Adding and Subtracting Functions

You can combine two functions into a new one just by adding or subtracting their outputs. This happens all the time in the real world: profit is revenue minus cost, and the sound of a chord or a phone’s dial tone is two waves added together. In this lesson you’ll add and subtract functions with equations, tables, and graphs, and see which properties of the pieces carry over to the result. Trig functions here use radians.

Key ideas

The sum and difference of two functions

For two functions $f$ and $g$:

$$(f + g)(x) = f(x) + g(x) \qquad\qquad (f - g)(x) = f(x) - g(x)$$

In words: put the same $x$ into both functions, then add (or subtract) the two outputs. For example, if $f(2) = 7$ and $g(2) = -3$, then $(f + g)(2) = 7 + (-3) = 4$ and $(f - g)(2) = 7 - (-3) = 10$.

Order matters for subtraction: $(g - f)(x) = -(f - g)(x)$, so the two differences are opposites.

The domain is the overlap

$(f + g)(x)$ only makes sense when both $f(x)$ and $g(x)$ exist. So the domain of $f + g$ (and of $f - g$) is the set of $x$-values that are in the domain of $f$ and in the domain of $g$: the intersection, or overlap, of the two domains.

From a table

Line up the $x$-values and add (or subtract) the outputs in each column. You can only combine at $x$-values that appear in both tables.

From a graph: add the y-values

To sketch $y = f(x) + g(x)$, add the heights of the two graphs at each $x$, not the $x$-values. Some points are especially easy:

• Where $g(x) = 0$, the sum equals $f(x)$, so the new graph meets the graph of $f$ there (and the same the other way round).
• Where $f(x)$ and $g(x)$ are opposites, the sum is $0$: a zero of $f + g$.
• Where $f(x) = g(x)$, the difference $f - g$ is $0$.
• Where both graphs are high, the sum is even higher; where one is positive and one is negative, they partly cancel.

Superposition of waves

Adding sinusoidal functions is called superposition. A musical chord, a phone’s dial tone (each key plays two pure tones at once), and noise-cancelling headphones all work by adding waves. The period of a sum like $\sin x + \sin 2x$ is the smallest length after which both pieces repeat: $\sin x$ repeats every $2\pi$ and $\sin 2x$ every $\pi$, so the sum repeats every $2\pi$. The sum is usually not a simple sinusoid: its shape can be quite different from either piece (see Example 3).

Properties of sums

From even and odd functions: $f$ is even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$.

$f$	$g$	$f + g$ and $f - g$
even	even	even
odd	odd	odd
even	odd	usually neither

For example, if $f$ and $g$ are both even, then $(f + g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f + g)(x)$.

Increasing behaviour adds up too: if $f$ and $g$ are both increasing on an interval, so is $f + g$. But the difference of two increasing functions can do anything (see Practice question 9).

Worked examples

Example 1: Combining equations

Let $f(x) = x^2 - 3x$ and $g(x) = 2x + 5$. Find $(f + g)(x)$ and $(f - g)(x)$, then evaluate $(f - g)(-1)$.

Solution.

$$(f + g)(x) = (x^2 - 3x) + (2x + 5) = x^2 - x + 5$$

For the difference, put brackets around $g(x)$ so the minus sign reaches every term:

$$\begin{aligned} (f - g)(x) &= (x^2 - 3x) - (2x + 5) \\ &= x^2 - 3x - 2x - 5 \\ &= x^2 - 5x - 5 \end{aligned}$$

Then $(f - g)(-1) = (-1)^2 - 5(-1) - 5 = 1 + 5 - 5 = 1$.

Check with the original functions: $f(-1) = 1 + 3 = 4$ and $g(-1) = -2 + 5 = 3$, so $(f - g)(-1) = 4 - 3 = 1$. ✓

Example 2: The domain of a sum

Let $f(x) = \sqrt{x - 1}$ and $g(x) = \sqrt{5 - x}$. Find the domain of $h(x) = (f + g)(x)$, and evaluate $h(1)$, $h(3)$ and $h(5)$.

Solution. $f(x)$ needs $x - 1 \ge 0$, so $x \ge 1$. $g(x)$ needs $5 - x \ge 0$, so $x \le 5$. The sum needs both, so the domain is the overlap:

$$\{x \in \mathbb{R} \mid 1 \le x \le 5\}$$

Now evaluate:

$$\begin{aligned} h(1) &= \sqrt{0} + \sqrt{4} = 2 \\ h(3) &= \sqrt{2} + \sqrt{2} = 2\sqrt{2} \approx 2.83 \\ h(5) &= \sqrt{4} + \sqrt{0} = 2 \end{aligned}$$

Something like $h(6)$ doesn’t exist: $g(6) = \sqrt{-1}$ isn’t a real number, even though $f(6) = \sqrt{5}$ is fine.

Example 3: Adding two waves

Let $h(x) = \sin x + \sin 2x$, with $x$ in radians. Make a table of values for $0 \le x \le 2\pi$, sketch the graph, and find the period and the zeros.

Solution. Add the outputs of $\sin x$ and $\sin 2x$ at multiples of $\tfrac{\pi}{4}$ (decimals rounded to 2 places):

$x$	$0$	$\frac{\pi}{4}$	$\frac{\pi}{2}$	$\frac{3\pi}{4}$	$\pi$	$\frac{5\pi}{4}$	$\frac{3\pi}{2}$	$\frac{7\pi}{4}$	$2\pi$
$\sin x$	$0$	$0.71$	$1$	$0.71$	$0$	$-0.71$	$-1$	$-0.71$	$0$
$\sin 2x$	$0$	$1$	$0$	$-1$	$0$	$1$	$0$	$-1$	$0$
$h(x)$	$0$	$1.71$	$1$	$-0.29$	$0$	$0.29$	$-1$	$-1.71$	$0$

The solid curve is the sum of the dashed and dotted waves: at every $x$, its height is the sum of their heights.

Period. $\sin x$ repeats every $2\pi$, and $\sin 2x$ repeats every $\pi$ (so also every $2\pi$). Both pieces are back to the start after $2\pi$, so the period of $h$ is $2\pi$.

Zeros. Use the double angle identity $\sin 2x = 2\sin x\cos x$ (you’ll see it in double angle formulas):

$$\sin x + 2\sin x\cos x = \sin x\,(1 + 2\cos x) = 0$$

So $\sin x = 0$, giving $x = 0, \pi, 2\pi$, or $\cos x = -\tfrac{1}{2}$, giving $x = \tfrac{2\pi}{3}, \tfrac{4\pi}{3}$. The zeros in $[0, 2\pi]$ are $0$, $\tfrac{2\pi}{3}$, $\pi$, $\tfrac{4\pi}{3}$ and $2\pi$.

Notice how different the sum looks from either wave. Its maximum, about $1.76$ near $x \approx 0.94$, is less than $1 + 1 = 2$, because $\sin x$ and $\sin 2x$ never reach their peaks at the same $x$.

Example 4: Even, odd, or neither?

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

• (a) $p(x) = x^4 + \cos x$
• (b) $q(x) = x^3 + \sin x$
• (c) $r(x) = x^2 + x$

Solution. Replace $x$ with $-x$ and compare.

(a) $x^4$ and $\cos x$ are both even: $p(-x) = (-x)^4 + \cos(-x) = x^4 + \cos x = p(x)$. Even.

(b) $x^3$ and $\sin x$ are both odd: $q(-x) = (-x)^3 + \sin(-x) = -x^3 - \sin x = -q(x)$. Odd.

(c) $r(-x) = (-x)^2 + (-x) = x^2 - x$. This is neither $r(x) = x^2 + x$ nor $-r(x) = -x^2 - x$. Neither. A quick numerical check: $r(1) = 2$ but $r(-1) = 0$, which is neither $2$ nor $-2$.

Common mistakes

Not distributing the minus sign. $(x^2 - 3x) - (2x + 5)$ is $x^2 - 5x - 5$, not $x^2 - 5x + 5$. Always put brackets around the function you’re subtracting.

Using every $x$ that works for either function. The domain of $f + g$ is where both are defined, not where at least one is. In Example 2, $x = 6$ works for $f$ but not for $g$, so it’s not in the domain of the sum.

Adding the x-values on a graph. Graphical addition adds heights at the same $x$. The point $(2, 3)$ on $f$ and $(2, 5)$ on $g$ give $(2, 8)$ on $f + g$, not $(4, 8)$.

Treating f − g and g − f as the same. They’re opposites: if $(f - g)(3) = 4$, then $(g - f)(3) = -4$.

Assuming the period of a sum is the smaller period. $\sin x + \sin 2x$ does not repeat every $\pi$: the $\sin x$ part hasn’t come back yet. Look for the first length after which both pieces repeat.

Assuming even + odd is even (or odd). It’s usually neither, as in $x^2 + x$. Test with $f(-x)$, or try one number like $x = 1$ and $x = -1$.

Practice

1. (Warm-up) Let $f(x) = 3x - 4$ and $g(x) = x^2 + 1$. Find $(f + g)(x)$, $(f - g)(x)$ and $(g - f)(x)$.

Solution

$$(f + g)(x) = 3x - 4 + x^2 + 1 = x^2 + 3x - 3$$$$(f - g)(x) = 3x - 4 - (x^2 + 1) = -x^2 + 3x - 5$$$$(g - f)(x) = x^2 + 1 - (3x - 4) = x^2 - 3x + 5$$

Notice that $g - f$ is the opposite of $f - g$.

2. (Warm-up) Use the table to complete a table of values for $f + g$ and $f - g$.

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

Solution

Add or subtract down each column:

$x$	$-2$	$-1$	$0$	$1$	$2$
$(f + g)(x)$	$2$	$1$	$2$	$5$	$10$
$(f - g)(x)$	$8$	$3$	$0$	$-1$	$0$

The difference is $0$ at $x = 0$ and $x = 2$, exactly where $f(x) = g(x)$.

3. (Warm-up) On the same axes, the graph of $f$ passes through $(1, 4)$ and the graph of $g$ passes through $(1, -6)$. Which point is on the graph of $y = f(x) + g(x)$? Which is on $y = f(x) - g(x)$?

Solution

Keep $x = 1$ and add or subtract the heights.

$f + g$: $4 + (-6) = -2$, so the point is $(1, -2)$.

$f - g$: $4 - (-6) = 10$, so the point is $(1, 10)$.

4. (Core) Let $f(x) = \sqrt{x + 3}$ and $g(x) = \dfrac{1}{x - 2}$. State the domain of $f + g$, and evaluate $(f + g)(1)$.

Solution

$f$ needs $x \ge -3$. $g$ needs $x \ne 2$. The sum needs both:

$$\{x \in \mathbb{R} \mid x \ge -3,\ x \ne 2\}$$$$(f + g)(1) = \sqrt{4} + \frac{1}{1 - 2} = 2 - 1 = 1$$

5. (Core) A bakery’s daily revenue from selling $n$ loaves is $R(n) = 15n$ dollars, and its daily cost is $C(n) = 0.02n^2 + 3n + 800$ dollars.

• (a) Find the profit function $P(n) = R(n) - C(n)$.
• (b) Find the profit when $200$ loaves are sold.
• (c) How many loaves give the maximum profit, and what is that profit?

Solution

(a)

$$P(n) = 15n - (0.02n^2 + 3n + 800) = -0.02n^2 + 12n - 800$$

(b) $P(200) = -0.02(40\,000) + 2400 - 800 = -800 + 2400 - 800 = 800$, so the profit is $800.

(c) $P$ is a parabola opening down. Its vertex is at $n = -\dfrac{b}{2a} = -\dfrac{12}{2(-0.02)} = 300$:

$$P(300) = -0.02(90\,000) + 3600 - 800 = -1800 + 3600 - 800 = 1000$$

Selling $300$ loaves gives the maximum profit, $1000.

6. (Core) Decide whether each function is even, odd, or neither. Show your reasoning.

• (a) $f(x) = x^3 + 2x$
• (b) $g(x) = x^2 - \cos x$
• (c) $h(x) = x^2 + \sin x$

Solution

(a) $f(-x) = -x^3 - 2x = -(x^3 + 2x) = -f(x)$. Odd (odd + odd).

(b) $g(-x) = (-x)^2 - \cos(-x) = x^2 - \cos x = g(x)$. Even (even − even).

(c) $h(-x) = x^2 - \sin x$, which is neither $h(x)$ nor $-h(x)$. Neither (even + odd). For example, $h\!\left(\tfrac{\pi}{2}\right) = \tfrac{\pi^2}{4} + 1 \approx 3.47$ but $h\!\left(-\tfrac{\pi}{2}\right) = \tfrac{\pi^2}{4} - 1 \approx 1.47$.

7. (Core) Let $h(x) = \sin x + \cos x$, with $x$ in radians.

• (a) Make a table of values at multiples of $\tfrac{\pi}{4}$ from $0$ to $2\pi$. Give exact values.
• (b) State the zeros in $[0, 2\pi]$, the period, and the largest value in your table.

Solution

(a) Use the special values $\sin\tfrac{\pi}{4} = \cos\tfrac{\pi}{4} = \tfrac{\sqrt{2}}{2}$:

$x$	$0$	$\frac{\pi}{4}$	$\frac{\pi}{2}$	$\frac{3\pi}{4}$	$\pi$	$\frac{5\pi}{4}$	$\frac{3\pi}{2}$	$\frac{7\pi}{4}$	$2\pi$
$h(x)$	$1$	$\sqrt{2}$	$1$	$0$	$-1$	$-\sqrt{2}$	$-1$	$0$	$1$

(b) The zeros are $x = \tfrac{3\pi}{4}$ and $x = \tfrac{7\pi}{4}$, where $\sin x$ and $\cos x$ are opposites. Both pieces have period $2\pi$, so $h$ has period $2\pi$. The largest value in the table is $\sqrt{2} \approx 1.41$ at $x = \tfrac{\pi}{4}$.

(In fact this sum is a sinusoid: $\sin x + \cos x = \sqrt{2}\sin\left(x + \tfrac{\pi}{4}\right)$, so $\sqrt{2}$ really is its maximum. Adding two waves with the same period always gives a sinusoid; adding waves with different periods, as in Example 3, usually doesn’t.)

8. (Challenge) Let $f(x) = 2^x + 2^{-x}$.

• (a) Show that $f$ is even.
• (b) Make a table of values for $x = -2, -1, 0, 1, 2$.
• (c) Show that $f(x) \ge 2$ for every $x$. (Hint: expand $\left(2^{x/2} - 2^{-x/2}\right)^2 \ge 0$.)

Solution

(a) $f(-x) = 2^{-x} + 2^{x} = f(x)$, so $f$ is even. (It’s the sum of $2^x$ and its mirror image $2^{-x}$.)

(b)

$x$	$-2$	$-1$	$0$	$1$	$2$
$f(x)$	$4.25$	$2.5$	$2$	$2.5$	$4.25$

(c) A square is never negative, and $2^{x/2} \cdot 2^{-x/2} = 2^0 = 1$, so

$$\begin{aligned} \left(2^{x/2} - 2^{-x/2}\right)^2 &\ge 0 \\ 2^{x} - 2\left(2^{x/2}\right)\left(2^{-x/2}\right) + 2^{-x} &\ge 0 \\ 2^{x} - 2 + 2^{-x} &\ge 0 \\ 2^{x} + 2^{-x} &\ge 2 \end{aligned}$$

So the minimum value of $f$ is $2$, reached at $x = 0$, matching the table.

9. (Challenge) $f$ and $g$ are both increasing for all real $x$.

• (a) Explain why $f + g$ must be increasing.
• (b) Give an example showing that $f - g$ does not have to be increasing.

Solution

(a) Take any $a \lt b$. Because both functions increase, $f(a) \lt f(b)$ and $g(a) \lt g(b)$. Adding these inequalities gives $f(a) + g(a) \lt f(b) + g(b)$, which says $(f + g)(a) \lt (f + g)(b)$. So $f + g$ is increasing.

(b) Let $f(x) = x$ and $g(x) = 2x$. Both are increasing, but $(f - g)(x) = x - 2x = -x$ is decreasing. (You can’t subtract inequalities the way you can add them.)