Multiplying and Dividing Functions

Just as you can add and subtract functions, you can multiply and divide them. Products and quotients build some of the most useful models there are: revenue is price times quantity, and a bouncing spring that slowly settles down is an exponential function times a sinusoidal one. Trig functions here use radians.

Key ideas

The product and quotient of two functions

$$(fg)(x) = f(x)\,g(x) \qquad\qquad \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}, \quad g(x) \ne 0$$

Put the same $x$ into both functions, then multiply or divide the outputs. Careful with notation: $(fg)(x)$ means the product $f(x) \cdot g(x)$. It is not the same as $f(g(x))$, which is a composition.

Domains

• $fg$: the overlap of the domains of $f$ and $g$ (just like $f + g$).
• $\dfrac{f}{g}$: the overlap of the domains, and also remove every $x$ where $g(x) = 0$.

Find the domain before you simplify. If a factor cancels, its zero is still excluded, and the graph has a hole there instead of an asymptote. This is the same idea as the restrictions on rational expressions.

Key features of a product

Think about the y-values of the two pieces at each $x$:

• Zeros: $(fg)(x) = 0$ exactly when $f(x) = 0$ or $g(x) = 0$ (inside the common domain). So the product’s zeros are the zeros of both factors put together.
• Sign: the product is positive where the factors have the same sign, and negative where they have opposite signs.
• Where a factor equals 1: if $g(x) = 1$, then $(fg)(x) = f(x)$, so the graphs meet there. If $g(x) = -1$, the product is $-f(x)$.
• Envelopes: since $-1 \le \sin x \le 1$, the product $f(x)\sin x$ always stays between $-f(x)$ and $f(x)$ (when $f(x) \ge 0$). The curves $y = \pm f(x)$ form an envelope that the product bounces between.

Damped oscillations

A spring, a pendulum, or a guitar string vibrates back and forth, but friction makes each swing a little smaller. That is a sinusoidal function multiplied by a decreasing exponential:

$$d(t) = A\,b^{t}\cos(kt) \qquad (0 \lt b \lt 1)$$

The cosine supplies the back-and-forth motion, and $A\,b^{t}$ is the envelope that shrinks the amplitude toward $0$.

Even and odd products and quotients

The rules work like the signs in multiplication (think of even as $+$ and odd as $-$):

$f$	$g$	$fg$ and $\dfrac{f}{g}$
even	even	even
odd	odd	even
even	odd	odd

For example, if $f$ and $g$ are both odd, then $(fg)(-x) = f(-x)\,g(-x) = \big(-f(x)\big)\big(-g(x)\big) = f(x)\,g(x)$, so $fg$ is even. And $\tan x = \dfrac{\sin x}{\cos x}$ is odd divided by even, which is odd.

Worked examples

Example 1: Product and quotient from equations

Let $f(x) = 2x - 1$ and $g(x) = x + 3$. Find $(fg)(x)$ and $\left(\dfrac{f}{g}\right)(x)$, and state the domain of each. Then evaluate $(fg)(2)$.

Solution.

$$(fg)(x) = (2x - 1)(x + 3) = 2x^2 + 5x - 3$$

Both pieces are defined for all real numbers, so the domain of $fg$ is $\{x \in \mathbb{R}\}$.

$$\left(\frac{f}{g}\right)(x) = \frac{2x - 1}{x + 3}$$

$g(x) = 0$ when $x = -3$, so the domain of $\dfrac{f}{g}$ is $\{x \in \mathbb{R} \mid x \ne -3\}$.

$(fg)(2) = 2(4) + 5(2) - 3 = 15$. Check: $f(2) = 3$ and $g(2) = 5$, and $3 \times 5 = 15$. ✓

Example 2: A quotient with a hole

Let $f(x) = x^2 - 9$ and $g(x) = x - 3$. Find $\left(\dfrac{f}{g}\right)(x)$, state its domain, and describe its graph.

Solution. First the domain: $g(x) = 0$ at $x = 3$, so $x = 3$ is excluded. Now simplify:

$$\left(\frac{f}{g}\right)(x) = \frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3, \qquad x \ne 3$$

The domain is $\{x \in \mathbb{R} \mid x \ne 3\}$. The graph is the line $y = x + 3$ with a hole at $x = 3$. The hole’s $y$-coordinate comes from the simplified form: $3 + 3 = 6$, so the hole is at $(3, 6)$.

$y = \dfrac{x^2 - 9}{x - 3}$ is the line $y = x + 3$ with the point $(3, 6)$ missing.

Compare $\left(\dfrac{g}{f}\right)(x) = \dfrac{x - 3}{(x - 3)(x + 3)} = \dfrac{1}{x + 3}$ for $x \ne \pm 3$. Here the factor $x + 3$ does not cancel, so $x = -3$ gives a vertical asymptote, while $x = 3$ gives a hole at $\left(3, \tfrac{1}{6}\right)$.

Example 3: A damped spring

A weight on a spring is pulled down and released. Its displacement below its resting position (positive means below), in centimetres, after $t$ seconds is modelled by

$$d(t) = 12(0.8)^t\cos(\pi t)$$

• (a) Find $d(0)$, $d(1)$, $d(2)$ and $d(3)$, and describe the motion.
• (b) Find the times in the first $4$ s when the weight passes through its resting position.
• (c) After how many seconds is the envelope less than $1$ cm?

Solution.

(a) $\cos(\pi t)$ is $1$ at even whole numbers and $-1$ at odd whole numbers:

$$\begin{aligned} d(0) &= 12(1)(1) = 12 \\ d(1) &= 12(0.8)(-1) = -9.6 \\ d(2) &= 12(0.64)(1) = 7.68 \\ d(3) &= 12(0.512)(-1) = -6.144 \end{aligned}$$

The weight swings from one side to the other every second, and each swing is $80\%$ as big as the one before. The cosine makes it oscillate with period $\tfrac{2\pi}{\pi} = 2$ s, and the factor $12(0.8)^t$ shrinks the amplitude.

The product bounces between the envelope curves $d = \pm 12(0.8)^t$.

(b) $d(t) = 0$ when either factor is $0$. The factor $12(0.8)^t$ is never $0$, so we need $\cos(\pi t) = 0$:

$$\pi t = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2} \quad\Rightarrow\quad t = 0.5,\ 1.5,\ 2.5,\ 3.5$$

The weight passes through its resting position at $0.5$, $1.5$, $2.5$ and $3.5$ s.

(c) Solve $12(0.8)^t \lt 1$, using logarithms:

$$\begin{aligned} (0.8)^t &\lt \frac{1}{12} \\ t \log 0.8 &\lt \log\frac{1}{12} \\ t &\gt \frac{\log(1/12)}{\log 0.8} \approx 11.14 \end{aligned}$$

(The inequality flips because $\log 0.8$ is negative.) After about $11.14$ s, every swing is less than $1$ cm.

Example 4: Even or odd products

Decide whether each function is even, odd, or neither.

• (a) $f(x) = x\sin x$
• (b) $g(x) = x^2\sin x$
• (c) $h(x) = \dfrac{\sin x}{x}$, $x \ne 0$

Solution. $x$ and $\sin x$ are odd; $x^2$ is even.

(a) Odd × odd is even: $f(-x) = (-x)\sin(-x) = (-x)(-\sin x) = x\sin x = f(x)$. Even.

(b) Even × odd is odd: $g(-x) = (-x)^2\sin(-x) = -x^2\sin x = -g(x)$. Odd.

(c) Odd ÷ odd is even: $h(-x) = \dfrac{\sin(-x)}{-x} = \dfrac{-\sin x}{-x} = h(x)$. Even.

Numerical check for (a): $f\!\left(\tfrac{\pi}{2}\right) = \tfrac{\pi}{2}(1) \approx 1.57$ and $f\!\left(-\tfrac{\pi}{2}\right) = \left(-\tfrac{\pi}{2}\right)(-1) \approx 1.57$. ✓

Common mistakes

Finding the domain after simplifying. $\dfrac{x^2 - 9}{x - 3}$ simplifies to $x + 3$, but $x = 3$ is still excluded. Always find the restrictions from the original quotient.

Confusing fg with f(g(x)). $(fg)(x)$ multiplies the outputs; $f(g(x))$ puts one function inside the other. For $f(x) = x + 1$ and $g(x) = x^2$, $(fg)(x) = x^3 + x^2$ but $f(g(x)) = x^2 + 1$.

Expecting zeros from an exponential factor. In $12(0.8)^t\cos(\pi t)$, only the cosine can be $0$: an exponential like $b^t$ is always positive.

Saying “odd × odd = odd”. The rule follows signs: two negatives make a positive, so odd × odd is even. Check with $x \cdot x = x^2$.

Forgetting the zeros of g are excluded even when f is also zero there. At $x = 3$ in Example 2, both $f$ and $g$ are $0$, and $\tfrac{0}{0}$ is undefined. That’s a hole, not a point on the graph.

Flipping the wrong way when dividing by a negative log. $\log 0.8 \lt 0$, so dividing by it reverses the inequality in Example 3(c).

Practice

1. (Warm-up) Let $f(x) = x + 4$ and $g(x) = x - 1$. Find $(fg)(x)$ and $\left(\dfrac{f}{g}\right)(x)$, and state the domain of $\dfrac{f}{g}$.

Solution

$$(fg)(x) = (x + 4)(x - 1) = x^2 + 3x - 4$$$$\left(\frac{f}{g}\right)(x) = \frac{x + 4}{x - 1}, \qquad \{x \in \mathbb{R} \mid x \ne 1\}$$

2. (Warm-up) If $f(3) = -2$ and $g(3) = 5$, find $(fg)(3)$, $\left(\dfrac{f}{g}\right)(3)$ and $\left(\dfrac{g}{f}\right)(3)$.

Solution

$(fg)(3) = (-2)(5) = -10$

$\left(\dfrac{f}{g}\right)(3) = \dfrac{-2}{5} = -0.4$

$\left(\dfrac{g}{f}\right)(3) = \dfrac{5}{-2} = -2.5$

3. (Warm-up) Use the table to find $(fg)(x)$ and $\left(\dfrac{f}{g}\right)(x)$ at each $x$. Where is $\dfrac{f}{g}$ undefined?

$x$	$0$	$1$	$2$	$3$
$f(x)$	$4$	$-3$	$6$	$0$
$g(x)$	$2$	$3$	$0$	$-5$

Solution

$x$	$0$	$1$	$2$	$3$
$(fg)(x)$	$8$	$-9$	$0$	$0$
$\left(\frac{f}{g}\right)(x)$	$2$	$-1$	undefined	$0$

$\dfrac{f}{g}$ is undefined at $x = 2$, because $g(2) = 0$. The product is $0$ at $x = 2$ and $x = 3$, where one of the factors is $0$.

4. (Core) Let $f(x) = x^2 - x - 6$ and $g(x) = x + 2$. Simplify $\left(\dfrac{f}{g}\right)(x)$, state its domain, and give the coordinates of any hole.

Solution

$g(x) = 0$ at $x = -2$, so that’s excluded. Factor and simplify:

$$\left(\frac{f}{g}\right)(x) = \frac{(x - 3)(x + 2)}{x + 2} = x - 3, \qquad x \ne -2$$

Domain: $\{x \in \mathbb{R} \mid x \ne -2\}$. The graph is the line $y = x - 3$ with a hole at $x = -2$, where $y = -2 - 3 = -5$: the hole is at $(-2, -5)$.

5. (Core) Let $f(x) = \sqrt{x + 1}$ and $g(x) = x^2 - 4$. State the domain of $\dfrac{f}{g}$.

Solution

$f$ needs $x \ge -1$. $g$ is defined everywhere, but $g(x) = 0$ at $x = \pm 2$. The value $x = -2$ is already outside $x \ge -1$, so only $x = 2$ needs to be removed:

$$\{x \in \mathbb{R} \mid x \ge -1,\ x \ne 2\}$$

6. (Core) Let $h(x) = (x - 1) \cdot 2^x$. Without graphing technology, find the zeros and the $y$-intercept of $h$, and state where $h(x)$ is positive and where it is negative.

Solution

$2^x$ is always positive, so it has no zeros. The only zero comes from $x - 1 = 0$: $x = 1$.

$y$-intercept: $h(0) = (0 - 1)(1) = -1$.

Sign: since $2^x \gt 0$, $h(x)$ has the same sign as $x - 1$. So $h(x) \lt 0$ for $x \lt 1$ and $h(x) \gt 0$ for $x \gt 1$.

7. (Core) A café sells $200$ muffins a day at $3.00 each. For every $0.10 decrease in price, it sells $20$ more muffins. After $x$ price decreases, the price is $p(x) = 3 - 0.1x$ dollars and the number sold is $q(x) = 200 + 20x$.

• (a) Find the revenue function $R(x) = (pq)(x)$.
• (b) What price gives the maximum revenue, and what is that revenue?

Solution

(a)

$$\begin{aligned} R(x) &= (3 - 0.1x)(200 + 20x) \\ &= 600 + 60x - 20x - 2x^2 \\ &= -2x^2 + 40x + 600 \end{aligned}$$

(b) The vertex is at $x = -\dfrac{40}{2(-2)} = 10$ price decreases:

$$R(10) = -200 + 400 + 600 = 800$$

The price is $3 - 0.1(10) = 2.00$, or $2.00, the café sells $200 + 200 = 400$ muffins, and the revenue is $800. Check: $2.00 \times 400 = 800$. ✓

8. (Challenge) Decide whether each function is even, odd, or neither, and prove the general rule behind (c).

• (a) $f(x) = x\cos x$
• (b) $g(x) = \sin^2 x$
• (c) $h(x) = \tan x$

Solution

(a) Odd × even: $f(-x) = (-x)\cos(-x) = -x\cos x = -f(x)$. Odd.

(b) $\sin^2 x = (\sin x)(\sin x)$ is odd × odd: $g(-x) = \big(\sin(-x)\big)^2 = (-\sin x)^2 = \sin^2 x = g(x)$. Even.

(c) $\tan x = \dfrac{\sin x}{\cos x}$ is odd ÷ even. Odd.

General rule: suppose $f$ is odd and $g$ is even. Then, wherever $g(x) \ne 0$,

$$\left(\frac{f}{g}\right)(-x) = \frac{f(-x)}{g(-x)} = \frac{-f(x)}{g(x)} = -\left(\frac{f}{g}\right)(x)$$

so $\dfrac{f}{g}$ is odd.

9. (Challenge) Water drains from a storm sewer into a lake. After $t$ seconds, the concentration of a pollutant in the water is $c(t) = 2t^2$ kilograms per cubic metre, and the water flows out at $w(t) = \dfrac{3}{t^4 + 16}$ cubic metres per second.

• (a) Show that the rate at which the pollutant enters the lake, in kilograms per second, is $r(t) = \dfrac{6t^2}{t^4 + 16}$.
• (b) Make a table of $r(t)$ for $t = 1, 1.5, 2, 2.5, 3$ (to 2 decimal places).
• (c) Show that $r(t) \le 0.75$ for every $t \gt 0$, so the maximum rate is $0.75$ kg/s at $t = 2$ s. (Hint: $(t^2 - 4)^2 \ge 0$.)

Solution

(a) (kg per m³) × (m³ per s) = kg per s, so the rate is the product:

$$r(t) = c(t)\,w(t) = 2t^2 \cdot \frac{3}{t^4 + 16} = \frac{6t^2}{t^4 + 16}$$

(b)

$t$ (s)	$1$	$1.5$	$2$	$2.5$	$3$
$r(t)$ (kg/s)	$0.35$	$0.64$	$0.75$	$0.68$	$0.56$

The table suggests a maximum near $t = 2$.

(c) Expand the square:

$$\begin{aligned} (t^2 - 4)^2 &\ge 0 \\ t^4 - 8t^2 + 16 &\ge 0 \\ t^4 + 16 &\ge 8t^2 \end{aligned}$$

Since $t^4 + 16 \ge 8t^2 \gt 0$, a bigger denominator gives a smaller fraction:

$$r(t) = \frac{6t^2}{t^4 + 16} \le \frac{6t^2}{8t^2} = 0.75$$

Equality happens only when $t^2 - 4 = 0$, so at $t = 2$. The maximum rate is $0.75$ kg/s, at $t = 2$ s. (Graphing technology shows the same peak.)