Algebraic Combinations of Functions

$(f+g)(x)=x^2+2x-2$, $(f-g)(x)=-x^2+2x+4$, $(f\cdot g)(x)=2x^3+x^2-6x-3$, all with domain $ℝ$; and $(f/g)(x)={\displaystyle \frac{2x+1}{x^2-3}}$ with domain $ℝ-\{-\sqrt{3},\sqrt{3}\}$.

Domains of the ingredients. Both $2x+1$ and $x^2-3$ are polynomials, and a polynomial can be evaluated at every real number. So

Sum.

Domain: $ℝ$.

Difference. Subtracting a bracket changes the sign of every term inside it:

Domain: $ℝ$. Note that $-(-3)=+3$; dropping that sign change is the most common error here.

Product. Multiply each term of the first bracket by each term of the second:

Domain: $ℝ$.

Quotient.

For the domain we must also remove the points where the denominator is zero:

So the domain is $ℝ-\{-\sqrt{3},\sqrt{3}\}$.

Check. Test $x=2$ in each result. $f(2)=5$ and $g(2)=1$. Then $f(2)+g(2)=6$, and the sum formula gives $4+4-2=6$ ✓. $f(2)-g(2)=4$, and the difference formula gives $-4+4+4=4$ ✓. $f(2)g(2)=5$, and the product formula gives $16+4-12-3=5$ ✓. Finally $f(2)/g(2)=5/1=5$, and the quotient formula gives $5/1=5$ ✓.

(a) $8$   (b) $-5$   (c) $-2$   (d) does not exist   (e) does not exist

The quickest route is to evaluate $f$ and $g$ separately at the given number, then combine.

(a) $f(3)=3^2=9$ and $g(3)=3-4=-1$. So

(b) $f(-1)=(-1{)}^2=1$ and $g(-1)=-1-4=-5$. So

Be careful with $(-1{)}^2$: an even power of a negative number is positive, so $f(-1)=1$, not $-1$.

(c) $f(2)=4$ and $g(2)=2-4=-2$. So

(d) $g(0)=0-4=-4$ and $f(0)=0^2=0$. The quotient $g/f$ requires dividing by $f(0)=0$:

which is undefined. So $(g/f)(0)$ does not exist; the input $0$ is excluded from the domain of $g/f$.

(e) $f(4)=16$ and $g(4)=4-4=0$. Again the denominator is zero, so

does not exist.

The lesson from (d) and (e). For a quotient, always evaluate the denominator function first and check that it is not zero. Both $f$ and $g$ here have domain $ℝ$, yet $f/g$ is undefined at $4$ and $g/f$ is undefined at $0$ — the exclusions come from the operation, not from the ingredients.

$f+g$ and $f\cdot g$: $[-1,4]$;   $f/g$: $[-1,4)$;   $g/f$: $(-1,4]$.

Domain of $f$. A square root needs a nonnegative quantity underneath:

Domain of $g$.

Take care when solving $4-x\ge 0$: adding $x$ to both sides gives $4\ge x$, which is the same as $x\le 4$.

The common domain. A number must satisfy both conditions:

$f+g$ and $f\cdot g$. For sums, differences and products the domain is exactly the common part:

$f/g$. Start from $[-1,4]$ and remove the points where the denominator $g$ is zero:

So $4$ must go, and

$g/f$. Now the denominator is $f$, and

So $-1$ must go, and

Check. At $x=4$: $g(4)=\sqrt{0}=0$, so $f(4)/g(4)=\sqrt{5}/0$ is undefined ✓, while $g(4)/f(4)=0/\sqrt{5}=0$ is a perfectly good value ✓. A zero in the numerator is harmless; only a zero in the denominator removes a point.

$(f+g)(x)={\displaystyle \frac{2x-2}{x(x-2)}}$, with domain $ℝ-\{0,2\}$.

Domains of the ingredients.

A number is allowed for the sum only if it is allowed for both, so

Combining the fractions. The two denominators are $x$ and $x-2$; they share no common factor, so the least common denominator is their product $x(x-2)$. Rewrite each fraction with that denominator:

Now add the numerators over the common denominator:

If you prefer the denominator expanded, $x(x-2)=x^2-2x$.

A warning about the domain. The single fraction ${\displaystyle \frac{2x-2}{x(x-2)}}$ happens to have exactly the same forbidden inputs, $0$ and $2$. But that is a lucky feature of this example, not a rule: the domain of $f+g$ is decided by the domains of $f$ and $g$ before any algebra, and it must be stated from there.

Check. Take $x=1$. Directly: $f(1)+g(1)={\displaystyle \frac{1}{1}}+{\displaystyle \frac{1}{-1}}=1-1=0$. From the single fraction: ${\displaystyle \frac{2-2}{1(1-2)}}={\displaystyle \frac{0}{-1}}=0$ ✓.

Take $x=4$. Directly: ${\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{4}}+{\displaystyle \frac{2}{4}}={\displaystyle \frac{3}{4}}$. From the fraction: ${\displaystyle \frac{8-2}{4\cdot 2}}={\displaystyle \frac{6}{8}}={\displaystyle \frac{3}{4}}$ ✓.

$(f/g)(x)=x-1$ for $x\ne -1$, with domain $ℝ-\{-1\}$. It is not equal to $h$, because the domains differ.

The domain first. Both $f$ and $g$ are polynomials, so both have domain $ℝ$ and the common domain is $ℝ$. For the quotient we must remove the zeros of the denominator:

Hence

The formula. Factor the numerator by the difference-of-squares formula:

For $x\ne -1$ the factor $x+1$ is not zero, so it may be cancelled:

Is $f/g$ equal to $h$? Two functions are equal only if they have the same domain and the same values. The values do agree wherever both are defined. But

and $-1$ belongs to the second but not the first. So the two functions are not equal.

The picture. The graph of $h$ is the whole line $y=x-1$; the graph of $f/g$ is that line with a hole at $(-1,-2)$.

Why the hole is real. At $x=-1$ the original quotient reads ${\displaystyle \frac{(-1{)}^2-1}{-1+1}}={\displaystyle \frac{0}{0}}$, which is not a number. Cancelling a factor can never legalise an input that was forbidden before the cancellation.

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

By definition $(f+g)(x)=f(x)+g(x)$, so

Subtract $f(x)$ from both sides:

Distribute the minus sign over every term of the second bracket:

Now combine like terms:

$x^2$-terms: $3x^2-x^2=2x^2$.
$x$-terms: $2x+x=3x$.
constants: $-1-4=-5$.

Check. Add $f$ and the answer:

The $x$-terms give $-x+3x=2x$ and the constants give $4-5=-1$, matching the given sum exactly.

$(f+g)(1)=3$, $(f-g)(2)=1$, $(f\cdot g)(0)=0$. The graph of $f+g$ is the horizontal line $y=3$.

Read the two values off the formulas (or off the picture) and then combine.

$(f+g)(1)$. $f(1)=1$ and $g(1)=3-1=2$, so

Geometrically: stack the height of $f$ above $x=1$ on top of the height of $g$ above $x=1$.

$(f-g)(2)$. $f(2)=2$ and $g(2)=3-2=1$, so

$(f\cdot g)(0)$. $f(0)=0$ and $g(0)=3-0=3$, so

Notice that a product is zero whenever either factor is zero, so the graph of $f\cdot g$ meets the $x$-axis at $x=0$ (a zero of $f$) and at $x=3$ (a zero of $g$).

The graph of $f+g$. Add the formulas:

The $x$-terms cancel, so $f+g$ is the constant function $3$, whose graph is the horizontal line $y=3$. Its domain is $ℝ$, since both $f$ and $g$ are defined everywhere.

This is visible in the picture: as $x$ increases, $f$ rises exactly as fast as $g$ falls, so their heights always total $3$. Check a few: at $x=0$, $0+3=3$; at $x=1$, $1+2=3$; at $x=3$, $3+0=3$ ✓.

(f+g)(x) = \begin{cases} 2x & \text{if } x \geq 0 \\ 0 & \text{if } x < 0 \end{cases}

Domain. Both formulas are defined for every real number, so $\mathrm{Dom}(f+g)=ℝ$.

The formula. By definition,

Now replace $|x|$ by its two-case definition.

If $x\ge 0$: then $|x|=x$, so

If : then $|x|=-x$, so

Hence

Reading the picture. For $x\ge 0$ the two graphs coincide, so adding them doubles the height. For the graph of $f$ is exactly as far below the axis as the graph of $g$ is above it, so the heights cancel and the sum sits on the axis.

Check. $x=3$: $3+|3|=3+3=6$, and the formula gives $2(3)=6$ ✓. $x=-3$: $-3+|-3|=-3+3=0$, and the formula gives $0$ ✓. $x=0$: $0+0=0$, and the first line gives $2(0)=0$ ✓, so the two pieces meet without a break.

(a) Equal; both have domain $[-2,2]$.   (b) Not equal: $\mathrm{Dom}(u_1)=[3,\mathrm{\infty })$ but $\mathrm{Dom}(u_2)=(-\mathrm{\infty },-1]\cup [3,\mathrm{\infty })$.

(a) Domain of $h_1$. This is a product of two square roots, so both quantities under the roots must be nonnegative:

Both conditions together give $-2\le x\le 2$, so $\mathrm{Dom}(h_1)=[-2,2]$.

Domain of $h_2$. Here only the product has to be nonnegative:

The factors are zero at $x=2$ and $x=-2$. Test one value in each of the three resulting stretches:

$x=-3$: .
$x=0$: .
$x=3$: .

So the product is nonnegative exactly on $[-2,2]$, and $\mathrm{Dom}(h_2)=[-2,2]$ too.

Values. On $[-2,2]$ both $2-x$ and $x+2$ are nonnegative, and for nonnegative $a$ and $b$ we have $\sqrt{a}\sqrt{b}=\sqrt{ab}$. So the two formulas agree at every point of the common domain.

Same domain and same values, so $h_1$ and $h_2$ are equal.

(b) Domain of $u_1$. Both radicands must be nonnegative:

Both hold exactly when $x\ge 3$, so $\mathrm{Dom}(u_1)=[3,\mathrm{\infty })$.

Domain of $u_2$. Only the product must be nonnegative:

The factors vanish at $x=3$ and $x=-1$. Testing:

$x=-2$: .
$x=0$: .
$x=4$: .

So $\mathrm{Dom}(u_2)=(-\mathrm{\infty },-1]\cup [3,\mathrm{\infty })$.

Conclusion. The number $-2$ belongs to $\mathrm{Dom}(u_2)$ but not to $\mathrm{Dom}(u_1)$: indeed $u_2(-2)=\sqrt{5}$, whereas $u_1(-2)$ would require $\sqrt{-5}$, which is not a real number. The domains differ, so $u_1$ and $u_2$ are not equal.

The moral. Two negative factors have a positive product. So $\sqrt{ab}$ can be a real number in situations where $\sqrt{a}$ and $\sqrt{b}$ separately are not. The rule $\sqrt{a}\sqrt{b}=\sqrt{ab}$ needs $a\ge 0$ and $b\ge 0$.

$(f/g)(x)={\displaystyle \frac{x+5}{x^2-4x+3}}$ with domain $ℝ-\{1,3\}$;   $(g/f)(x)={\displaystyle \frac{x^2-4x+3}{x+5}}$ with domain $ℝ-\{-5\}$.

Both $f$ and $g$ are polynomials, so each has domain $ℝ$ and the common domain is $ℝ$. Only the zeros of whichever function sits in the denominator need to be removed.

$f/g$. The formula is

Find the zeros of $g$. Factor the quadratic: we need two numbers with product $3$ and sum $-4$, namely $-1$ and $-3$:

A product is zero exactly when a factor is zero, so $g(x)=0$ at $x=1$ and $x=3$. Hence

$g/f$. The formula is

Now the denominator is $f$, and

Hence

Check. At $x=1$: $g(1)=1-4+3=0$, so $f(1)/g(1)=6/0$ is undefined ✓, whereas $g(1)/f(1)=0/6=0$ is fine ✓. At $x=-5$: $f(-5)=0$, so $g(-5)/f(-5)$ is undefined ✓, while $f(-5)/g(-5)=0/48=0$ is fine ✓.

The general point. $\mathrm{Dom}(f/g)$ and $\mathrm{Dom}(g/f)$ are usually different sets. Swapping numerator and denominator swaps which zeros must be excluded.

(a) $P(x)=40x-2000$   (b) $P(30)=-800$ (a loss of $800), $P(80)=1200$ (a profit of $1200)   (c) $50$ chairs

(a) By the definition of a difference of functions,

Distribute the minus sign over both terms of the bracket:

The number $40$ is the profit per chair (each chair sells for $85 and costs $45 to make), and $2000$ is the fixed cost that has to be covered whatever happens.

(b)

A negative profit is a loss: selling only $30$ chairs leaves the workshop $800 short, because the $1200 earned above cost does not cover the $2000 of fixed costs.

Selling $80$ chairs yields a profit of $1200.

(c) Set $P(x)=0$:

Add $2000$ to both sides:

Divide both sides by $40$:

So $50$ chairs must be sold to break even.

Check. At $x=50$: $R(50)=85(50)=4250$ and $C(50)=45(50)+2000=2250+2000=4250$. Revenue equals cost, so the profit is indeed $0$ ✓.

A remark on the domain. Only whole, nonnegative numbers of chairs make sense, so in practice we would use the formula for $x=0,1,2,\dots$ even though the algebraic expression $40x-2000$ is defined for every real $x$.

The correct domain is $[0,1)\cup (1,\mathrm{\infty })$.

What the definition requires. The domain of $f/g$ is

There are three conditions, and the student checked only the third.

Condition 1: $x$ must be in $\mathrm{Dom}(f)$. Since $f(x)=\sqrt{x}$ requires a nonnegative radicand,

This is what the student missed: numbers like $-4$ are excluded because $\sqrt{-4}$ is not a real number, quite apart from any division.

Condition 2: $x$ must be in $\mathrm{Dom}(g)$. Since $g$ is a polynomial, $\mathrm{Dom}(g)=ℝ$, so this condition excludes nothing.

Condition 3: $g(x)\ne 0$.

so $1$ must be removed.

Putting the three together.

Check. At $x=-4$ the student's answer would allow the input, but

has no meaning, since $\sqrt{-4}$ is not a real number. At $x=0$: ${\displaystyle \frac{\sqrt{0}}{-1}}={\displaystyle \frac{0}{-1}}=0$, a perfectly good value, so $0$ correctly belongs to the domain — a zero in the numerator causes no trouble.

The habit to build. Before worrying about the operation, list the domains of the ingredients. The operation can only take points away, never add them back.

Sometimes true: the two domains agree exactly when $g$ has no zeros in $\mathrm{Dom}(f)\cap \mathrm{Dom}(g)$.

Write $D=\mathrm{Dom}(f)\cap \mathrm{Dom}(g)$. By the definitions,

So the second set is the first with the zeros of $g$ removed. The two are equal exactly when there are no such zeros to remove.

A case where they are equal. Let $f(x)=x$ and $g(x)=x^2+1$. Here $D=ℝ$, and for every real $x$, so $g$ is never zero. Then

A case where they are not equal. Let $f(x)=1$ and $g(x)=x-4$. Again $D=ℝ$, and $g(4)=0$. Then

and these are different sets.

Conclusion. Both situations occur, so the statement is sometimes true.

One direction is always true. Whatever $f$ and $g$ are, $\mathrm{Dom}(f/g)$ is contained in $\mathrm{Dom}(f\cdot g)$, since dividing can only impose an extra requirement. The quotient's domain is never larger.

For example $f(x)=1$ and $g(x)=x-2$. Many other answers work.

Work out what each requirement forces.

Requirement 1: $\mathrm{Dom}(f)=\mathrm{Dom}(g)=ℝ$. Choosing both to be polynomials guarantees this, since a polynomial can be evaluated at every real number. So $D=\mathrm{Dom}(f)\cap \mathrm{Dom}(g)=ℝ$.

Requirement 2: $\mathrm{Dom}(f/g)=ℝ-\{2\}$. Since $\mathrm{Dom}(f/g)=D$ with the zeros of $g$ removed, we need $g$ to be zero at $x=2$ and nowhere else. The simplest such polynomial is

which is zero exactly when $x=2$.

Requirement 3: $\mathrm{Dom}(g/f)=ℝ$. Here the denominator is $f$, and nothing is to be removed, so $f$ must have no zeros at all. The simplest such choice is the constant

which is never zero.

Verification.

Other correct answers. Any $f$ with no real zeros and any $g$ vanishing only at $2$ will do — for instance $f(x)=x^2+5$ together with $g(x)=3x-6$, or $f(x)=7$ together with $g(x)=(x-2{)}^3$. What is not allowed is a choice like $g(x)=x^2-4$, which is zero at $-2$ as well and would give the domain $ℝ-\{-2,2\}$.