Dividing functions and the domain of a combination
Two quantities depend on the same input, and you want one divided by the other at every input. Where does that ratio exist, and what happens where the bottom quantity is zero?
Where you are 5 in Combining and composing functions
When you reach for it
A workshop’s weekly cost and the number of items it makes both depend on how many items it makes; the cost per item is one divided by the other. A town’s population and its area both change with time; the number of people per square kilometer is one divided by the other. Whenever two functions share an input and the question is about their ratio at that input, you want a quotient of functions.
Adding, subtracting and multiplying functions built sums, differences and products output by output. Division works the same way, with one new difficulty: a number cannot be divided by zero. So a quotient of functions loses every input where the bottom output is zero, and finding exactly which inputs survive is half the work. The same question, which inputs does a combined function accept, is asked of sums and products too, and this lesson answers it for all of them.
The procedure
The quotient is defined output by output
For two functions f and g whose outputs are numbers, the quotient function \displaystyle \frac{f}{g} is defined by what it does to each input x:
This holds at every input where f(x) and g(x) both exist and g(x) \ne 0, and nowhere else. As in Function notation, the bracket \displaystyle \left(\frac{f}{g}\right) is the name of the new function and the (x) after it is the input.
The condition g(x) \ne 0 is not a convention; it comes from what division means. To divide f(x) by g(x) is to find the one number that, multiplied by g(x), gives f(x). If g(x) = 0 and f(x) \ne 0, no number works, since every number times 0 is 0. If both are 0, every number works, so there is no single answer to call the output. Either way the quotient has no value there. The top output being zero causes no trouble: \displaystyle \frac{0}{g(x)} = 0 whenever g(x) \ne 0.
The domain of any combination
A combined function has a value at x only when every output it is built from has one. So:
- the domain of f + g, of f - g and of fg is the set of inputs in the domain of f and in the domain of g, written (domain of f) \cap (domain of g);
- the domain of \displaystyle \frac{f}{g} is that same set with every input where g(x) = 0 removed.
Each separate domain comes from its own rule, as in The implied domain of a formula: a square root needs the quantity under it to be at least 0, and a fraction needs its bottom to be nonzero. The inputs to remove for a quotient are the solutions of g(x) = 0, found by factoring and the Solving quadratics by factoring when g is a polynomial. Only solutions that lie in the shared domain matter, since the others were never in it. The answer is usually written in Interval and set-builder notation, or as “every real number except” a short list.
Record the domain before you simplify
When f and g are polynomials, \displaystyle \frac{f(x)}{g(x)} is a rational expression, and it simplifies by factoring and canceling, as in Multiplying and dividing rational expressions. Canceling a common factor x - a rests on \displaystyle \frac{x - a}{x - a} = 1, which is true only when x - a \ne 0. So the simplified formula gives the same output as \displaystyle \frac{f}{g} at every input in the domain, but it also happily accepts x = a, where \displaystyle \frac{f}{g} has no value. The simplified formula is therefore written with its exclusions beside it, for instance x + 1, \; x \ne 3.
On a graph the lost input shows up in one of two ways. Where the factor that made g(x) = 0 canceled, the graph is the graph of the simplified formula with one point removed, drawn as an open circle and called a hole. Where that factor is still in the bottom after canceling, the graph instead runs off upward or downward beside a vertical asymptote.
When f and g are themselves fractions, dividing by g(x) is multiplying by its reciprocal. The exclusions then pile up from three places: where the bottom of f(x) is zero, where the bottom of g(x) is zero, and where g(x) itself is zero, which for a fraction means where its top is zero.
Tables and graphs
The definition asks only for f(x) and g(x) at the same input, so it works on any description that gives outputs. From a table, divide the two entries in the same column; a column where the entry of g is 0 gives no value. From a graph, read each height at the same input and divide. Each input where the graph of g meets the horizontal axis is an input missing from the domain of \displaystyle \frac{f}{g}. Where the two graphs cross, f(x) = g(x), so the quotient is 1 there, unless the crossing is on the horizontal axis, where it is \displaystyle \frac{0}{0} and has no value.
Two routes to one value
As with sums, \displaystyle \left(\frac{f}{g}\right)(2) can be found by working out f(2) and g(2) and dividing, or by putting 2 into a simplified formula. The two agree at every input in the domain, because each simplifying step is true wherever the quotient exists. Outside the domain they part company: the simplified formula still produces a number, but that number is not an output of \displaystyle \frac{f}{g}.
Worked examples
A quotient of polynomials, and the hole it leaves
Let f(x) = 2x^2 + 7x - 15 and g(x) = x + 5. Find \displaystyle \left(\frac{f}{g}\right)(x) and its domain, find \displaystyle \left(\frac{f}{g}\right)(2) both ways, and say what happens at x = -5. Then find \displaystyle \left(\frac{g}{f}\right)(x) and its domain.
SolutionBoth rules are polynomials, so f and g accept every real number. The only inputs the quotient loses are those where g(x) = 0, that is x + 5 = 0, so x = -5. The domain of \displaystyle \frac{f}{g} is every real number except -5, which is (-\infty, -5) \cup (-5, \infty).
To simplify, factor the top. Two numbers with product 2 \cdot (-15) = -30 and sum 7 are 10 and -3, so splitting the middle term as in Factoring when the leading coefficient is not one gives 2x^2 + 10x - 3x - 15, which is 2x(x + 5) - 3(x + 5), which is (2x - 3)(x + 5). Then
The canceling uses \displaystyle \frac{x + 5}{x + 5} = 1, which holds at every input in the domain because x + 5 \ne 0 there.
Evaluating first: f(2) = 8 + 14 - 15 = 7 and g(2) = 7, so \displaystyle \left(\frac{f}{g}\right)(2) = \frac{7}{7} = 1. From the formula: 2(2) - 3 = 1. The routes agree, as they must at an input in the domain.
At x = -5, g(-5) = 0 and also f(-5) = 50 - 35 - 15 = 0, so the quotient would be \displaystyle \frac{0}{0}: no value. The line y = 2x - 3 passes through (-5, -13), but that point is not on the graph of \displaystyle \frac{f}{g}. The graph is the line with an open circle there.
Dividing the other way round,
for x \ne -5 and \displaystyle x \ne \frac{3}{2}. Now the bottom is f, which is zero at both \displaystyle x = \frac{3}{2} and x = -5, so both are excluded. The domains of \displaystyle \frac{f}{g} and \displaystyle \frac{g}{f} are different, and so are the functions.
Domains of four combinations with a square root
Let \displaystyle f(x) = \sqrt{x + 6} and g(x) = x^2 - 16. Find the domains of f + g, fg, \displaystyle \frac{f}{g} and \displaystyle \frac{g}{f}. Then find \displaystyle \left(\frac{f}{g}\right)(3) and \displaystyle \left(\frac{g}{f}\right)(-2).
SolutionThe square root needs x + 6 \ge 0, so the domain of f is [-6, \infty). The polynomial g accepts every real number. The inputs both accept are [-6, \infty), and that is the domain of f + g and of fg.
For \displaystyle \frac{f}{g}, remove the zeros of g. Factoring, x^2 - 16 = (x - 4)(x + 4), which is zero at x = 4 and x = -4, and both lie in [-6, \infty). So the domain of \displaystyle \frac{f}{g} is
For \displaystyle \frac{g}{f}, remove the zeros of f instead. \displaystyle \sqrt{x + 6} = 0 only when x + 6 = 0, so at x = -6. The domain of \displaystyle \frac{g}{f} is (-6, \infty). The inputs 4 and -4 stay in: there the top is zero and the quotient is 0.
The values: \displaystyle \left(\frac{f}{g}\right)(3) = \frac{\sqrt{9}}{9 - 16}, which is \displaystyle \frac{3}{-7} = -\frac{3}{7}. And \displaystyle \left(\frac{g}{f}\right)(-2) = \frac{4 - 16}{\sqrt{4}}, which is \displaystyle \frac{-12}{2} = -6. Here no common factor cancels, so the formula \displaystyle \frac{\sqrt{x + 6}}{x^2 - 16} is already as simple as it gets, and the domain must still be stated beside it.
A square root in the bottom
Let f(x) = x + 1 and \displaystyle g(x) = \sqrt{x} - 3. Find the domain of \displaystyle \frac{f}{g}, and find \displaystyle \left(\frac{f}{g}\right)(16) and \displaystyle \left(\frac{f}{g}\right)(4).
SolutionThe function f accepts every real number and g needs x \ge 0, so the shared domain is [0, \infty). Now solve g(x) = 0: \displaystyle \sqrt{x} = 3. Squaring both sides gives x = 9, and squaring can bring in false solutions, so check: \displaystyle \sqrt{9} = 3, true. The domain of \displaystyle \frac{f}{g} is [0, 9) \cup (9, \infty).
The values: \displaystyle \left(\frac{f}{g}\right)(16) = \frac{17}{4 - 3} = 17, and \displaystyle \left(\frac{f}{g}\right)(4) = \frac{5}{2 - 3} = -5. Notice that f(-1) = 0, but -1 was never in the domain, because \displaystyle g(-1) = \sqrt{-1} - 3 is not a real number.
Dividing one fraction by another
Let \displaystyle f(x) = \frac{x^2 - 4}{x - 3} and \displaystyle g(x) = \frac{x + 2}{x^2 - 9}. Find the domain of \displaystyle \frac{f}{g}, simplify \displaystyle \left(\frac{f}{g}\right)(x), and check at x = 0.
SolutionCollect the exclusions from all three places. The bottom of f(x) is zero at x = 3. The bottom of g(x) is x^2 - 9 = (x - 3)(x + 3), zero at x = 3 and x = -3. And g(x) itself is zero where its top is zero and its bottom is not, so at x = -2. The domain of \displaystyle \frac{f}{g} is every real number except -3, -2 and 3.
Dividing by a fraction is multiplying by its reciprocal. Then factor everything and cancel, which is allowed because none of the canceled factors is zero on the domain:
for x \ne -3, x \ne -2 and x \ne 3. The final formula is a polynomial with no restriction of its own; every exclusion came from the pieces. Its graph is a parabola with three holes, at (-3, 0), (-2, -4) and (3, 6), the points the parabola would have there.
Check at 0: \displaystyle f(0) = \frac{-4}{-3} = \frac{4}{3} and \displaystyle g(0) = \frac{2}{-9} = -\frac{2}{9}. Dividing is multiplying by the reciprocal, so \displaystyle \left(\frac{f}{g}\right)(0) is \displaystyle \frac{4}{3} \cdot \left(-\frac{9}{2}\right) = -6. The formula gives 0^2 + 0 - 6 = -6.
Dividing two tables
Two functions c and d are known only from a table.
| x | -2 | -1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|
| c(x) | 6 | -4 | 0 | 9 | 5 | -3 |
| d(x) | 3 | 2 | -5 | 0 | -2 | 6 |
Make a table of \displaystyle \frac{c}{d} and of \displaystyle \frac{d}{c}.
SolutionEach entry uses only the two entries in its own column, and a column where the bottom entry is 0 gives no value. With one row per input:
| x | \displaystyle \left(\frac{c}{d}\right)(x) | \displaystyle \left(\frac{d}{c}\right)(x) |
|---|---|---|
| -2 | 2 | \displaystyle \frac{1}{2} |
| -1 | -2 | \displaystyle -\frac{1}{2} |
| 0 | 0 | no value |
| 1 | no value | 0 |
| 2 | \displaystyle -\frac{5}{2} | \displaystyle -\frac{2}{5} |
| 3 | \displaystyle -\frac{1}{2} | -2 |
At x = 0 the top of \displaystyle \frac{c}{d} is 0 and its bottom is -5, so \displaystyle \frac{0}{-5} = 0; the same column gives \displaystyle \frac{d}{c} a zero bottom, so no value. At x = 1 the roles swap. In every column where both quotients exist, each is the reciprocal of the other.
Average cost at a pottery
Brannoc’s pottery spends C(n) = 2n + 450 dollars in a week in which it makes n mugs: 2 dollars of clay and glaze per mug, and 450 dollars of rent and kiln power whatever the output. Build the average cost per mug A(n), give its domain, find A(50) and A(150), and find how many mugs bring the average cost down to 3 dollars.
SolutionAverage cost is total cost divided by the number of mugs. The number of mugs is the function I(n) = n, so A is the quotient \displaystyle \frac{C}{I}:
The units are dollars divided by mugs, that is dollars per mug. As a formula, A excludes only n = 0, the zero of the bottom; in the setting, n counts mugs made, so the domain is n = 1, 2, 3 and so on, up to whatever the kiln can fire in a week.
\displaystyle A(50) = \frac{100 + 450}{50} = \frac{550}{50} = 11 dollars per mug, and \displaystyle A(150) = \frac{300 + 450}{150} = \frac{750}{150} = 5 dollars per mug.
Dividing each term of the top by n, which is allowed because n \ne 0, gives \displaystyle A(n) = 2 + \frac{450}{n}. The fixed 450 dollars is shared among more mugs as n grows, so the average falls toward 2 dollars, and it never reaches 2 because \displaystyle \frac{450}{n} > 0. For an average of 3: \displaystyle 2 + \frac{450}{n} = 3 gives \displaystyle \frac{450}{n} = 1, so n = 450 mugs. Check: C(450) = 900 + 450 = 1350, and \displaystyle \frac{1350}{450} = 3. This leveling off is the horizontal asymptote of a Rational functions in context.
Working backwards from a quotient
A function g(x) = x^2 + 2x - 8 and an unknown polynomial f give the quotient
for x \ne -4 and x \ne 2. Find f(x), and explain why 2 is excluded even though the simplified formula is fine at x = 2.
SolutionAt every input in the domain, \displaystyle \frac{f(x)}{g(x)} = \frac{x - 1}{x + 4}, so multiplying both sides by g(x) gives \displaystyle f(x) = \frac{x - 1}{x + 4} \cdot g(x). Factor g: two numbers with product -8 and sum 2 are 4 and -2, so g(x) = (x + 4)(x - 2). Then
This agrees with f on the domain, and since both sides are polynomials, which agree everywhere if they agree at infinitely many inputs, f(x) = x^2 - 3x + 2 at every input.
The exclusions are exactly the zeros of g, namely -4 and 2. At x = 2 the factor x - 2 sat in both the top and the bottom of \displaystyle \frac{f(x)}{g(x)} and canceled, so the simplified formula no longer shows it. But g(2) = 0, so \displaystyle \frac{f}{g} has no value at 2, and its graph has a hole at \displaystyle \left(2, \frac{1}{6}\right). At x = -4 the factor x + 4 is still in the bottom, so the graph has a vertical asymptote there instead. Check: f(3) = 2 and g(3) = 7, and \displaystyle \frac{3 - 1}{3 + 4} = \frac{2}{7}.
Boundaries
A zero bottom output has no quotient. At an input where g(x) = 0 the quotient \displaystyle \frac{f}{g} has no value, whether f(x) is zero there or not. No amount of simplifying gives it one: a simplified formula that produces a number there is describing a different function.
The domain belongs to the function, not to the formula. In Example 1, \displaystyle \left(\frac{f}{g}\right)(x) = 2x - 3 for x \ne -5, and the function h(x) = 2x - 3 uses the same rule on every real number. They are different functions, because they have different domains. The same caution applies to products and sums: with \displaystyle s(x) = \sqrt{x - 2}, the product (ss)(x) simplifies to x - 2, but only for x \ge 2, since s(0) is not a real number.
Order matters. \displaystyle \frac{f}{g} and \displaystyle \frac{g}{f} are reciprocals at the inputs where both exist, but their domains differ: one loses the zeros of g, the other the zeros of f.
Units may differ. Unlike a sum, a quotient of two quantities with different units is meaningful and makes a new unit, such as dollars per mug or people per square kilometer.
Not composition, not an inverse. \displaystyle \left(\frac{f}{g}\right)(x) divides two outputs. Feeding one output into the other function is composition, taught in Composing functions algebraically. And the reciprocal \displaystyle \frac{1}{f}, which divides 1 by each output, is a different thing from the inverse function f^{-1} that undoes f, even though for a number a^{-1} does mean \displaystyle \frac{1}{a}.
Where this goes next
A rational function is exactly a quotient of two polynomial functions, and everything about its graph starts from this lesson: its domain is every input where the bottom is nonzero, a canceled zero of the bottom makes a hole, and a zero that stays makes a vertical asymptote.
The same domain question returns for composition, where an input survives only if the inner function accepts it and the outer function accepts the inner output; that is Domain of a composition. In calculus, the rate of change of a quotient has its own rule, the quotient rule, stated for exactly the functions built here.
Exercises
Quotients of polynomials
Let p(x) = 3x^2 + x - 10 and q(x) = x + 2.
- Find \displaystyle \left(\frac{p}{q}\right)(x) and its domain.
- Find \displaystyle \left(\frac{p}{q}\right)(4) in two ways: from p(4) and q(4), and from your formula in Exercise 1.
- Find \displaystyle \left(\frac{q}{p}\right)(x) and its domain.
- Does \displaystyle \left(\frac{p}{q}\right)(-2) have a value? Name the point that the line y = 3x - 5 has and the graph of \displaystyle \frac{p}{q} lacks.
- Let f(x) = x^2 - 36 and g(x) = x^2 + 2x - 24. Find \displaystyle \left(\frac{f}{g}\right)(x) in simplest form, with its domain.
- Let f(x) = 4x and g(x) = x^2 + 1. Find the domain of \displaystyle \frac{f}{g} and find \displaystyle \left(\frac{f}{g}\right)(1).
Domains of combinations
- Let \displaystyle f(x) = \sqrt{x - 1} and g(x) = x - 5. Find the domains of f + g, fg, \displaystyle \frac{f}{g} and \displaystyle \frac{g}{f}.
- Let \displaystyle f(x) = \sqrt{7 - x} and \displaystyle g(x) = \sqrt{x + 1}. Find the domains of f - g and \displaystyle \frac{f}{g}, and find \displaystyle \left(\frac{f}{g}\right)(3).
- Let \displaystyle f(x) = \frac{1}{x - 4} and \displaystyle g(x) = \frac{x}{x + 1}. Find the domain of f + g. Then simplify \displaystyle \left(\frac{f}{g}\right)(x) and give its domain.
- Let f(x) = x^2 - 1 and let g(x) = 5 for every x. Find \displaystyle \left(\frac{f}{g}\right)(x) and \displaystyle \left(\frac{g}{f}\right)(x), each with its domain.
- Let f(x) = x + 3 and \displaystyle g(x) = \sqrt{x} - 1. Find the domain of \displaystyle \frac{f}{g} and find \displaystyle \left(\frac{f}{g}\right)(9).
Dividing tables
The functions F and G are given by this table.
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| F(x) | 8 | -6 | 0 | 5 | 12 |
| G(x) | 4 | 3 | -7 | 0 | -3 |
- Make a table of \displaystyle \left(\frac{F}{G}\right)(x) for x = 0, 1, 2, 3, 4.
- At which input in the table does \displaystyle \frac{G}{F} have no value? Find \displaystyle \left(\frac{G}{F}\right)(4).
- Find every input in the table at which \displaystyle \left(\frac{F}{G}\right)(x) is negative.
Dividing graphs
Use the graphs of j and w.
- Find \displaystyle \left(\frac{j}{w}\right)(0) and \displaystyle \left(\frac{j}{w}\right)(3).
- Find \displaystyle \left(\frac{w}{j}\right)(1) and \displaystyle \left(\frac{w}{j}\right)(-1).
- Which inputs are missing from the domain of \displaystyle \frac{j}{w}, and which from the domain of \displaystyle \frac{w}{j}?
- Find every x with \displaystyle \left(\frac{j}{w}\right)(x) = 1.
In context
- The town of Harrowmere has P(t) = 4000 + 150t people and covers L(t) = 20 + 0.5t square kilometers, t years from now, as it takes in farmland. Build the population density \displaystyle D = \frac{P}{L}, give its units, and find D(0) and D(10).
- Sela’s print shop pays C(n) = 1200 + 4n dollars to set up and print n posters. Find the average cost per poster A(n), find A(300), and find how many posters bring the average cost down to 6 dollars.
Reasoning
- Explain why an input where g(x) = 0 is removed from the domain of \displaystyle \frac{f}{g}, but an input where f(x) = 0 is not. What happens at an input where both are zero?
- Let f(x) = x^2 + 3x and g(x) = x. Ilsabet says that \displaystyle \frac{f}{g} and h(x) = x + 3 are the same function. Explain why they are not, and name the one input where they differ.
- Give polynomials f and g for which the domain of \displaystyle \frac{f}{g} is every real number except 1 and 4, and \displaystyle \left(\frac{f}{g}\right)(0) = 2.
- Explain why, at every input where both \displaystyle \frac{f}{g} and \displaystyle \frac{g}{f} have values, the two outputs are reciprocals. Then, with f and g from Example 1, give an input where \displaystyle \frac{f}{g} has a value and \displaystyle \frac{g}{f} does not.
Answers
- \displaystyle \left(\frac{p}{q}\right)(x) = 3x - 5, x \ne -2. Splitting the middle term with 6 and -5, 3x^2 + x - 10 = (3x - 5)(x + 2), and \displaystyle \frac{x + 2}{x + 2} = 1 wherever x \ne -2.
- \displaystyle \left(\frac{p}{q}\right)(4) = 7. Evaluating first: p(4) = 48 + 4 - 10 = 42 and q(4) = 6, so \displaystyle \frac{42}{6} = 7. From the formula: 3(4) - 5 = 7.
- \displaystyle \left(\frac{q}{p}\right)(x) = \frac{1}{3x - 5}, for x \ne -2 and \displaystyle x \ne \frac{5}{3}. The bottom is now p(x) = (3x - 5)(x + 2), zero at both of those inputs.
- No: q(-2) = 0 (and p(-2) = 12 - 2 - 10 = 0 too, so it would be \displaystyle \frac{0}{0}). The line y = 3x - 5 passes through (-2, -11); the graph of \displaystyle \frac{p}{q} has a hole there.
- \displaystyle \left(\frac{f}{g}\right)(x) = \frac{x - 6}{x - 4}, for x \ne -6 and x \ne 4. Factor: \displaystyle \frac{(x - 6)(x + 6)}{(x + 6)(x - 4)}; both zeros of g are excluded, including the canceled one.
- The domain is every real number, since x^2 + 1 \ge 1 is never zero. \displaystyle \left(\frac{f}{g}\right)(1) = \frac{4}{2} = 2.
- f + g and fg: [1, \infty). \displaystyle \frac{f}{g}: [1, 5) \cup (5, \infty), removing the zero of g. \displaystyle \frac{g}{f}: (1, \infty), removing x = 1, where \displaystyle \sqrt{x - 1} = 0.
- f - g: [-1, 7], the inputs with 7 - x \ge 0 and x + 1 \ge 0. \displaystyle \frac{f}{g}: (-1, 7], since g(-1) = 0. \displaystyle \left(\frac{f}{g}\right)(3) = \frac{\sqrt{4}}{\sqrt{4}} = 1.
- f + g: every real number except -1 and 4. Multiplying by the reciprocal, \displaystyle \frac{1}{x - 4} \cdot \frac{x + 1}{x} gives \displaystyle \left(\frac{f}{g}\right)(x) = \frac{x + 1}{x(x - 4)} for x \ne -1, x \ne 0 and x \ne 4: the two bottoms exclude 4 and -1, and g(0) = 0 excludes 0.
- \displaystyle \left(\frac{f}{g}\right)(x) = \frac{x^2 - 1}{5}, every real number, because g is never zero. \displaystyle \left(\frac{g}{f}\right)(x) = \frac{5}{x^2 - 1}, for x \ne -1 and x \ne 1, the zeros of x^2 - 1 = (x - 1)(x + 1).
- [0, 1) \cup (1, \infty), since g needs x \ge 0 and \displaystyle \sqrt{x} = 1 at x = 1. \displaystyle \left(\frac{f}{g}\right)(9) = \frac{12}{3 - 1} = 6.
- \displaystyle \left(\frac{F}{G}\right)(x) is 2, -2, 0, no value, -4 for x = 0, 1, 2, 3, 4. Column by column: \displaystyle \frac{8}{4}, \displaystyle \frac{-6}{3}, \displaystyle \frac{0}{-7}, then G(3) = 0, then \displaystyle \frac{12}{-3}.
- At x = 2, where F(2) = 0 is the bottom. \displaystyle \left(\frac{G}{F}\right)(4) = \frac{-3}{12} = -\frac{1}{4}.
- x = 1 and x = 4, where F and G have opposite signs: the quotients there are -2 and -4.
- \displaystyle \left(\frac{j}{w}\right)(0) = -\frac{1}{2} and \displaystyle \left(\frac{j}{w}\right)(3) = 1. From the graphs j(0) = 2, w(0) = -4, and j(3) = w(3) = 5.
- \displaystyle \left(\frac{w}{j}\right)(1) = -1 and \displaystyle \left(\frac{w}{j}\right)(-1) = -3. From the graphs w(1) = -3, j(1) = 3, and w(-1) = -3, j(-1) = 1.
- \displaystyle \frac{j}{w} loses x = -2 and x = 2, where the parabola meets the horizontal axis. \displaystyle \frac{w}{j} loses only x = -2, where the line meets it.
- x = 3 only. The graphs cross at x = 3 and x = -2, but at x = -2 both heights are 0, so the quotient there is \displaystyle \frac{0}{0} and has no value.
- \displaystyle D(t) = \frac{4000 + 150t}{20 + 0.5t} people per square kilometer. \displaystyle D(0) = \frac{4000}{20} = 200 and \displaystyle D(10) = \frac{5500}{25} = 220 people per square kilometer.
- \displaystyle A(n) = \frac{1200 + 4n}{n} dollars per poster, for n = 1, 2, 3 and so on. \displaystyle A(300) = \frac{2400}{300} = 8 dollars. An average of 6 needs 1200 + 4n = 6n, so n = 600 posters.
- The output \displaystyle \frac{f(x)}{g(x)} is the number that times g(x) gives f(x). If g(x) = 0 and f(x) \ne 0, no number does, since every number times 0 is 0. If f(x) = 0 and g(x) \ne 0, the number 0 does, and it is the only one, so the quotient is 0. If both are zero, every number does, so there is no single output, and the input is still removed.
- \displaystyle \left(\frac{f}{g}\right)(x) = \frac{x(x + 3)}{x} = x + 3 only for x \ne 0, because g(0) = 0. The function h has the value h(0) = 3, while \displaystyle \frac{f}{g} has no value at 0. Same rule, different domains, so different functions; they differ only at x = 0.
- One choice: f(x) = 8 and g(x) = (x - 1)(x - 4) = x^2 - 5x + 4. The bottom is zero exactly at 1 and 4, and \displaystyle \left(\frac{f}{g}\right)(0) = \frac{8}{4} = 2.
- Where both exist, f(x) \ne 0 and g(x) \ne 0, and \displaystyle \frac{f(x)}{g(x)} \cdot \frac{g(x)}{f(x)} = 1, so each is the reciprocal of the other. With Example 1’s functions, \displaystyle x = \frac{3}{2}: \displaystyle f\left(\frac{3}{2}\right) = 0 and \displaystyle g\left(\frac{3}{2}\right) = \frac{13}{2}, so \displaystyle \left(\frac{f}{g}\right)\left(\frac{3}{2}\right) = 0, while \displaystyle \frac{g}{f} has a zero bottom there.