Combine two functions with arithmetic

Explain it like I am five

Imagine two pay envelopes for the same year. One person earned f(x) dollars and another earned g(x) dollars. To find their household income, you open both envelopes and add the amounts. You use the same year for both people. Algebraic operations on functions work this way: get two answers at the same input, then add, subtract, multiply, or divide those answers. You are making a new function from two old ones. A sum of functions adds the outputs. A difference of functions subtracts them. A product of functions multiplies them. A quotient of functions divides them, provided the bottom answer is not zero.

At shared input 2, the solid line gives height 3 and the dashed line gives height 4; their sum is 7.

Reminder

Substitution. Replace the input letter everywhere: $x^{2}$ − 1 at x = 3 gives $3 squared$ − 1 = 8.
Subtracting parentheses. A minus outside reverses every sign: −(x − 1) = −x + 1.
Factoring. $x^{2}$ − 1 = (x − 1)(x + 1), because the middle terms cancel when multiplied.
Domain. A denominator cannot be zero: $1 over x − 1$ excludes x = 1.
Shared factor. $x^{2}$ − x = x·x − x·1 = x(x − 1). Distributing x rebuilds the two terms.
Multiplying powers. x· $x^{2}$ = x·x·x = $x^{3}$ , because three copies of x are multiplied.

Why it works. Each original function gives exactly one output at an allowed input. Arithmetic then combines those two numbers into one new answer, so the result is another function. Both original answers must exist: an unavailable answer cannot be added or multiplied. Division needs one more check, because no number multiplied by zero can give a nonzero numerator. Units also matter. Two incomes measured in dollars can be added. Dollars and hours describe different things, so their sum would not represent a meaningful total.

Rule(f + g)(x) = f(x) + g(x); (f − g)(x) = f(x) − g(x); (fg)(x) = f(x)·g(x); ()(x) = .
Use inputs allowed by both functions; for the quotient also require g(x) ≠ 0.

The same idea, five ways

Say it

f plus g of x; f times g of x; f divided by g of x

Write it

Put one input into both recipes, then combine their two outputs.

In math

(f + g)(x) = f(x) + g(x)
(f − g)(x) = f(x) − g(x)
(fg)(x) = f(x)·g(x)
( $\frac{f}{g}$ )(x) = $f(x) over g(x)$
f + g names a recipe; (f + g)(2) names its output at 2

Two pay envelopes for the same year.

See it

(f + g)(2) = f(2) + g(2)

(fg)(2) = f(2)·g(2)

The final parentheses hold the input 2

The operation combines functions; the final parentheses give their shared input.

The same idea, other ways

As two envelopes

At one year, open both pay envelopes. Adding their amounts gives the total for that year; using different years would answer a different question. At year 2, the two incomes are 38400 and 43600 dollars. Together they give 82000 dollars in the pictured household column.

Each household output adds the two incomes in that year.

With two small outputs

If f(2) = 3 and g(2) = 4, the sum at 2 is 7 and the product at 2 is 12. You combine 3 and 4 after evaluating, rather than putting 3 through g.

Multiplying output 3 by output 4 gives 12; a product uses both outputs independently.

.1Sum of functions

A sum collects two amounts. If f and g represent incomes in the same currency and year, their sum is household income. Both amounts must exist before you can add them.

(f + g)(x) = f(x) + g(x).
The domain is the intersection of the original domains: inputs accepted by both.
The shared input may be a date or an item name instead of a number; both outputs must be real numbers.
Addition and subtraction need matching units or two unitless outputs. If two price rules for the same book give 12 dollars and 8 dollars, their sum is 20 dollars because the quantities match.

f(x) = x + 1

g(x) = 2x

(f + g)(x) = 3x + 1

Combine the two x amounts and keep the extra 1.

Reminder

Function notation. The name chooses a recipe and the parentheses supply its input: f(3) is the f output at input 3.

The same idea, five ways

Say it

Sum of functions

Write it

A sum collects two amounts. If f and g represent incomes in the same currency and year, their sum is household income. Both amounts must exist before you can add them.

In math

(f + g)(x) = f(x) + g(x).
The domain is the intersection of the original domains: inputs accepted by both.
The shared input may be a date or an item name instead of a number; both outputs must be real numbers.
Addition and subtraction need matching units or two unitless outputs. If two price rules for the same book give 12 dollars and 8 dollars, their sum is 20 dollars because the quantities match.

A sum collects two amounts. If f and g represent incomes in the same currency and year, their sum is household income. Both amounts must exist before you can add them.

See it

f(x) = x + 1

g(x) = 2x

(f + g)(x) = 3x + 1

Combine the two x amounts and keep the extra 1.

Worked exampleAdd formulas

For f(x) = x + 1 and g(x) = 2x, find (f + g)(x). You want one formula for the total output at any input. Plan: use the quantity or input named in the question, write the first result, then finish the stated operation.

The sum formula matches adding the separate outputs.

(f + g)(x) = (x + 1) + 2x.A sum adds the original function outputs.
x + 2x + 1 = 3x + 1.One x and two x amounts make three x amounts.

Answer

(f + g)(x) = 3x + 1, for every real x.

Check At x = 2, f(2) + g(2) = 3 + 4 = 7 and 3(2) + 1 = 7.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: f(2) + g(3) is (f + g)(2).

The g function received a different input.

✓ Instead: (f + g)(2) uses f(2) + g(2).

Tips and tricks

Write the input beside both function names before adding.

.2Difference of functions

A difference compares two outputs at one input. Think of money earned minus money spent in one month. Subtract the whole second amount, including every term in its formula. In a cup-selling model, profit is revenue minus cost. If R(x) = 3x and C(x) = 0.75x + 18 dollars, profit is 2.25x − 18 dollars. At 24 cups, 72 dollars earned minus 36 dollars spent leaves 36 dollars.

(g − f)(x) = g(x) − f(x).
Subtraction order matters: a − b and b − a usually differ.

(g − f)(x) = ( − 1) − (x − 1)

= − 1 − x + 1

= − x = x(x − 1)

The subtraction changes both signs inside the second parentheses.

Reminder

Pulling out a factor. $x^{2}$ − x = x·x − x·1 = x(x − 1), because both terms contain x.

The same idea, five ways

Say it

Difference of functions

Write it

In math

(g − f)(x) = g(x) − f(x).
Subtraction order matters: a − b and b − a usually differ.

See it

(g − f)(x) = ( − 1) − (x − 1)

= − 1 − x + 1

= − x = x(x − 1)

The subtraction changes both signs inside the second parentheses.

Worked exampleSubtract the whole linear formula

Let f(x) = x − 1 and g(x) = $x^{2}$ − 1. Find (g − f)(x). You want the g output minus the whole f output. Plan: keep the second output in parentheses, change both of its signs, then collect matching terms.

( − 1) − (x − 1)

= − 1 − x + 1

= − x

Subtracting the whole second output changes both of its signs.

(g − f)(x) = ( $x^{2}$ − 1) − (x − 1).The requested order is g minus f.
= $x^{2}$ − 1 − x + 1.Subtracting x − 1 means adding its opposite, −x + 1.
= $x^{2}$ − x = x(x − 1).The constants −1 and +1 cancel, and both remaining terms contain a factor x.

Answer

(g − f)(x) = − x = x(x − 1), for every real x.

Check At x = 3, g(3) − f(3) = 8 − 2 = 6; − 3 = 6.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: ( − 1) − (x − 1) = − x − 2.

The final −1 was subtracted, so it becomes +1.

✓ Instead: − 1 − x + 1 = − x.

Tips and tricks

Put parentheses around the second function before distributing the subtraction.

.3Product of functions

A product multiplies two outputs at the same input. If one output is a price per ticket and the other is the number of tickets, their product is the total price. Multiplication may change the units.

(fg)(x) = f(x)·g(x).
The product needs both functions defined, but an output of zero is allowed.

Every term of x + 2 multiplies .

Reminder

Function notation. The name chooses a recipe and the parentheses supply its input: f(3) is the f output at input 3.

The same idea, five ways

Say it

Product of functions

Write it

A product multiplies two outputs at the same input. If one output is a price per ticket and the other is the number of tickets, their product is the total price. Multiplication may change the units.

In math

(fg)(x) = f(x)·g(x).
The product needs both functions defined, but an output of zero is allowed.

A product multiplies two outputs at the same input. If one output is a price per ticket and the other is the number of tickets, their product is the total price. Multiplication may change the units.

See it

Every term of x + 2 multiplies .

Worked exampleMultiply a line by a square

Let f(x) = x + 2 and g(x) = $x^{2}$ . Find (fg)(x). You want to multiply their outputs at the same x. Plan: use the quantity or input named in the question, write the first result, then finish the stated operation.

(x + 2)

= + 2

A product distributes the entire factor.

(fg)(x) = (x + 2) $x^{2}$ .The product multiplies f(x) by g(x).
= x· $x^{2}$ + 2· $x^{2}$ = $x^{3}$ + 2 $x^{2}$ .Distribution multiplies both terms, and x times $x^{2}$ contains three factors x.

Answer

(fg)(x) = + 2, for every real x.

Check At x = 2, f(2)g(2) = 4 × 4 = 16; + 2() = 8 + 8 = 16.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: (fg)(x) means f(g(x)).

The product multiplies outputs; composition passes one output into another function.

✓ Instead: (fg)(x) = (x + 2).

Tips and tricks

A multiplication dot belongs between the two output expressions.

.4Quotient of functions

A quotient is a ratio: one output divided by another at the same input. The denominator is the bottom expression. It must not be zero, even if later cancellation makes the formula look harmless. The same cost model gives average cost per cup as $0.75 x + 18 over x$ , for x > 0 cups. At 24 cups, $36 over 24$ = $3 over 2$ dollars per cup. Zero cups cannot supply a cost per cup because division by zero fails.

( $\frac{g}{f}$ )(x) = $g(x) over f(x)$ , with both functions defined and f(x) ≠ 0.
Equal functions need matching domains and matching outputs, not only matching simplified formulas.

The canceled quotient still leaves a hole at input 1.

Reminder

Function notation. The name chooses a recipe and the parentheses supply its input: f(3) is the f output at input 3.

The same idea, five ways

Say it

Quotient of functions

Write it

In math

( $\frac{g}{f}$ )(x) = $g(x) over f(x)$ , with both functions defined and f(x) ≠ 0.
Equal functions need matching domains and matching outputs, not only matching simplified formulas.

See it

The canceled quotient still leaves a hole at input 1.

Worked exampleCancel without restoring the forbidden input

For f(x) = x − 1 and g(x) = $x^{2}$ − 1, find ( $\frac{g}{f}$ )(x). You want the g output divided by the f output and the inputs where that division works. Plan: use the quantity or input named in the question, write the first result, then finish the stated operation.

= x + 1 only when x ≠ 1

Cancellation changes the written formula, not its original accepted inputs.

( $\frac{g}{f}$ )(x) = $x squared − 1 over x − 1$ . Require x ≠ 1.The original bottom x − 1 is zero at 1.
$x^{2}$ − 1 = (x + 1)(x − 1).Multiplying these factors gives $x^{2}$ − x + x − 1, so the middle terms cancel.
$(x + 1)(x − 1) over x − 1$ = x + 1, while keeping x ≠ 1.A common nonzero factor can be divided out; at 1 that factor is zero, so cancellation was never legal there.

Answer

( $\frac{g}{f}$ )(x) = x + 1, with x ≠ 1.
Domain: (−∞, 1) ∪ (1, ∞).

Check At x = 3, = = 4 and x + 1 = 4. At x = 1 the original expression is , so it has no defined quotient.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: The quotient is the unrestricted function x + 1.

That line accepts input 1, but the original quotient does not.

✓ Instead: Use x + 1 with x ≠ 1; the domain is part of the function.

Tips and tricks

Write excluded inputs before factoring.

Strategy: step by step

1. Find the requested operation and its order. g − f means the g answer minus the f answer.
2. Use the same input in both original functions, because their outputs must refer to the same starting value.
3. Record where both formulas work and, for division, where the bottom output is nonzero.
4. Put each whole expression in parentheses before combining it. Simplify while keeping every original domain restriction.
5. Check one permitted input directly in the original functions.

Strategy

Combine two functions and keep their domains

Is the operation division?

YesAlso remove every input giving a zero bottom output before canceling.

NoUse the overlap of the original domains.

↓

Did cancellation hide a forbidden input?

YesKeep it excluded; a shorter formula cannot restore a missing original output.

NoKeep the domain you recorded.

Write the accepted inputs of each original function. Keep inputs accepted by both.
For a quotient, solve bottom function = 0 to locate the additional inputs to remove. Plug each proposed excluded input back into that bottom.
Substitute the formulas with parentheses around the whole second expression.
Simplify and keep all original exclusions. Check using the two original outputs.

Worked exampleThe same input, four different combinations

Let f(x) = x + 1 and g(x) = 2x. Find their sum, difference f − g, product, and quotient f divided by g at input 2. The question asks you to get both outputs at 2, then combine those outputs in four ways. Plan: evaluate both functions at the same input 2, combine their outputs in four ways, then derive the four combined formulas.

Same input: 2

f(2) = 3; g(2) = 4

3 + 4 = 7; 3 − 4 = −1

3 × 4 = 12; 3 ÷ 4 =

Arithmetic combines two outputs made from the same input.

f(2) = 2 + 1 = 3 and g(2) = 2 × 2 = 4.Both functions receive input 2, so both answers describe that same input.
(f + g)(2) = 3 + 4 = 7.The sum adds the two outputs.
(f − g)(2) = 3 − 4 = −1.The written order puts the f output first.
(fg)(2) = 3 × 4 = 12.The product multiplies the outputs.
( $\frac{f}{g}$ )(2) = $3 over 4$ .The quotient divides the f output by the nonzero g output.
As formulas, (f + g)(x) = (x + 1) + 2x = 3x + 1 and (f − g)(x) = (x + 1) − 2x = 1 − x.Only matching x terms combine; the written subtraction order stays fixed.
(fg)(x) = (x + 1)(2x) = 2 $x^{2}$ + 2x, and ( $\frac{f}{g}$ )(x) = $x + 1 over 2 x$ , requiring x ≠ 0.Distribution multiplies both terms. The quotient removes the input 0 that makes 2x zero.

Answer

Sum: 7.
Difference: −1.
Product: 12.
Quotient: $3 over 4$ .

Check Undo each output operation at input 2: 7 − 4 = 3 checks the sum; −1 + 4 = 3 checks the difference; 12 ÷ 4 = 3 checks the product; × 4 = 3 checks the quotient. The four combined formulas derived above also give these answers at 2.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: subtract one constant output

Let p(x) = x + 5 and q(x) = 2. This is a constant function: it returns the same output 2 for every input. Find (p − q)(x). You want the first output minus 2. Plan: write the original outputs with parentheses, record any forbidden input, then simplify and check.

Subtracting a constant shifts the output by that amount.

(p − q)(x) = (x + 5) − 2.Subtraction acts on the two outputs.
= x + 3.5 − 2 = 3.

Answer

x + 3, for every real x.

Check At 0, 5 − 2 = 3, matching 0 + 3.

Rung 2Rung 2: subtract an entire expression

Let p(x) = $x^{2}$ + 4 and q(x) = 2x − 3. Find (p − q)(x). You must subtract both terms of q. Plan: write the original outputs with parentheses, record any forbidden input, then simplify and check.

−(2x − 3) = −2x + 3

+ 4 − 2x + 3 = − 2x + 7

The subtraction changes each sign inside the parentheses.

(p − q)(x) = $x^{2}$ + 4 − (2x − 3).Parentheses keep the second formula together.
= $x^{2}$ + 4 − 2x + 3 = $x^{2}$ − 2x + 7.Subtracting −3 adds 3.

Answer

− 2x + 7.

Check At 1, p(1) − q(1) = 5 − (−1) = 6; 1 − 2 + 7 = 6.

Rung 3Rung 3: factor a difference of squares

Rewrite $x^{2}$ − 16 as a product. You want two factors whose multiplication recreates the original expression. Plan: recognize $x^{2}$ − $4 squared$ , use the difference and sum of x and 4, then multiply to check.

The opposite middle terms disappear when added. Negative labels are signed multiplication amounts, not physical side lengths; the grid records the products. Negative labels are bookkeeping amounts, not physical side lengths.

16 = $4 squared$ , so $x^{2}$ − 16 = $x^{2}$ − $4 squared$ .This is a difference of two squares.
$x^{2}$ − $4 squared$ = (x − 4)(x + 4).Expanding gives $x^{2}$ + 4x − 4x − 16; the middle terms cancel.

Answer

(x − 4)(x + 4).

Check At x = 6, 36 − 16 = 20 and (6 − 4)(6 + 4) = 2 × 10 = 20.

Rung 4Rung 4: factor and retain the hole

Let p(x) = $x^{2}$ − 16 and q(x) = x − 4. Find ( $\frac{p}{q}$ )(x) and its domain. You want a quotient formula and every permitted input. Plan: write the original outputs with parentheses, record any forbidden input, then simplify and check.

The open dot keeps the original excluded input visible.

Start with $x squared − 16 over x − 4$ and record x ≠ 4.The original denominator must be nonzero.
Factor the top: $(x − 4)(x + 4) over x − 4$ . $x^{2}$ − 16 is a difference of squares.
Cancel the factor x − 4 to obtain x + 4, still with x ≠ 4.The cancellation divides by a factor known to be nonzero only on that restricted domain.

Answer

( $\frac{p}{q}$ )(x) = x + 4, x ≠ 4.
Domain: (−∞, 4) ∪ (4, ∞).

Check At 6 the original quotient is = 10, matching 6 + 4. Input 4 still produces in the original quotient.

Rung 5Rung 5: a product and a quotient with two original exclusions

Let p(x) = x − 3 and q(x) = $x^{2}$ − 9. Find (pq)(x), ( $\frac{p}{q}$ )(x), their domains, and whether they are the same function. You want to multiply the outputs, then divide them. Plan: identify the original bottom zeros before factoring or canceling.

Both original bottom zeros, −3 and 3, remain open after cancellation.

(pq)(x) = (x − 3)( $x^{2}$ − 9).A product multiplies outputs at one shared input.
= x· $x^{2}$ + x(−9) + (−3) $x^{2}$ + (−3)(−9).Every term in the first group multiplies every term in the second.
= $x^{3}$ − 9x − 3 $x^{2}$ + 27 = $x^{3}$ − 3 $x^{2}$ − 9x + 27.x· $x^{2}$ contains three x factors, and two negative factors make positive 27.
For the quotient, start with $x − 3 over x squared − 9$ . Solve $x^{2}$ − 9 = 0, so $x^{2}$ = 9 and x = 3 or x = −3.This finds both inputs that would make the original bottom zero, so both must be removed.
Check: $3 squared$ − 9 = 0 and (−3 $) squared$ − 9 = 0. Exclude both 3 and −3.Substitution confirms that both original divisions fail.
Factor $x^{2}$ − 9 = (x − 3)(x + 3). Then $x − 3 over (x − 3)(x + 3)$ = $1 over x + 3$ , with both exclusions retained.Canceling a common nonzero factor preserves outputs only where the original quotient existed.
At x = 4, (pq)(4) = 1 × 7 = 7 and ( $\frac{p}{q}$ )(4) = $1 over 7$ .Two different outputs at one shared allowed input prove the functions are different.

Answer

(pq)(x) = $x^{3}$ − 3 $x^{2}$ − 9x + 27, domain (−∞, ∞).
( $\frac{p}{q}$ )(x) = $1 over x + 3$ , domain (−∞, −3) ∪ (−3, 3) ∪ (3, ∞).
They are different functions.

Check At 4, the expanded product gives 64 − 48 − 36 + 27 = 7. The shorter quotient gives = . At 3 its original form is , so 3 stays excluded.

Rung 6Rung 6: the domains of the two ingredients differ

Let f(x) = $square root of x + 6$ and g(x) = $2 over x − 5$ . Find f + g, f − g, fg, f/g and g/f with their domains. You want formulas and the inputs each operation can accept. Plan: overlap the two original domains, then remove any zero bottom output for each quotient.

The overlap includes the root boundary −6 but excludes the fraction input 5.

f needs x + 6 ≥ 0, so x ≥ −6. g needs x − 5 ≠ 0, so x ≠ 5.The root requires a nonnegative inside, and the fraction requires a nonzero bottom.
The overlap is [−6, 5) ∪ (5, ∞).Arithmetic needs both original outputs at the same starting input.
(f + g)(x) = $square root of x + 6$ + $2 over x − 5$ ; (f − g)(x) = $square root of x + 6$ − $2 over x − 5$ ; (fg)(x) = $2 square root of x + 6 over x − 5$ .Apply the requested operation to the two complete outputs.
For f/g, g(x) never equals zero on its domain, since its top is 2. Thus ( $\frac{f}{g}$ )(x) = $square root of x + 6$ × $x − 5 over 2$ = $(x − 5) square root of x + 6 over 2$ .Division multiplies by the reciprocal of the nonzero bottom output.
For g/f, also require $square root of x + 6$ ≠ 0. It equals zero at x = −6, because $square root of −6 + 6$ = 0. Remove −6.This finds and confirms the extra starting input that would make the quotient bottom zero.
( $\frac{g}{f}$ )(x) = $2 over (x − 5) square root of x + 6$ , with domain (−6, 5) ∪ (5, ∞).The bottom combines the original fraction bottom and the root output; both must stay nonzero.

Answer

(f + g)(x) = $square root of x + 6$ + $2 over x − 5$ , domain [−6, 5) ∪ (5, ∞).
(f − g)(x) = $square root of x + 6$ − $2 over x − 5$ , domain [−6, 5) ∪ (5, ∞).
(fg)(x) = $2 square root of x + 6 over x − 5$ , domain [−6, 5) ∪ (5, ∞).
( $\frac{f}{g}$ )(x) = $(x − 5) square root of x + 6 over 2$ , domain [−6, 5) ∪ (5, ∞).
( $\frac{g}{f}$ )(x) = $2 over (x − 5) square root of x + 6$ , domain (−6, 5) ∪ (5, ∞).

Check At −6, f gives 0 and g gives −, so their sum exists and equals −. The quotient g/f fails because its bottom output is 0. At 5, g fails before any arithmetic, so even the shorter f/g formula cannot use 5.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: A product must exclude inputs where either output is zero.

Multiplication by zero has the valid answer zero. Only division forbids a zero bottom.

✓ Instead: A product requires both functions to be defined; a quotient additionally requires its denominator output to be nonzero.

✗ Not this: (f + g)(2) = 2f + 2g.

f and g name recipes. The final parentheses supply input 2 to both recipes rather than multiplying the recipes.

✓ Instead: (f + g)(2) = f(2) + g(2) = 3 + 4 = 7 in the main example.

✗ Not this: After becomes , only −3 is excluded.

At x = 3, the original expression is . Canceling has hidden that failed input.

✓ Instead: Keep x ≠ −3 and x ≠ 3 in the simplified quotient.

✗ Not this: The domain of f + g for the last rung is [−6, ∞).

At x = 5, g gives , so there is no second output to add.

✓ Instead: Both original domains must hold: [−6, 5) ∪ (5, ∞).

Tips and tricks

Know cold.
Arithmetic gives both functions the same input. Memory cue: same input, two answers, one operation.
Composition runs the inner function first. Memory cue: read ∘ as after, so f ∘ g means f after g.
Replace every input slot with the whole expression. Memory cue: every slot gets the whole package.
Check both domain gates. Memory cue: gate 1 checks x, gate 2 checks the first answer.
An ordinary root permits zero; a denominator forbids it. Memory cue: root zero can stay, bottom zero goes away.
When decomposing, identify the last operation. Memory cue: the last calculator step is the outer function.
Understand, then rebuild when needed.
Expanded sums, products, factorizations, and composition formulas can be recreated with substitution and distribution. Do not memorize each final formula.
Rebuild a table or graph path by carrying the inner output to the outer input.
Rebuild domain intervals from the allowed points and the endpoint checks.
Rebuild a decomposition by identifying the final operation, then recomposing to check.
Put on the cheat sheet for study.
Arithmetic: (f + g)(x) = f(x) + g(x); (f − g)(x) = f(x) − g(x); (fg)(x) = f(x)·g(x); ( $\frac{f}{g}$ )(x) = $f(x) over g(x)$ . Use the shared domain, with g(x) ≠ 0 for the quotient.
Composition: (f ∘ g)(x) = f(g(x)). Its domain keeps x allowed by g and g(x) allowed by f.
Roots: $\sqrt{A}$ needs A ≥ 0; $1 over square root of A$ needs A > 0.
Square expansion: (a + b $) squared$ = $a^{2}$ + 2ab + $b^{2}$ . Difference of squares: $a^{2}$ − $b^{2}$ = (a − b)(a + b).
Intervals: brackets include finite endpoints, parentheses exclude them, and infinity always takes parentheses.
This is a study sheet. For a closed-book exam, practice recalling the short definitions and rebuilding the rest without looking.
A function name before parentheses means input: f(2) reads f of 2. A number before parentheses means multiplication: 3(2) = 6.

Trap. Subtracting only the first term of a function. ( − 1) − (x − 1) subtracts the entire second expression, so both of its signs change.