is large and $x$ is close to zero, $h'(x) \approx ab$ (since $e^0 = 1$), so $h'(x)$ is relatively large near $x = 0\text{.}$ At the same time, for large $b\text{,}$ $e^{-bx}$ approaches zero quickly as $x$ increases, so the curve's slope will quickly approach zero as $x$ increases. If $b$ is small, the graph is less steep near $x = 0$ and its slope goes to zero less quickly as $x$ increases.

5. Observing that $h(0) = 0$ and $\lim_{x \to \infty} h(x) = a\text{,}$ along with the facts that $h$ is always increasing and always concave down, we see that a typical member of this family looks like the following graph. ###### Example3.46

Let $L(t) = \frac{A}{1+ce^{-kt}}\text{,}$ where $A\text{,}$ $c\text{,}$ and $k$ are all positive real numbers.

1. Observe that we can equivalently write $L(t) = A(1+ce^{-kt})^{-1}\text{.}$ Find $L'(t)$ and explain why $L$ has no critical numbers. Is $L$ always increasing or always decreasing? Why?

2. Given the fact that

\begin{equation*} L''(t) = Ack^2e^{-kt} \frac{ce^{-kt}-1}{(1+ce^{-kt})^3}\text{,} \end{equation*}

find all values of $t$ such that $L''(t) = 0$ and then construct a second derivative sign chart. For which values of $t$ is a function in this family concave up? Concave down?

3. What is the value of $\lim_{t \to \infty} \frac{A}{1+ce^{-kt}}\text{?}$ $\lim_{t \to -\infty} \frac{A}{1+ce^{-kt}}\text{?}$

4. Find the value of $L(t)$ at the inflection point found in (b).

5. Without using a graphing utility, sketch the graph of a typical member of this family. Write several sentences to describe the overall behavior of a typical function $L$ and how this behavior depends on the values of the parameters $A\text{,}$ $c\text{,}$ and $k\text{.}$

6. Explain why it is reasonable to think that the function $L(t)$ models the growth of a population over time, in a setting where the surrounding environment cannot support a population larger than $A\text{.}$

Hint
1. Use the chain rule, treating $A\text{,}$ $c\text{,}$ and $k$ as constants.

2. Note that the only way $L''(t) = 0$ is if $ce^{-kt}-1 = 0\text{.}$

3. Remember that $e^{-t} \to 0$ as $t \to \infty$ and $e^{-t} \to \infty$ as $t \to -\infty\text{.}$

4. Note that at the inflection point $t_0\text{,}$ we have $ce^{-kt_0}=1\text{.}$

5. Think about horizontal asymptotes, where $L$ is increasing and decreasing, and concavity.

6. Intuitively, if an environment can only support a population of a certain size, how should the population be growing when it is well below the limit? When it is approaching the limit? Does this intuition describe what you found to be true for $L\text{?}$

1. $L$ is an always increasing function.

2. $L$ is concave up for all $t \lt -\frac{1}{k} \ln \big(\frac{1}{c}\big)$ and concave down for all $t\gt-\frac1k\ln\big(\frac1c\big)\text{.}$

3. $\lim_{t \to \infty} \frac{A}{1+ce^{-kt}} = A\text{,}$ and

\begin{equation*} \lim_{t \to -\infty} \frac{A}{1+ce^{-kt}} = 0\text{.} \end{equation*}
4. The inflection point on the graph of $L$ is $\big( -\frac{1}{k} \ln \big(\frac{1}{c}\big), \frac{A}{2}\big)\text{.}$

5. 6. The population grows rapidly at first, but its growth rate decreases to near zero as the population approaches the limiting size of $A\text{.}$

Solution
1. By the chain rule and treating $A\text{,}$ $c\text{,}$ and $k$ as constants, we find that

\begin{equation*} L'(t) = A(-1)(1+ce^{-kt})^{-2} ce^{-kt}(-k) = Acke^{-kt}(1+ce^{-kt})^{-2}\text{.} \end{equation*}

Since $A\text{,}$ $c\text{,}$ and $k$ are all positive and $e^{-kt} \gt 0$ for all values of $t\text{,}$ it is apparent that $L'(t)$ is never zero (nor undefined), and indeed is positive for every value of $t\text{.}$ Thus, $L$ is an always increasing function.

2. Given that

\begin{equation*} L''(t) = Ack^2e^{-kt} \frac{ce^{-kt}-1}{(1+ce^{-kt})^3}\text{,} \end{equation*}

the only way $L''(t) = 0$ is if $ce^{-kt}-1 = 0\text{.}$ Solving $ce^{-kt}-1 = 0$ for $t\text{,}$ we first write $e^{-kt} = \frac{1}{c}\text{.}$ Taking the natural logarithm of both sides yields $-kt = \ln\big(\frac{1}{c}\big)\text{,}$ so that

\begin{equation*} t = -\frac{1}{k} \ln \left(\frac{1}{c}\right) \end{equation*}

is the only value of $t$ for which $L''(t) = 0\text{.}$ Now, observe that since $ce^{-kt} \to 0$ as $t \to \infty$ and $ce^{-kt}\to\infty$ as $t\to-\infty\text{,}$ it follows that the quantity $ce^{-kt} - 1$ will be positive to the left of where it is zero and negative to the right of where it is zero. Since this is the only term in $L''(t)$ that can change sign, it follows that $L''(t) \gt 0$ for $t \lt -\frac{1}{k} \ln \big(\frac{1}{c}\big)$ and $L''(t) \lt 0$ for $t \gt -\frac{1}{k} \ln \big(\frac{1}{c}\big)\text{,}$ making $L$ concave up to the left of the noted inflection point and concave down thereafter.

3. Recalling that $e^{-kt} \to 0$ as $t \to \infty\text{,}$ we observe that

\begin{equation*} \lim_{t \to \infty} \frac{A}{1+ce^{-kt}} = \frac{A}{1+0} = A\text{,} \end{equation*}

so $L$ has a horizontal asymptote of $y = A$ as $t \to \infty\text{.}$ On the other hand, since $e^{-kt} \to \infty$ as $t \to -\infty\text{,}$ this causes the denominator of $L$ to grow without bound (while the numerator remains constant), and therefore

\begin{equation*} \lim_{t \to -\infty} \frac{A}{1+ce^{-kt}} = 0\text{,} \end{equation*}

which means $L$ has a horizontal asymptote of $y = 0$ as $t \to -\infty\text{.}$

4. From (b), we know that $t = -\frac{1}{k} \ln \big(\frac{1}{c}\big)$ is the location of the inflection point of $L\text{.}$ We evaluate $L\big( -\frac{1}{k} \ln \big(\frac{1}{c}\big) \big)\text{,}$11If you prefer to evaluate the function without this apparent shortcut, be careful with parentheses as you plug in and evaluate. In particular, notice that we do indeed have $ce^{-k\left[-\frac{\ln(\frac{1}{c})}{k}\right]}=ce^{k\frac{\ln(\frac{1}{c})}{k}}=ce^{\ln(\frac{1}{c})}=c(\frac{1}{c})=1\text{.}$ The remainder of the calculation should match what is shown below. recalling that at this $t$-value we have the equality $ce^{-kt}=1\text{,}$ thus

\begin{equation*} L\left( -\frac{1}{k} \ln \left(\frac{1}{c}\right) \right) = \frac{A}{1+1} = \frac{A}{2}\text{.} \end{equation*}

Thus, the inflection point on the graph of $L$ is located at $\big( -\frac{1}{k} \ln \big(\frac{1}{c}\big), \frac{A}{2}\big)\text{.}$

5. We have shown that $L$ is an always increasing function that has horizontal asymptotes at $y =0$ and $y = A\text{,}$ as well as an inflection point at $\big( -\frac{1}{k} \ln \big(\frac{1}{c}\big), \frac{A}{2}\big)\text{,}$ which we note lies vertically halfway between the asymptotes. In addition, we see that $L(0) = \frac{A}{1+c}\text{.}$ The combination of all of this information shows us that a typical graph in this family of functions is given by the following figure. 6. The population grows rapidly at first, but its growth rate decreases to near zero as the population approaches the limiting size of $A\text{.}$ This makes sense when environmental factors would affect the population to keep it at a sustainable size.

### SubsectionSummary

• Given a family of functions that depends on one or more parameters, we can often accurately describe the shape of the function in terms of the parameters by investigating how critical numbers and locations where the second derivative is zero depend on the values of these parameters.

• In particular, just as we can creat first and second derivative sign charts for a single function, we can often do so for entire families of functions where critical numbers and possible inflection points depend on arbitrary constants. These sign charts then reveal where members of the family are increasing or decreasing, concave up or concave down, and help us to identify relative extrema and inflection points.

### SubsectionExercises

Consider the one-parameter family of functions given by $p(x) = x^3-ax^2\text{,}$ where $a \gt 0\text{.}$

1. Sketch a plot of a typical member of the family, using the fact that each is a cubic polynomial with a repeated zero at $x = 0$ and another zero at $x = a\text{.}$

2. Find all critical numbers of $p\text{.}$

3. Compute $p''$ and find all values for which $p''(x) = 0\text{.}$ Hence construct a second derivative sign chart for $p\text{.}$

4. Describe how the location of the critical numbers and the inflection point of $p$ change as $a$ changes. That is, if the value of $a$ is increased, what happens to the critical numbers and inflection point?

Let $q(x) = \frac{e^{-x}}{x-c}$ be a one-parameter family of functions where $c \gt 0\text{.}$

1. Explain why $q$ has a vertical asymptote at $x = c\text{.}$

2. Determine $\lim_{x \to \infty} q(x)$ and $\lim_{x \to -\infty} q(x)\text{.}$

3. Compute $q'(x)$ and find all critical numbers of $q\text{.}$

4. Construct a first derivative sign chart for $q$ and determine whether each critical number leads to a local minimum, local maximum, or neither for the function $q\text{.}$

5. Sketch a typical member of this family of functions with important behaviors clearly labeled.

Let $E(x) = e^{-\frac{(x-m)^2}{2s^2}}\text{,}$ where $m$ is any real number and $s$ is a positive real number.

1. Compute $E'(x)$ and hence find all critical numbers of $E\text{.}$

2. Construct a first derivative sign chart for $E$ and classify each critical number of the function as a local minimum, local maximum, or neither.

3. It can be shown that $E''(x)$ is given by the formula

\begin{equation*} E''(x) = e^{-\frac{(x-m)^2}{2s^2}} \left(\frac{(x-m)^2 - s^2}{s^4} \right)\text{.} \end{equation*}

Find all values of $x$ for which $E''(x) = 0\text{.}$

4. Determine $\lim_{x \to \infty} E(x)$ and $\lim_{x \to -\infty} E(x)\text{.}$

5. Construct a labeled graph of a typical function $E$ that clearly shows how important points on the graph of $y = E(x)$ depend on $m$ and $s\text{.}$