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Using derivatives to describe families of functions

Mathematicians are often interested in making general observations to describe patterns that hold in a large number of related situations.

Introduction

Mathematicians are often interested in making general observations to describe patterns that hold in a large number of related situations. 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.

You are already familiar with certain families of functions. For example, \[\begin{aligned}\end{aligned}\] is a stretched and shifted version of the sine function with amplitude \(a\), period \(\frac{2\pi}{b}\), phase shift \(c\), and vertical shift \(d\). We know that \(a\) affects the size of the oscillation, \(b\) the rapidity of oscillation, and \(c\) where the oscillation starts, as shown in Figure, while \(d\) affects the vertical positioning of the graph.

As another example, every function of the form \(y = mx + b\) is a line with slope \(m\) and \(y\)-intercept \((0,b)\). 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: where members of the family are increasing or decreasing, concave up or concave down, where relative extremes 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.

Exploration

Describing 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.

The work we've completed in Example 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 \(x \to \infty\). 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. The following activities offer several key examples where we see that the values of the parameters substantially affect the behavior of individual functions within a given family.

Condensed — the full section is in Boelkins, Active Calculus.

Summary

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

Practice (7)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

In the Desmos window below, you can experiment with the graph of the function shown by adjusting the slider for the value of the parameter \(b\):
1. With \(b = 1\), estimate the numerical coordinates of the relative minimum.
2. With \(b = 1\), estimate the numerical coordinates of the inflection point.
3. Play with the slider for \(b\). What are two specific things you notice about how the graph of the function changes?
4. As you change the value of \(b\), how do the coordinates of the local minimum change?
5. As you change the value of \(b\), how do the coordinates of the inflection point change?
Consider the one-parameter family of functions given by \(p(x) = x^3-ax^2\), where \(a \gt 0\).
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 \(x = a\).
2. Find all critical numbers of \(p\).
3. Compute \(p''\) and find all values for which \(p''(x) = 0\). Hence construct a second derivative sign chart for \(p\).
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?
Cevabı açıkla.
1. Writing \(p(x) = x^3 - ax^2 = x^2(x-a)\), we see that \(p(x) = 0\) at \(x=0\) and \(x=a\). Further, since \(x^2 \ge 0\) for all \(x\), we note that \(p(x) \le 0\) whenever \(x \lt a\) and \(p(x) \gt 0\) for all \(x \gt a\). A typical such function is plotted below.
2. To find the critical numbers of \(p\), we first compute \(p'(x)\). Treating \(a\) as constant, we have \(p'(x) = 3x^2 - 2ax = x(3x - 2a)\). Thus, setting \(p'(x) = 0\), we find that \(x(3x-2a) = 0\) and thus \(x = 0\) and \(x = \frac{2a}{3}\) are the critical numbers of \(p\).
3. Observe that \(p''(x) = 6x - 2a\) and thus \(p''(x) = 0\) implies that \(x = \frac{a}{3}\). Since \(p''(x)\) is linear with positive slope, we see that \(p''(x)\) changes sign from negative to positive at \(x = \frac{a}{3}\), and thus \(p\) is concave down for \(x \lt \frac{a}{3}\) and concave up for \(x \gt \frac{a}{3}\) and \(x = \frac{a}{3}\) is the only inflection point of \(p\).
4. There is always a critical number of \(p\) at \(x = 0\). But the other critical number at \(x = \frac{2a}{3}\) and the inflection point at \(x = \frac{a}{3}\) both obviously depend directly on \(a\): as we increase the value of \(a\), both the location of the critical number and the inflection point move to the right along with \(a\).
Let \(q(x) = \frac{e^{-x}}{x-c}\) be a one-parameter family of functions where \(c \gt 0\).
1. Explain why \(q\) has a vertical asymptote at \(x = c\).
2. Determine \(\lim_{x \to \infty} q(x)\) and \(\lim_{x \to -\infty} q(x)\).
3. Compute \(q'(x)\) and find all critical numbers of \(q\).
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 \(q\).
5. Sketch a typical member of this family of functions with important behaviors clearly labeled.
Cevabı açıkla.
1. The value of the numerator at \(x=c\) is \(e^{-c} \ne 0\), while the value of the denominator is \(0\) at \(x=c\). Thus, \(\lim_{x \to c^+} \frac{e^{-x}}{x-c} = \infty\) and \(\lim_{x \to c^-} \frac{e^{-x}}{x-c} = -\infty\). This makes the line \(x=c\) a vertical asymptote for \(q\).
2. We can write \(q(x)\) as \(q(x) = \frac{1}{(x-c)e^x}\) and thus see the denominator increases without bound as \(x \to \infty\). Therefore, \(\lim_{x \to \infty} \frac{e^{-x}}{x-c} = 0\). This makes the line \(y=0\) a horizontal asymptote for \(q\). As \(x \to -\infty\), \(e^{-x}\) increases without bound, while \((x-c)\) decreases without bound. Thus, \(q\) has an indeterminate form as \(x \to -infty\). Applying L'Hopital's Rule gives us \[\begin{aligned}\end{aligned}\].
3. Using \(q(x) = \frac{1}{(x-c)e^x}\), the quotient and product rules along with algebraic simplification show that \[\begin{aligned}\end{aligned}\]. The function \(q\) has critical points in its domain where \(q'(x) = 0\) or \(q'(x)\) doesn't exist. Note that \(q'(x)\) is undefined at \(x=c\), but \(c\) is not in the domain of mq. Thus, the only critical points of \(q\) occur when \(q'(x)=0\). The only way the fraction \(q'(x)\) can be \(0\) is if the numerator is \(0\). Since \(x-c+1=0\) when \(x=c-1\), the only critical number for \(q\) is \(x=c-1\).
4. The values of the exponential function \(e^x\) and the quadratic \((x-c)^2\) are positive for \(x \neq c\), so the sign of \(q'(x) = -\frac{x-c+1}{(x-c)^2e^x}\) depends entirely on the sign of \(x-(c-1)\). When \(x \lt c-1\) we see that \(q'(x) \gt 0\) and when \(x \gt c-1\) we have \(q'(x) \lt 0\). Thus, \(q\) is increasing on \((-\infty, c-1)\) and is decreasing on \((c-1,c)\), so it follows that \(q(c-1)\) is a relative maximum value. This value is a local maximum, since the value of \(q\) can be made as large as we like as shown in part (b).
5. We've shown that a typical member of the family of functions given by \(q(x) = \frac{1}{(x-c)e^x}\) has a vertical asymptote at \(x = c\) and a critical number at which a local maximum occurs one unit to the left of \(c\) at \(x = c-1\). A typical such graph is shown below. Note how \(q\) is increasing and decreasing according to our earlier work in (d).
Let \(E(x) = e^{-\frac{(x-m)^2}{2s^2}}\), where \(m\) is any real number and \(s\) is a positive real number.
1. Compute \(E'(x)\) and hence find all critical numbers of \(E\).
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 \[\begin{aligned}\end{aligned}\]. Find all values of \(x\) for which \(E''(x) = 0\).
4. Determine \(\lim_{x \to \infty} E(x)\) and \(\lim_{x \to -\infty} E(x)\).
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 \(s\).
Cevabı açıkla.
1. Treating \(m\) and \(s\) as constants, by the chain rule, \[\begin{aligned}\end{aligned}\] Note that \(E'(x)\) is defined for every value of \(x\), thus the only critical numbers occur where \(E'(x) = 0\). Since \(e\) raised to any power is never zero, the only way that \(E'(x) = 0\) is for \(\left[ -\frac{2(x-m)}{2s^2} \right] = 0\), which implies that \(x = m\). Thus, \(x = m\) is the only critical number of \(E\).
2. As we noted in (a), \(x = m\) is the only critical number of \(E\). More over, since \(e\) to any power is always positive, the sign of \(E'(x)\) depends entirely on the expression \(-\frac{2(x-m)}{2s^2} = -\frac{1}{2}(x-m)\). This linear function of \(x\) has negative slope, and thus is positive for \(x \lt m\) and negative for \(x \gt m\), and thus \(E'(x)\) is positive for \(x \lt m\) and negative for \(x \gt m\). Therefore, \(E\) is increasing for \(x \lt m\) and decreasing for \(x \gt m\), with a local maximum at \(x = m\).
3. From the given formula for \(E''(x)\) and our knowledge that \(e\) to any power is positive, the only way \(E''(x) = 0\) is for \[\begin{aligned}\end{aligned}\], which implies that \((x-m)^2 = s^2\), and therefore \(x-m = \pm s\), or \(x = m \pm s\). These are the two values for which \(E''(x) = 0\).
4. As \(x \to \pm \infty\), we know that the quadratic function \(-\frac{(x-m)^2}{2s^2}\) must decrease without bound because the quadratic function opens down. Thus, the power to which we are raising \(e\) in the formula for \(E(x)\) is decreasing without bound, which means that \(E(x)\) must approach \(0\) since raising \(e\) to more and more negative numbers drives the result to \(0\). Thus, \(\lim_{x \to \infty} E(x) = \lim_{x \to -\infty} E(x) = 0\).
5. We've shown that \(E\) has a single critical number at \(x = m\) and two potential points of inflection at \(x = m \pm s\), as well as that \(E(x) \to 0\) as \(x \to \pm \infty\). All of these properties are reflected in the following figure.
Consider the one-parameter family of functions given by \(p(x) = x^3-ax^2\), where \(a \gt 0\).
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 \(x = a\).
2. Find all critical numbers of \(p\).
3. Compute \(p''\) and find all values for which \(p''(x) = 0\). Hence construct a second derivative sign chart for \(p\).
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?
Cevabı açıkla.
1. Writing \(p(x) = x^3 - ax^2 = x^2(x-a)\), we see that \(p(x) = 0\) at \(x=0\) and \(x=a\). Further, since \(x^2 \ge 0\) for all \(x\), we note that \(p(x) \le 0\) whenever \(x \lt a\) and \(p(x) \gt 0\) for all \(x \gt a\). A typical such function is plotted below.
2. To find the critical numbers of \(p\), we first compute \(p'(x)\). Treating \(a\) as constant, we have \(p'(x) = 3x^2 - 2ax = x(3x - 2a)\). Thus, setting \(p'(x) = 0\), we find that \(x(3x-2a) = 0\) and thus \(x = 0\) and \(x = \frac{2a}{3}\) are the critical numbers of \(p\).
3. Observe that \(p''(x) = 6x - 2a\) and thus \(p''(x) = 0\) implies that \(x = \frac{a}{3}\). Since \(p''(x)\) is linear with positive slope, we see that \(p''(x)\) changes sign from negative to positive at \(x = \frac{a}{3}\), and thus \(p\) is concave down for \(x \lt \frac{a}{3}\) and concave up for \(x \gt \frac{a}{3}\) and \(x = \frac{a}{3}\) is the only inflection point of \(p\).
4. There is always a critical number of \(p\) at \(x = 0\). But the other critical number at \(x = \frac{2a}{3}\) and the inflection point at \(x = \frac{a}{3}\) both obviously depend directly on \(a\): as we increase the value of \(a\), both the location of the critical number and the inflection point move to the right along with \(a\).
Let \(q(x) = \frac{e^{-x}}{x-c}\) be a one-parameter family of functions where \(c \gt 0\).
1. Explain why \(q\) has a vertical asymptote at \(x = c\).
2. Determine \(\lim_{x \to \infty} q(x)\) and \(\lim_{x \to -\infty} q(x)\).
3. Compute \(q'(x)\) and find all critical numbers of \(q\).
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 \(q\).
5. Sketch a typical member of this family of functions with important behaviors clearly labeled.
Cevabı açıkla.
1. The value of the numerator at \(x=c\) is \(e^{-c} \ne 0\), while the value of the denominator is \(0\) at \(x=c\). Thus, \(\lim_{x \to c^+} \frac{e^{-x}}{x-c} = \infty\) and \(\lim_{x \to c^-} \frac{e^{-x}}{x-c} = -\infty\). This makes the line \(x=c\) a vertical asymptote for \(q\).
2. We can write \(q(x)\) as \(q(x) = \frac{1}{(x-c)e^x}\) and thus see the denominator increases without bound as \(x \to \infty\). Therefore, \(\lim_{x \to \infty} \frac{e^{-x}}{x-c} = 0\). This makes the line \(y=0\) a horizontal asymptote for \(q\). As \(x \to -\infty\), \(e^{-x}\) increases without bound, while \((x-c)\) decreases without bound. Thus, \(q\) has an indeterminate form as \(x \to -infty\). Applying L'Hopital's Rule gives us \[\begin{aligned}\end{aligned}\].
3. Using \(q(x) = \frac{1}{(x-c)e^x}\), the quotient and product rules along with algebraic simplification show that \[\begin{aligned}\end{aligned}\]. The function \(q\) has critical points in its domain where \(q'(x) = 0\) or \(q'(x)\) doesn't exist. Note that \(q'(x)\) is undefined at \(x=c\), but \(c\) is not in the domain of mq. Thus, the only critical points of \(q\) occur when \(q'(x)=0\). The only way the fraction \(q'(x)\) can be \(0\) is if the numerator is \(0\). Since \(x-c+1=0\) when \(x=c-1\), the only critical number for \(q\) is \(x=c-1\).
4. The values of the exponential function \(e^x\) and the quadratic \((x-c)^2\) are positive for \(x \neq c\), so the sign of \(q'(x) = -\frac{x-c+1}{(x-c)^2e^x}\) depends entirely on the sign of \(x-(c-1)\). When \(x \lt c-1\) we see that \(q'(x) \gt 0\) and when \(x \gt c-1\) we have \(q'(x) \lt 0\). Thus, \(q\) is increasing on \((-\infty, c-1)\) and is decreasing on \((c-1,c)\), so it follows that \(q(c-1)\) is a relative maximum value. This value is a local maximum, since the value of \(q\) can be made as large as we like as shown in part (b).
5. We've shown that a typical member of the family of functions given by \(q(x) = \frac{1}{(x-c)e^x}\) has a vertical asymptote at \(x = c\) and a critical number at which a local maximum occurs one unit to the left of \(c\) at \(x = c-1\). A typical such graph is shown below. Note how \(q\) is increasing and decreasing according to our earlier work in (d).
Let \(E(x) = e^{-\frac{(x-m)^2}{2s^2}}\), where \(m\) is any real number and \(s\) is a positive real number.
1. Compute \(E'(x)\) and hence find all critical numbers of \(E\).
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 \[\begin{aligned}\end{aligned}\]. Find all values of \(x\) for which \(E''(x) = 0\).
4. Determine \(\lim_{x \to \infty} E(x)\) and \(\lim_{x \to -\infty} E(x)\).
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 \(s\).
Cevabı açıkla.
1. Treating \(m\) and \(s\) as constants, by the chain rule, \[\begin{aligned}\end{aligned}\] Note that \(E'(x)\) is defined for every value of \(x\), thus the only critical numbers occur where \(E'(x) = 0\). Since \(e\) raised to any power is never zero, the only way that \(E'(x) = 0\) is for \(\left[ -\frac{2(x-m)}{2s^2} \right] = 0\), which implies that \(x = m\). Thus, \(x = m\) is the only critical number of \(E\).
2. As we noted in (a), \(x = m\) is the only critical number of \(E\). More over, since \(e\) to any power is always positive, the sign of \(E'(x)\) depends entirely on the expression \(-\frac{2(x-m)}{2s^2} = -\frac{1}{2}(x-m)\). This linear function of \(x\) has negative slope, and thus is positive for \(x \lt m\) and negative for \(x \gt m\), and thus \(E'(x)\) is positive for \(x \lt m\) and negative for \(x \gt m\). Therefore, \(E\) is increasing for \(x \lt m\) and decreasing for \(x \gt m\), with a local maximum at \(x = m\).
3. From the given formula for \(E''(x)\) and our knowledge that \(e\) to any power is positive, the only way \(E''(x) = 0\) is for \[\begin{aligned}\end{aligned}\], which implies that \((x-m)^2 = s^2\), and therefore \(x-m = \pm s\), or \(x = m \pm s\). These are the two values for which \(E''(x) = 0\).
4. As \(x \to \pm \infty\), we know that the quadratic function \(-\frac{(x-m)^2}{2s^2}\) must decrease without bound because the quadratic function opens down. Thus, the power to which we are raising \(e\) in the formula for \(E(x)\) is decreasing without bound, which means that \(E(x)\) must approach \(0\) since raising \(e\) to more and more negative numbers drives the result to \(0\). Thus, \(\lim_{x \to \infty} E(x) = \lim_{x \to -\infty} E(x) = 0\).
5. We've shown that \(E\) has a single critical number at \(x = m\) and two potential points of inflection at \(x = m \pm s\), as well as that \(E(x) \to 0\) as \(x \to \pm \infty\). All of these properties are reflected in the following figure.

Symbols used here

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Using derivatives to describe families of functions

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?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

Kendini dene.

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

Daha fazlası Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests