Chapter 5

Functions

Functions

Classical geometry handles a number world of day.

Function theory is the genuine mathematic of night.

  • Oswald Spengler, Decline of the West, Chapter VI, part I.

We begin this chapter with one of the most important definitions in all of mathematics.

Definition. A function is a rule that transforms numbers into numbers in an unambiguous way.

By unambiguous, I mean that for each possible input into the function, there is only one possible output. “One in, one out” is a function’s motto. The following examples will clarify this idea.

Example 1. Consider this rule: When given any number, we square it. Is this rule a function?

Solution. Yes. For each possible input, there is only one possible output. Put in 12, and only 144 can come out. Put in -8 and only 64 can come out. One in, one out. No ambiguity.

Example 2. Given any number, we return the number 7. Is this rule a function?

Solution. Yes it is. There’s no ambiguity here either. It’s a boring function, but it’s a function.

Example 3. Given any nonnegative number, we return one of its square roots. Is this a function?

Solution. This is not a function, because for each positive input, there are two possible outputs. (Example: If the input is 9, the output could be either 3 or -3.*) This ambiguity means that the rule is not a function.

Example 4. Given any nonnegative number, we return its nonnegative square root. Is it a function?

Solution. Yes it is. By specifying that we want the nonnegative square root, we remove that pesky ambiguity marring the previous example. This function is called the square root function.

A function’s domain is the set of numbers it accepts as inputs. The domain of many functions (including the “squaring function” from Example 1) is the set of all real numbers. Other functions have smaller domains. For example, negative numbers can’t go into the square root function, because negatives lack square roots. $ ^{†} $ Consequently, the square root function’s domain consists of all nonnegative real numbers.

A function’s range is the set of all numbers it can produce as outputs. For example, 100 is in the squaring function’s range, but -100 isn’t. Indeed, the squaring function’s range is all nonnegative real numbers. (This also happens to be the square root function’s range.)

Function Notation

Do not laugh at notations; they are powerful. In fact,

mathematics is, to a large extent, the invention of better notations.

  • Richard Feynman, The Feynman Lectures on Physics, Chapter 17, section 5.

We usually denote a function by a letter (often f or g). For example, if we decide to call the square root function g, we might say that “g’s domain is all nonnegative numbers” or that “g turns 400 into 20.” The following notation builds on this convention, and is crucial for all that follows.

Notation. If f is a function, and x is an input, then $ f(x) $ denotes the corresponding output. We read $ f(x) $ aloud as “f of x”.

Example. Let g be the square root function. Since $ g(81) $ represents $ g’ $ s output when we input 81, it follows that $ g(81) = 9 $ . Similarly, $ g(169) = 13 $ , and $ g(12) = = 2 $ .

A word of warning: Confused beginners sometimes think that, say, $ f(40) $ means f multiplied by 40. It doesn’t. To test your understanding, answer the following questions about our square root function g. Do not read on until you have answered them.

  1. Find the value of $ g(9) $ .

  2. Find the value of $ 9g(9) $ .

  3. Find the value of $ g(g(9)) $ .

  4. Find the value of $ g(9g(9)) $ .

  5. Solve the equation $ g(x) = 9 $ .

If you missed #1 (its answer is 3), then you aren’t reading attentively. If you missed #2 (its answer is 27), observe that $ g(9) = 9 $ . Following the customary order of operations will take care of the next two: their answers are $ $ and $ $ (or $ 3 $ ). Finally, the fifth question asks, “g of what is 9?” In other words, what number does g turn into 9? The only such number is 81, so the equation’s only solution is x = 81.

Applied Examples

In some situations, one variable’s value completely determines the value of some other variable. Whenever this happens, a function is at work.

Example 1. A square’s side determines its perimeter. (Ex: If a square’s side is 3 cm, its perimeter must be 12 cm.) Thus, a function is at work here, unambiguously turning lengths into perimeters.

Terminology.

If X’s value completely determines Y’s value, we say that “Y is a function of X.” In such a functional relationship, we call Y the dependent variable (for it depends on X), and X the independent variable.

Thus, we can summarize example 1 by saying, “A square’s perimeter is a function of its side length.” Once the independent variable (side length) is fixed, the dependent variable (perimeter) is fixed, too.

In applied contexts, we typically name functions after their outputs. Hence, we might call the function in the previous example P, since it turns lengths into perimeters. We could then use functional notation to write such compact statements as $ P(3) = 12 $ .

The problem of recognizing when a functional relationship exists is distinct from the problem of actually producing a formula for the function. The next few examples should clarify this.

Example 2. All equilateral triangles have the same shape. They differ only in size. Once we specify an equilateral triangle’s side length, it should be clear that the triangle’s area will be determined. (We may or may not be able to find that area, but it surely has some definite value!) In other words, an equilateral triangle’s area is a function of its side length.

If we call this function A, then notation such as $ A(3) $ acquires a precise geometrical meaning and can be used in mathematical “sentences.” For example, $ A(3) < A(4) $ is just a compact way of stating an obvious fact: An equilateral triangle whose sides are 3 units long encloses less space than an equilateral triangle whose sides are 4 units long.

Example 3. If a ball is dropped from 144 feet above the ground, its height (in feet) is a function of the time elapsed (in seconds) since it was dropped. If we call this function h, then notation such as $ h(1.2) $ acquires physical meaning, even when we don’t know its numerical value; it represents the ball’s height 1.2 seconds after being dropped. Despite our ignorance of the precise value of $ h(1.2) $ , we can still make statements about it. For example, it is certainly true that $ h(1.2) < 144 $ .

Sometimes, we can swap the roles of the independent and dependent variables. We could, for example, do this in Example 1, by thinking of the square’s perimeter as the independent variable that determines the square’s side length. (Ex: If we choose a square whose perimeter is 40 cm, its sides must be 10 cm.) Be careful, though: Not all functional relationships are reversible. Here is an irreversible one.

Example 4. An American’s birth year is a function of his or her Social Security number. This sounds peculiar, but a little thought reveals that it is true: Given any specific SSN (say, 345217212), it belongs to a unique American, who was, naturally, born in some particular year (say, 1980). This meets the “one in, one out” criterion, so birth year is indeed a function of SSN, as claimed. $ ^{*} $

It is not, however, a not reversible function; that is, SSN is not a function of birth year. To see why, note that in any particular year we care to specify (say 1999), many thousands of Americans were born, each with a distinct SSN. Which SSN should be the output of the alleged function? Because of this ambiguity, SSN is not a function of birth year.

Exercises

  1. If h is the function defined in Example 3, what is the numerical value of $ h(0) $ ?

  2. The ball in Example 3 will hit the ground exactly three seconds after being dropped. (You’ll learn why in physics.) What is the value of $ h(h(3)) $ ?

  3. Let A be the function in Example 2.

  1. What does $ A(10) $ represent geometrically?

  2. How many solutions does the equation $ A(x) = 10 $ have? What do they represent geometrically?

  1. Let k be the function that divides numbers by two, squares the result, and finally adds 1.

  1. Find the value of $ k(5) $

  2. Find the value of $ 5 + k(6) $

  3. Find the value of $ k(k(-2)) $

  4. Find the value of $ k(k(0) + k(1)) $

  5. What is the domain of $ k? $

  6. Is 5 in the range of $ k? $

  7. What is the smallest number in the range of $ k? $

  8. What is the largest number in the range of $ k? $

  1. Some terminology: To evaluate a function at a number is to put the number in the function, and get the output. Thus, to evaluate k (from the previous exercise) at 8 is to observe that $ k(8) = 17 $ .

  1. Evaluate k at 0

  2. Evaluate k at 10

  3. Evaluate k at 1

  1. As we’ve discussed, the area of an equilateral triangle is a function of its side length. Is the converse statement also true? (That is, is the side length of an equilateral triangle a function of its area?)

  2. Is the area of an isosceles triangle a function of the length of its base?

  3. Is a sphere’s volume a function of its diameter?

Functions Described by Formulas

We can describe many functions by formulas. Consider, for example, the function that adds its input to its input’s square. We can describe this function concisely by the formula

\[ f(x)=x^{2}+x\;. \]

There’s nothing special about the letter x. The formula $ f(t) = t^{2} + t $ describes this function just as well, as does $ f() = ()^{2} + $ , for that matter. The variable is just a “dummy variable”, a mere placeholder. The dummy’s symbol is as insignificant as the color of ink in which we write it.

Evaluating a function described by a formula

To evaluate a function at a given input (either a number or an algebraic expression $ ^{*} $ ), we substitute the input for the dummy variable everywhere that the dummy occurs in the function’s formula.

Note: Since we think of the input as a single package, we must enclose it in parentheses (at least mentally!) when we substitute it for the dummy.

Let’s look at a few examples.

Example 1. If $ f(x) = x^{2} + x $ , find $ f(-5) $ .

Solution. $ f(-5) = (-5)^{2} + (-5) $

\[ = 25 - 5 = 20. \]

Observe the parentheses in Example 1. Had we omitted them, we might have reached the erroneous conclusion that $ f(-5) = -30 $ , since $ -5^{2} + (-5) = -30 $ . This is a common mistake. Don’t be common.

Example 2. If $ (x) = x^{2} + x $ , find $ f(3 - 1) $ .

\[ \begin{aligned}f\big(3\sqrt{r}-1\big)&=\big(3\sqrt{r}-\mathbf{1}\big)^{2}+\big(3\sqrt{r}-\mathbf{1}\big)\\&=\big(9r+1-6\sqrt{r}\big)+\big(3\sqrt{r}-1\big)\\&=9r-3\sqrt{r}.\end{aligned} \]

In Example 2, the input was an algebraic expression rather than a number, but the process was the same: We substituted the expression – enclosed in parentheses because it is a single package – for the dummy x everywhere that it occurred in the formula for f. The rest was just algebra. The input can even be an expression involving the dummy variable itself. The process remains the same:

Example 3. If $ k(x)=+x $ , find $ k(-) $ and express the result as a single fraction.

\[ \begin{aligned}Solution.\ k\left(-\frac{1}{x}\right)&=\frac{1}{\left(-\frac{1}{x}\right)^{2}}+\left(-\frac{1}{x}\right)\quad&(evaluating the function)\\&=\frac{1}{\frac{1}{x^{2}}}-\frac{1}{x}\ =x^{2}-\frac{1}{x}\ =x^{3}-1\text{.}\quad&(algebra)\end{aligned} \]

Note that the first step was just straight substitution: We substituted $ (-1/x) $ for the dummy x in both places where it occurs in the formula that defines function k.

Exercises

  1. Let $ f(x) = x^{3} - 1 $ .

  1. Compute $ f(1 + 2) $ b) Compute $ f(1) + f(2) $

  2. There’s an important moral to be learned here: Function notation does not distribute over addition! (You just saw a specific instance, after all, in which $ f(1 + 2) f(1) + f(2) $ .)

  1. Compute $ $ , then compute $ + $ , and state the appropriate moral.

  2. Compute $ (4 + 5)^{2} $ , then compute $ 4^{2} + 5^{2} $ , and state the appropriate moral.

  3. Compute $ |7 + (-2)| $ , then compute $ |7| + |(-2)| $ , and state the appropriate moral.

  4. Let $ f(x) = x^{2} - x + 1 $ .

  1. State $ f’ $ s domain.

  2. Is 0 in $ f’ $ s range?

  3. Solve $ f(x) = 1 $ .

Now use $ f’ $ s formula to evaluate the following, simplifying your answers:

  1. $ f(3) $

  2. $ f(-3) $

  3. $ f(-) $

  4. $ f(-) $

  5. $ f(1 + 2 + 3) $

  6. $ f(3c) $

  7. $ f(a + b) $

  8. $ f(-x) $

  1. Let $ g(x) = x - $ .

  1. State $ g’ $ s domain.

  2. Is 10 in $ g’ $ s range?

  3. Solve $ g(x) + 3 = 4 $ .

Evaluate and simplify, ultimately expressing each result as a number or a ratio of polynomials:

  1. $ g(1) $ e) $ g() $ f) $ g(-x) $ g) $ g(t+) $ h) $ g(x-) $
  1. If $ h(x) = 2x^{2} - x $ and $ k(x) = x - 3 $ ,
  1. Solve the equation $ 2h(x) = k(5x) $ .

Compute and simplify:

  1. $ h(k(0)) $ c) $ k(h(0)) $ d) $ h(k(x)) $ e) $ k(h(x)) $ f) $ h(x)k(x) $

  2. $ h(x) + k(x) $ h) $ $ i) $ h(k(3))h(k(h(k(x)))) $

  1. Determine the domains of the following functions by thinking about what they cannot accept as inputs:

  1. $ f(x) = $

  2. $ g(x) = $

  3. $ h(x) = $

  4. $ k(x) = $

Playing With the Difference Quotient

Expressions of the sort we’ve been considering might seem arcane, but they often occur in mathematics, including in an expression lying at the heart of calculus, the so-called “difference quotient” of a function. I do not propose to explain the difference quotient’s significance here. I’m introducing it only to give you a sense of how important it is that you master functional notation and basic algebra if you intend to study calculus in the future. There are plenty of challenging ideas in calculus itself; trying to master them with an inadequate grasp of algebra is like trying to read Shakespeare’s sonnets while struggling with the rudiments of the English language. It doesn’t work. With that in mind, I offer the following definition. There is no need to memorize it. For the purposes of this class, the difference quotient merely provides a playground in which to practice your skills.

Given a function f, its difference quotient is defined to be the expression

\[ \frac{f(x+h)-f(x)}{h}. \]

Here, for example, is the difference quotient of the function $ f(x)=+1 $ :

\[ \begin{aligned}\frac{f(x+h)-f(x)}{h}=&\frac{\overbrace{\left(\frac{1}{x+h}+1\right)}^{\frac{f(x+h)}{}}\overbrace{\left(\frac{1}{x}+1\right)}^{\frac{f(x)}{}}h}{h}=\frac{\left(\frac{1+x+h}{x+h}\right)-\left(\frac{1+x}{x}\right)}{h}\\=&\frac{\frac{(x+x^{2}+xh)-(x+h+x^{2}+hx)}{(x+h)x}}{h}=\frac{-h}{(x+h)xh}=\frac{-1}{(x+h)x}.\end{aligned} \]

Be sure you understand why each and every one of those equalities hold. This is precisely the sort of calculation you’ll be expected to carry out with ease on the first day of calculus. Yes, I repeat: with ease. The difference quotient itself is only a part of a larger (more interesting) expression called a derivative. The goal in the first days of calculus is to understand the derivative’s significance. If you are still fumbling over basic algebra and function notation at that point, well, it’s likely that you won’t be moving on to Calculus 2 the next term. If you still have trouble with algebra, now is the time to clear it up before it is too late. Master it by working through books, working with others, and in general, doing whatever it takes. Your teachers will help, but no one can put algebra into your head. The hard work of learning is up to you.

Exercise

  1. Compute and simplify the difference quotients of the following functions as much as possible.

  1. $ f(x) = x $

  2. $ f(x) = x^{2} $

\[ \mathsf{c})f(x)=-x^{2} \]

\[ \mathsf{d})f(x)=x^{3} \]

  1. $ f(x) = -x^{2} + 3x $

$$ f) f(x) = - \

  1. f(x) = $$

Expressing A as a Function of B

When two variables are related by a function, it’s often worthwhile to link them with an explicit formula. To “express one variable as a function of another” simply means to find such a formula. More precisely, to express A as a function of B means to find a formula of the form

$ A = ( B ). $

Example 1. Express a circle’s area, A, as a function of its radius, r.

Solution. Here, we need only write down a familiar friend: $ A = r^{2} $

[It’s acceptable, but unnecessary, to write $ A(r)=r^{2} $ .]

Example 2. Express a circle’s radius, r, as a function of its area, A.

Solution. We need something of the form $ r = ( A ). $ We already have a relationship linking these variables ( $ A = r^{2} $ ), so to write r as a function of A, we simply rearrange this relationship, isolating r on one side. Doing so, we obtain

\[ \boldsymbol{r}=\sqrt{A/\pi}. \]

Example 3. Express an equilateral triangle’s area, A, as a function of its side length, x.

Solution. We seek something of the form $ A = ( x ). $ There’s no well-known formula linking A and x, so we must forge our own.

But how? Well, start with what we know about area: Any triangle’s area is half its base times its height. Thus, if we call the triangle’s height h (as shown in the figure), we know that

\[ A=\frac{1}{2}(base)(height)=\frac{1}{2}xh \]

We’ve now written A as a function of two variables, x and h, which is a start, but we can’t have h in our final formula. To eliminate it, we must find a way to write h in terms of x. Very well… How are h and x related to one another? The figure shows us how: h, x, and x/2 are the sides of a right triangle. Applying the Pythagorean Theorem establishes our link:

\[ x^{2}=(x/2)^{2}+h^{2}. \]

Solving this last equation for h, we obtain, as you should verify, $ h = x $ .

Substituting this expression for h into (★) eliminates h from that equation, yielding

\[ A=\frac{1}{2}x\left(\frac{\sqrt{3}}{2}x\right)=\frac{\sqrt{3}}{4}x^{2}. \]

Our problem is solved. We’ve now expressed A as a function of x: $ A = x^{2} $ .

Observe that this last example really required just two ideas: First, we wrote A in terms of x and h (using a familiar formula), then we used the Pythagorean Theorem to eliminate h. The rest was just algebra.

As is so often the case, our figures guided us to the solution. Draw pictures!

A cube’s body diagonal is one that runs through its “body”, while a face diagonal lies within one of its square faces. (In the figures below, $ AC’ $ is a body diagonal; AC and $ A’B $ are face diagonals.)

Example 4. Express a cube’s body diagonal, d, as a function of its surface area, S.

Solution. There’s no obvious link between d and S, so we’ll try an old strategy: Relate both variables to a third variable. This will often help us bridge the gap. Here, we’ll relate our two variables to the cube’s edge length, which we’ll call x. Linking S to x is easy: Each of the cube’s six faces has area $ x^{2} $ , so

\[ S=6x^{2}. \]

A picture will help us link x to d. Draw the face diagonal AC as shown at right. This introduces two right triangles, which are always a welcome sight, since they set the stage for the Pythagorean Theorem. We apply it first to $ ABC $ , obtaining $ AC = x $ . (You should supply the details.) Next, we apply it to $ ACC’ $ , and find that $ d^{2} = x^{2} + (x)^{2} $ . After simplification, this becomes, as you should verify,

\[ d=\sqrt{3}x. \]

Having linked S to x and x to d, we can now cross the bridge.

Solving for x in equation (1) and substituting the result into (2) yields, as you should verify,

\[ \boldsymbol{d}=\sqrt{S/2}, \]

which expresses the body diagonal of a cube as a function of its surface area, as required.

Now it’s your turn.

Exercises

  1. Express the side length of a square as a function of its area.

  2. Express the perimeter of a square as a function of its area.

  3. Express the area of a square as a function of its perimeter.

  4. Express area of a circle as a function of its circumference.

  5. Express the circumference of a circle as a function of its area.

  6. Express area of a semicircle as a function of its perimeter.

  7. Express the perimeter of a semicircle as a function of its area.

  8. Express the body diagonal of a cube as a function of its face diagonal.

  9. Express the volume of a cube as a function of its surface area.

  10. Express the volume of a sphere as a function of its diameter.

[You’ll need the formula for a sphere’s volume, $ V = ( ) r^{3} $ , which you should memorize.]

  1. Express the radius of a sphere as a function of its volume.

  2. Express the volume of a sphere as a function of its surface area. [Recall that a sphere’s surface area is $ 4r^{2} $ .]

  3. A regular tetrahedron is a triangular-based pyramid, all four faces of which are congruent equilateral triangles. Express the surface area of a regular tetrahedron as a function of its edge length.

  4. A regular icosahedron is a solid composed of twenty congruent equilateral triangles. (Search for a picture of one.) Express the edge length of a regular icosahedron as a function of its surface area.

  5. Express the area of an equilateral triangle as a function of its height.

Functions Defined by Graphs

A function, as you know, is an unambiguous rule that transforms numbers into numbers. Functions are often defined by formulas, but they can also be defined by graphs.

For example, we can interpret the graph at right as a function (which I’ll call f) as follows: Given any input a, slide your finger along the vertical line x = a until it hits the graph at a point. Let this point’s y-coordinate be the function’s output, $ f(a) $ .

Thus, for example, we see that $ f(-2) = -1 $ , $ f(0) = 1 $ , and $ f(1) = 2 $ .

I’ve just described the standard way to read a graph as a function, but this won’t work for every graph. Namely, it won’t work for any graph that is cut more than once by a vertical line. Consider, for example, the graph at left. If we attempt to read it as a function g in the standard way, we’ll fail. What, for instance, would be the value of $ g(4) $ ? Is it 1? Or -4? This ambiguity immediately disqualifies the graph from functionhood. Never forget the function’s motto: one in, one out.

… and Graphs Defined by Functions

We know that the graph of an equation involving both x and y is the set of all points $ (x, y) $ that satisfy it. But what about the graph of an equation of the form

\[ f(x)=(stuff involving x), \]

which features only one variable? In that case, we think of $ f(x) $ , the function’s output, as our y variable, so we define the graph of such an equation to be the graph of the related equation

\[ \boldsymbol{y}=(stuff involving x). \]

For example, what is the graph of the function $ f(x) = x^{2} $ ? It’s simply the graph of $ y = x^{2} $ , the familiar parabola we all know and love.

Each point on a function’s graph corresponds to an input/output pair.

That is, each point on the graph of a function f has the form $ (a, f(a)) $ .

For example, the graph of the squaring function $ f(x) = x^{2} $ contains the point $ (2, f(2)) = (2, 4) $ , a fact which will surprise no one familiar with the graph of $ y = x^{2} $ . Indeed, apart from notation, there’s nothing new here. Graphs of functions are just graphs of equations that happen to have a special form: $ = ( x) $ .

A Preview of Your First Day of Calculus

For an example of how we use functional notation in mathematics (and for a preview of calculus), consider the figure at right, which depicts part of a curve called a hyperbola. This particular hyperbola is the graph of $ f(x) = (1/x) + 1 $ .

Observe that for every x within $ f’ $ s domain, there is a unique point $ (x, f(x)) $ on the hyperbola; at each such point, the hyperbola has a unique tangent line; each tangent line has a particular slope. Putting all this together, we see that the slope of a tangent line is a function of the x-coordinate of its point of tangency. That last sentence is densely packed with mathematical ideas; please dwell on it a bit until you are sure that you fully understand it. This brings us to our calculus problem, whose solution will weave together several of this chapter’s threads.

Our problem: Express the slope of the tangent at $ (x, f(x)) $ as a function of x.

That is, we want an explicit formula of the form

Tangent line’s slope = (some expression whose only variable is x).

Before we solve this problem, let us identify what makes it unlike any problem we’ve considered before. Finding a line’s slope is a trivial “rise over run” affair if we know the coordinates of any two of its points.

This won’t work here; we know only one point on the tangent line: $ (x, f(x)) $ .

From our lone point, it seems we have nowhere to run. Still, we will rise to this challenge. We’ll begin by approximating the tangent with another line. We do this by picking a second point on the hyperbola near $ (x, f(x)) $ , then drawing the line through it and $ (x, f(x)) $ . (See the dashed line in the figure at right.) Such a line is a kind of “tangent impersonator”. Its slope is somewhat close to the slope we want. Moreover – and this is the key insight! – the closer the second point is to $ (x, f(x)) $ , the better the “impersonation” of the tangent becomes. And eventually, when the distance between the two points is infinitesimally small, the tangent impersonator will be indistinguishable from the genuine tangent itself.

Now let’s translate this geometric idea into algebraic language.

If we let h stand for a very small number, we can label our second point $ (x + h, f(x + h)) $ . The smaller h is, the better job the “impersonator” does. Moreover, we now have the coordinates of two of the impersonator’s points Using them, we can express its slope as follows.

\[ \mathrm{T h e i m p e r s o n a t o r^{\prime}s s l o p e}=\frac{\Delta y}{\Delta x}=\frac{f(x+h)-f(x)}{(x+h)-x}=\frac{f(x+h)-f(x)}{h}. \]

This last expression should look familiar: It is $ f’ $ s difference quotient! As luck (or authorial design) would have it, we’ve already computed $ f’ $ s difference quotient earlier in this chapter. Using that result, we have

\[ The impersonator^{\prime}s slope=\frac{-1}{(x+h)x}. \]

Now, if we imagine h getting smaller and smaller, essentially becoming zero, two things happen. Geometrically, the impersonator gets closer and closer to the tangent line. Algebraically, the expression on the right-hand side of $ $ approaches $ -1/x^{2} $ . Combining these two pieces of information, we see that as the impersonator line gets closer and closer to the tangent, its slope gets closer and closer to $ -1/x^{2} $ . It follows that the slope of the tangent line itself must be $ -1/x^{2} $ . We have therefore solved our problem!

Our answer: The slope of the tangent line to the hyperbola at $ (x, f(x)) $ is $ - $ .

Thus, to take a specific instance, the slope of the hyperbola’s tangent at $ (3, 4/3) $ is -1/9.

Similarly, the slope of the tangent line at $ (1/2,3) $ is $ - = -4 $ .

To follow mathematical arguments such as the one that just unfolded over the last page and a half (and which, moreover, are typical in calculus), you need to be fully comfortable with functional notation, mathematical terminology, coordinate geometry, and algebra. Learn this material well. You will need it.

Functions in Science

Your next few mathematics courses will concentrate doggedly on functions. One justification for this is that a goal – arguably the goal – in the great game of science is to discover functional relationships between variables. Scientific laws, however, often take the form of functions of several variables: functions with multiple inputs. Isaac Newton’s law of universal gravitation, for example, describes a function of three independent variables: Given a trio of numbers (the masses of two objects and the distance between them), Newton’s function provides, as its unambiguous output, the gravitational force with which the two objects attract one another. His law is “universal” in the sense that it applies to any two objects in the universe, and thus it describes the force that pulls an apple down from the tree (and onto Newton’s head) as surely as the force that keeps the moon perpetually “falling around” the earth.

For budding scientists, engineers, economists, and others, a major goal of studying precalculus is to become thoroughly comfortable with functions of one variable, particularly those that appear so often throughout science (polynomials, exponential functions, logarithmic functions, trigonometric functions, and so forth). Next comes calculus, which, to a large extent, is the mathematics of change; we use calculus to study how functions change on a “microscopic” scale, and conversely, how an accumulation of tiny changes can be summed up into a functional relationship. Calculus is crucial for understanding the ever-changing world scientifically. Multivariable calculus (usually the third or fourth calculus course one takes) then applies all the ideas of calculus to functions of several variables – the types of functions that physicists, computer scientists, astronomers, and economists deal with regularly.

Exercises

  1. In terms of a, b, and the function f graphed at right…

  1. What are point P’s coordinates?

  2. What is the slope of the line through $ (b, f(b)) $ and P?

  3. Find the equation of the line through P and the origin.

In applications, we must often determine where a function’s graph intersects other objects (axes, other graphs, etc.). Since the graph of a function f is just the graph of the equation $ y = f(x) $ , there’s nothing new here; after all, you learned how to find the intersections of equations’ graphs with other objects in Chapter 4. The next few exercises will reinforce this idea, albeit in the context of functions.

  1. At right is the graph of $ f(x) = ( x - )^{2} - 1 $ .

  1. Because $ f(x) $ represents the graph’s “height” at x, finding the points where the graph crosses the x-axis amounts to finding solutions to $ f(x) = 0 $ . Use this idea to find the coordinates of the points where $ f’ $ s graph crosses the x-axis.

  2. Where does $ f’ $ ’s graph cross the y-axis?

  3. Where does $ f’ $ s graph intersect the line x = 2?

  4. Where does $ f’ $ s graph intersect the line y = 2?

  1. Where does \(f\)’s graph intersect the graph of \(k(x) = (1/2)x + 1\)?

[It may help to draw the graph of \(k\) on the same set of axes as \(f\).]

  1. Where does \(f\)’s graph cross the graph of \(g(x) = -x^{2}?\)

[Both graphs are shown at right.]

  1. Find the slope of the line through the two intersection points you found in part (f).

  2. Does the point (100, 4950) lie on $ f’ $ s graph?

  3. Does the line through $ (2,0) $ with slope 1 intersect $ f’ $ s graph? If so, where? If not, how can you be certain that it doesn’t?

  1. Depicted at right is the graph of $ k(x) = x^{3} - 2x $ .

  1. Points $ (-1, 1) $ and $ (1, -1) $ appear to lie on the graph. Do they really?

  2. Find the points where the graph crosses the x-axis.

  3. Find the point where the graph crosses the line x = 1/2.

  4. To find the points at which the graph crosses the line y = 1/2, we’d need to solve a certain equation. Find that equation, but don’t solve it.

  5. Why do you think I asked you not to solve the equation you found in part (d)?

  1. Sketch a graph of the function $ f(x) = -(983/612)x + 1 $ . Find the coordinates of the graph’s intersections with the axes, and label them on the graph.

  2. For each of the two curves at right, answer the following two questions:

  1. Could the curve be the graph of an equation?

  2. Could the curve be the graph of a function?

  1. If a ball falls from a cliff 144 feet high, then its height h after t seconds will be given by $ h(t) = 144 - 16t^{2} $ . This function’s graph is shown at right.

  1. How many seconds will pass before the ball hits the ground?

  2. How high will the ball be after 1 second? After 2 seconds?

  3. When will the ball be exactly 100 feet high? Give an exact answer and an approximation to the nearest hundredth of a second.

  4. When will the ball be exactly 10 feet high? Give an exact answer and an approximation to the nearest hundredth of a second.

  5. As the ball begins its fall, a remarkably nimble insect begins scaling the cliff at a constant speed of 6 feet per second. How high will the insect be when the ball passes it? (Give your answer to the nearest inch.)

  1. The figure at right shows the graph of a function f, and a line through two points on the graph.

  1. Find the coordinates of P and Q (in terms of x and h).

  2. Use your answers from part (a) to write an expression for the slope of line PQ. This should look familiar. (If it doesn’t, try multiplying the top and bottom by -1.)

  1. The equation of the unit circle, as we’ve seen earlier, is $ x^{2} + y^{2} = 1 $ .
  1. The following reasoning may initially look correct, but it is subtly flawed. Identify the flaw:

If we solve for y in the unit circle’s equation, we get $ y = $ . Hence, the graph of $ f(x) = $ must be the unit circle.

  1. What actually is the graph of $ (x) = $ ?

  2. Let $ (x,0) $ be a variable point on the x-axis. Let PQ be the chord of the circle that passes through this variable point and is perpendicular to the x-axis. (See the figure at right.) Express this chord’s length as a function of x.

  3. Let N be the fixed point $ (0,1) $ , and think of P as a variable point that can slide around on the top half of the unit circle. If you’ve studied enough geometry to do this, prove that NP’s perpendicular bisector must pass through the origin.

[Hint: Start with the line through the origin and NP’s midpoint, and prove that it is NP’s perpendicular bisector.]

  1. Express the slope of NP’s perpendicular bisector as a function of x. Then find this function’s domain and range.
  1. Given an equation involving x and y, we can sometimes “disentangle” the two variables to express y as a function of x. (For example, we can rewrite the equation xy = 7 as y = 7/x.) But often we can’t do this. As a particular example, I claim that it is impossible to rewrite the equation

\[ x^{3}+4x^{2}y+9xy^{2}-36x-9y^{3}+36y=0 \]

in an equivalent form in which y is expressed as a function of x. Not merely difficult, mind you, but impossible. Your problem: Explain why this is so.

Hint: You can’t graph this equation by hand, but a computer can do it for you. The insight it gives you will help you solve this problem. (And remember, equivalent equations have the same graph.)