D. Characteristics of Functions · Combining Functions and ModellingProducts of Functions
D2.1About 70 minutes
Learning goals- I can find the product f · g of two functions from equations, tables and graphs.
- I can use the zeros and signs of each factor to sketch the graph of a product.
- I can state the domain of a product and find its range, maximum and minimum points and number of zeros.
- I can use one factor as an envelope to describe a product such as y = x sin x.
Words to know
product of functions
The function (fg)(x) = f(x) · g(x). At each x, multiply the two outputs.
sign chart
A number line split at the zeros of the factors, showing whether each factor (and the product) is positive or negative on each interval.
envelope
A pair of curves that a graph stays between, touching them at some points. For y = x sin x the envelope is y = x and y = −x.
double zero
A zero that comes from a squared factor, like (x − 1)². The graph touches the x-axis there without crossing.
Price times quantity
A school sells tickets to a winter hockey tournament. If it raises the price, fewer people buy tickets. The money it takes in, the revenue, is
revenue = (price per ticket) × (number of tickets sold).
Both the price and the number sold depend on the same decision, so the revenue is the product of two functions. In the sibling unit Polynomial Functions and Equations you met polynomials in factored form like (x + 3)(x − 1)(x − 4). Each of those is already a product of simple functions. In this lesson you will multiply any two functions and predict the graph from the pieces.
Multiplying functions
The product of f and g is
(fg)(x) = f(x) · g(x)
and its domain is again the intersection of the domains of f and g.
Four facts make products easy to sketch:
- Zeros: the product is 0 exactly where f(x) = 0 or g(x) = 0 (as long as both are defined there).
- Signs: the product is positive where f and g have the same sign, and negative where their signs are different.
- Where one factor is 1, the product equals the other factor. Where one factor is −1, the product is the opposite of the other factor.
- Polynomials: the degree of the product is the sum of the degrees (a line times a line is a quadratic; a line times a quadratic is a cubic).
Worked exampleExample 1: A product of two lines
Let f(x) = x + 1 and g(x) = x − 2. Find (fg)(x), make a sign chart, and state the key features.
- Multiply: (fg)(x) = (x + 1)(x − 2) = x² − x − 2.
- Zeros: f(x) = 0 at x = −1 and g(x) = 0 at x = 2. So fg has two zeros, x = −1 and x = 2.
- Sign chart. For x < −1: f < 0 and g < 0, so fg > 0. For −1 < x < 2: f > 0 and g < 0, so fg < 0. For x > 2: f > 0 and g > 0, so fg > 0.
- The vertex is halfway between the zeros: x = (−1 + 2)/2 = 0.5, and (fg)(0.5) = (1.5)(−1.5) = −2.25.
- Domain: ℝ. Range: y ≥ −2.25.
(fg)(x) = x² − x − 2: zeros −1 and 2, minimum point (0.5, −2.25), range y ≥ −2.25, negative only on −1 < x < 2.
| Interval | x < −1 | −1 < x < 2 | x > 2 |
|---|
| f(x) = x + 1 | − | + | + |
| g(x) = x − 2 | − | − | + |
| (fg)(x) | + | − | + |
Worked exampleExample 2: A line times a square
Let f(x) = x + 2 and g(x) = (x − 1)². Sketch y = (fg)(x) and state its zeros, y-intercept and end behaviour.
- Zeros: x = −2 (from f) and x = 1 (from g). Because (x − 1)² is a square, x = 1 is a double zero.
- Signs: (x − 1)² ≥ 0 everywhere, so the product has the same sign as x + 2, except at x = 1 where it is 0. It is negative for x < −2 and positive for x > −2 (x ≠ 1).
- So the graph crosses the x-axis at x = −2 and only touches it at x = 1.
- y-intercept: (0 + 2)(0 − 1)² = 2.
- Degree 1 + 2 = 3 with positive leading coefficient, so as x → −∞, y → −∞ and as x → ∞, y → ∞.
A cubic that rises from the lower left, crosses at (−2, 0), passes through (0, 2), turns down to touch the axis at (1, 0), and rises to the upper right.
Products with a trigonometric factor: envelopes
What does y = x sin x look like? Since −1 ≤ sin x ≤ 1, multiplying by x gives
−|x| ≤ x sin x ≤ |x|.
So the graph is trapped between the lines y = x and y = −x. These lines are the envelope. The graph touches the envelope where sin x = ±1 (x = π/2, 3π/2, ...), and it has a zero wherever x = 0 or sin x = 0, that is at every multiple of π. The wiggles get taller as you move away from the origin because the factor x gets bigger.
In the lesson Real-World Combinations: Damped Motion, Tones and Profit you will use an exponential envelope that shrinks, which models a pendulum slowly coming to rest.
Worked exampleExample 3: Key features of y = x sin x on −2π ≤ x ≤ 2π
State the domain, the zeros and the y-values at x = π/2 and x = −π/2 for h(x) = x sin x, −2π ≤ x ≤ 2π.
- Domain: given as −2π ≤ x ≤ 2π (both factors are defined everywhere).
- Zeros: x = 0 (from the factor x) and sin x = 0 at x = −2π, −π, 0, π, 2π. So there are five zeros: −2π, −π, 0, π, 2π. (x = 0 is a zero of both factors, so the graph only touches the axis there.)
- h(π/2) = (π/2)(1) = π/2 ≈ 1.57.
- h(−π/2) = (−π/2)(−1) = π/2 ≈ 1.57. Same value: this is no accident (see the lesson Properties of Combined Functions).
Five zeros on this interval; h(±π/2) = π/2. Both points lie on the envelope, h(π/2) on y = x and h(−π/2) on y = −x.
Domains of products with roots
The product √x · √(4 − x) needs both roots to exist, so 0 ≤ x ≤ 4. On that interval you may combine the roots: √x · √(4 − x) = √(4x − x²). Squaring y = √(4x − x²) gives y² = 4x − x², or (x − 2)² + y² = 4: a circle of radius 2 with centre (2, 0). Since y ≥ 0, the product is the top half of that circle.
Careful: combining roots can change the domain
√(x − 1) · √(x − 3) needs x ≥ 1 and x ≥ 3, so its domain is x ≥ 3. But √((x − 1)(x − 3)) is defined whenever the product inside is ≥ 0, which is x ≤ 1 or x ≥ 3. They agree only on x ≥ 3. Find the domain from the original factors.
Just for funRiddle: why do two negative functions get along so well?
Wherever they are both down, their product is up.
Hands-on activityA rectangle made of string
You need: a piece of string (or a shoelace) exactly 24 cm long, a ruler, a pencil and paper.
- Tie or tape the ends of the string together to make a loop with perimeter 24 cm.
- Stretch it into a rectangle on the table with width x = 1 cm. The length is then 12 − x = 11 cm. Record the area x(12 − x).
- Repeat for x = 2, 3, ..., 11. Make a table with columns x, 12 − x, and A = x(12 − x).
- Plot the two factors, w(x) = x and l(x) = 12 − x, and the product A(x), on one set of axes.
What to look for: the areas are 11, 20, 27, 32, 35, 36, 35, 32, 27, 20, 11 cm². The product of an increasing line and a decreasing line is a parabola opening down, with zeros at x = 0 and x = 12 (the zeros of the factors) and its maximum, 36 cm², at x = 6, where the rectangle is a square. The two lines cross at x = 6 too: the largest area happens when the factors are equal.
Your turn
1f(x) = 2x − 1 and g(x) = x + 3. Find (fg)(x) and its zeros.
(fg)(x) = 2x² + 5x − 3. Zeros: x = 1/2 and x = −3.
2Make a sign chart for (fg)(x) in the previous question.
x < −3: (−)(−) = +. −3 < x < 1/2: (−)(+) = −. x > 1/2: (+)(+) = +.
3f(x) = x and g(x) = x² − 9. State the degree and zeros of fg and its end behaviour.
x(x − 3)(x + 3) = x³ − 9x: degree 3, zeros −3, 0, 3. As x → −∞, y → −∞; as x → ∞, y → ∞.
4f(x) = √(x + 1) and g(x) = log x. State the domain of fg.
√(x + 1) needs x ≥ −1; log x needs x > 0. Domain: x > 0.
5Between which two lines does y = 3 sin x · x lie? Where does it touch the upper line for x > 0?
Between y = 3x and y = −3x (since −3 ≤ 3 sin x ≤ 3). For x > 0 it touches y = 3x where sin x = 1: x = π/2 + 2nπ (n = 0, 1, 2, ...).
6g(4) = 1 and f(4) = 7. Find (fg)(4) without any equations.
(fg)(4) = 7 × 1 = 7: where one factor is 1, the product equals the other factor.
More practice: warm-up
1f(x) = x and g(x) = x − 5. Find (fg)(x).
2f(2) = −3 and g(2) = 4. Find (fg)(2).
3What is the degree of the product of a quadratic and a cubic?
4If f(x) < 0 and g(x) < 0 at x = 1, is (fg)(1) positive or negative?
5Name the zeros of y = (x − 4)(x + 7).
More practice: on your own
1f(x) = x² and g(x) = x − 3. Sketch fg and explain what happens at x = 0.
x²(x − 3) = x³ − 3x²: zeros 0 (double) and 3. Since x² ≥ 0, the product has the sign of x − 3: negative for x < 3 (except 0 at x = 0), positive for x > 3. At x = 0 the graph touches the axis from below and turns back down. It crosses at x = 3.
2Find the domain and range of y = √x · √(16 − x).
Domain 0 ≤ x ≤ 16. y = √(16x − x²), the top half of the circle (x − 8)² + y² = 64. Range 0 ≤ y ≤ 8 (maximum at x = 8).
3f(x) = 2ˣ and g(x) = x − 1. Where is fg negative? How many zeros does it have?
2ˣ > 0 always, so fg has the sign of x − 1: negative for x < 1. It has one zero, x = 1.
4A ticket costs p(x) = 10 + x dollars and n(x) = 300 − 20x tickets are sold, where x is the number of $1 price increases. Find the revenue function and the price that gives the most revenue.
R(x) = (10 + x)(300 − 20x). Zeros x = −10 and x = 15; the maximum is halfway, x = 2.5. R(2.5) = 12.5 × 250 = $3125. The best price is $12.50 (revenue $3125). If only whole-dollar increases are allowed, x = 2 or 3 both give 12 × 260 = 13 × 240 = $3120.
5Use a table with x = 0, 1, 2, 3, 4 to describe y = x(0.5)ˣ for x ≥ 0.
Values: 0, 0.5, 0.5, 0.375, 0.25. It starts at 0, rises to about 0.5 between x = 1 and 2, then decreases toward 0: the decaying factor 0.5ˣ eventually beats the growing factor x.
Challenge
1Estimate the maximum of y = x(0.5)ˣ, x ≥ 0, to two decimal places, using a table with step 0.1 near x = 1.5.
y(1.3) ≈ 0.528, y(1.4) ≈ 0.531, y(1.5) ≈ 0.530. The maximum is about 0.53 near x ≈ 1.4. (In calculus the exact location turns out to be x = 1/ln 2 ≈ 1.443.)
2Find two different functions f and g, neither of them constant, whose product is (fg)(x) = x² − 1 with domain x ≥ 1 only.
For example, f(x) = (x + 1)√(x − 1) and g(x) = √(x − 1). Then fg = (x + 1)(x − 1) = x² − 1, and the domain of fg is x ≥ 1 because both factors need x ≥ 1. (Many answers are possible; the key is to build the restriction into a factor.)
3For y = x² sin x on −2π ≤ x ≤ 2π, state the envelope, the zeros, and explain why the graph is flatter near the origin than y = x sin x.
Envelope y = x² and y = −x². Zeros: −2π, −π, 0, π, 2π. Near 0, sin x ≈ x, so x² sin x ≈ x³ and x sin x ≈ x². Since |x³| < |x²| when |x| < 1, the graph of x² sin x hugs the x-axis more closely there.
Quiz
1. Which x-values are zeros of (fg)(x) if f(x) = x − 3 and g(x) = x² + 1?
2. f(x) > 0 and g(x) < 0 on the interval 2 < x < 5. What is true of fg on that interval?
3. What is the degree of (fg)(x) if f is a cubic and g is a linear function?
4. Between which lines does the graph of y = x cos x lie?
5. What is the domain of y = √(x − 2) · √(7 − x)?
6. At x = 6, g(6) = 1. Which statement must be true?
Big idea: products are the second part of D2.1. The tools are the zeros of the factors, a sign chart, and (for trigonometric factors) an envelope. The domain is still the intersection of the domains.
How to help: ask your teen to explain why the product is negative on an interval ("one factor is positive and the other is negative"). For y = x sin x, ask what the graph must look like at x = π/2 and why.
Common difficulties:
- Multiplying the graphs' x-values or adding the y-values instead of multiplying the y-values.
- Missing that a squared factor gives a double zero (touch, not cross).
- Combining √a · √b into √(ab) before finding the domain.
- Expecting the envelope lines to be asymptotes: the graph touches them over and over.
In the Challenge, the second question has many correct answers; check that the product simplifies to x² − 1 and that the domain is exactly x ≥ 1.