Digital SAT Math · Advanced Math

Function notation and evaluation

Digital SAT Math · Advanced Math

Nothing on this skill is hard arithmetic. $f(g(2))$ , $f(x+1)$ , $f(2x)$ , $h(-3) = 8$ — every one is a two-line computation once you have decoded it, and the test knows it. What it charges you for is the decoding: which symbol goes in, which comes out, what the parentheses enclose, and in which order two functions act. Get the notation reflexes right and this becomes the fastest-scoring skill in Advanced Math; get them wrong and you lose points on questions whose math you could do in your sleep.

On the test


Domain	Advanced Math (score report)
What it looks like	A rule such as $f(x) = 3x^2 - 5$ , a table headed $x$ , $f(x)$ , $g(x)$ , a graph labeled $y = f(x)$ , or a described model
Often asked	“What is $f(g(4))$ ?”, “Which expression is equivalent to $f(2x)$ ?”, “If $f(k) = 24$ , what is $k$ ?”, “Which point must lie on the graph?”
Format	Multiple choice and student-produced response
Calculator	Desmos evaluates and composes for you if you type the notation correctly — and does nothing at all for the questions this skill really tests

Recognition cues: a letter immediately followed by a parenthesis — $f(\,)$ , $g(\,)$ , $h(\,)$ , $P(\,)$ , $R(\,)$ ; nested parentheses; a table with two output columns; the phrases the function f is defined by, for all values of x, which expression is equivalent to.

What this page owns

Function notation is a language, and it is spoken on every function page in this domain. Here we drill the language itself: evaluating, composing, substituting expressions, and reading a value out of a table or a graph. What a parabola’s vertex means, how to pick growth factors, how to read an exponential curve — those live on quadratic functions, exponential functions and interpreting nonlinear graphs. If you can already do the algebra on those pages but keep losing points, the leak is almost always here.

Pattern recognition

Six shapes cover essentially every item:

Number in, number out — $f(-2)$ with a rule you can substitute into.
Number out, input unknown — $f(k) = 24$ ; this one is an equation to solve, not a substitution.
Composition — $f(g(4))$ , where the answer comes from two lookups or two substitutions, never one.
Expression in — $f(2x)$ , $f(x+2)$ , $f(a+h)$ ; the input is an algebraic package, and it goes in whole.
Statement to point — $h(-3) = 8$ rewritten as $(-3, 8)$ on the graph of $y = h(x)$ .
Table or graph lookup — the numbers exist only in the figure, and the reading direction is what is being tested.

Method

Name the direction. Is the unknown an input or an output? Inside the parentheses means input; alone on the far side of the equals sign means output. Inputs given → substitute. Output given → set the rule equal to it and solve.
Work strictly inside-out. In $f(g(4))$ the function written closest to the number acts first. Compute $g(4)$ , write the number down, and rewrite the expression as $f(\text{that number})$ before continuing. Writing the intermediate value is not optional bookkeeping — it is what prevents the order swap.
Substitute with parentheses, every time. Replace every $x$ in the rule with the entire input wrapped in parentheses: $f(2x) = 3(2x)^2 - 5$ , not $3 \cdot 2x^2 - 5$ . The parentheses are what make the exponent and the sign apply to the whole input.
Read a figure in one direction only. Enter a table or a graph at the input, then move out to the output column or up to the curve. If a question hands you an output instead, you are running the lookup backwards on purpose — say so out loud, because that is a different and slower job.
Check by feeding your answer back in. Substitute the number you found into the original rule and confirm it produces the stated output. Every wrong answer on this skill survives its own bad method and dies instantly under this check.

Notation	Says	Common misreading
$f(3)$	the output when the input is 3	the input when the output is 3
$f(x) = 3$	the output is 3; the input is unknown	evaluate at 3
$f(g(2))$	evaluate $g$ at 2, then $f$ at that result	evaluate $f$ at 2 first
$f(x+1)$	shift the input, then apply the rule	$f(x) + 1$
$f(2x)$	double the input, then apply the rule	$2f(x)$
$f(a) = b$	the point $(a, b)$ is on $y = f(x)$	the point $(b, a)$

Desmos on this skill. It genuinely helps in two places. Type f(x) = 3x^2 - 5 on one line and f(2) on the next, and Desmos prints the output — useful when the rule is ugly and you only need a number. Composition works the same way: define both functions on separate lines and type f(g(4)), and the nesting is evaluated correctly, which is a real safeguard on an item where the order is the whole question. Where it will not save you: anything where the answer is an expression ( $f(2x)$ , $f(x+2)$ ), anything where the numbers live in a printed table, and — most importantly — the moment you mistype the very thing being tested. Typing f(x)+1 when the stem said $f(x+1)$ produces a confident, beautifully formatted wrong answer. The calculator will not catch a notation error; it will only compute it faster.

Worked example 1 — a composition, inside-out

Stem. The functions $f$ and $g$ are defined by $f(x) = x^2 + 4x$ and $g(x) = 2x - 7$ . What is the value of $f(g(5))$ ?

Step 1 — find the inner function. $g$ is written next to the 5, so $g$ acts first.

g(5) = 2(5) - 7 = 3

Step 2 — rewrite before continuing. The expression is now $f(3)$ . Write that down; do not carry it in your head.

Step 3 — evaluate the outer function.

f(3) = 3^2 + 4(3) = 9 + 12 = 21

Check. Rebuild the chain: input 5 → $g$ → 3 → $f$ → 21. Each arrow uses one function once. Answer: 21.

Trap watch. Reversing the order gives $g(f(5))$ : $f(5) = 25 + 20 = 45$ , then $g(45) = 83$ — a clean-looking number produced by the wrong machine order. Stopping after step 1 gives 3, the intermediate value. Ignoring $g$ altogether gives $f(5) = 45$ . All three appear as choices on real items of this archetype.

Worked example 2 — $f(x+1)$ is not $f(x)+1$

Stem. The function $f$ is defined by $f(x) = x^2 - 3x$ . Write $f(x+1)$ and $f(x)+1$ as expressions in $x$ , and evaluate both at $x = 4$ .

Step 1 — $f(x+1)$ : the package goes through the door. Every $x$ in the rule becomes $(x+1)$ :

f(x+1) = (x+1)^2 - 3(x+1) = x^2 + 2x + 1 - 3x - 3 = x^2 - x - 2

Step 2 — $f(x)+1$ : the rule runs first, then 1 is added.

f(x) + 1 = (x^2 - 3x) + 1 = x^2 - 3x + 1

Step 3 — evaluate both at $x = 4$ . From the expressions: $16 - 4 - 2 = 10$ and $16 - 12 + 1 = 5$ .

Check them independently, straight from the definitions: $f(5) = 25 - 15 = 10$ ✓ and $f(4) + 1 = (16 - 12) + 1 = 5$ ✓. Two different expressions, two different values, one keystroke apart. Answers: $x^2 - x - 2$ and $x^2 - 3x + 1$ ; at $x = 4$ , 10 and 5.

Trap watch. The most expensive slip here is distributing the shift only into the first $x$ — writing $(x+1)^2 - 3x$ — because the rule contains $x$ twice and both copies must change. The second is dropping the parentheses in $(x+1)^2$ and expanding it as $x^2 + 1$ ; squaring a sum is never the sum of the squares, and this misstep costs points on the equivalent-expressions skill too.

Worked example 3 — one table, read in both directions

Stem. The table gives values of the function $f$ .

$x$	0	1	2	3	4
$f(x)$	7	3	0	5	2

(a) What is $f(3)$ ? (b) For what value of $x$ is $f(x) = 3$ ? (c) Name one point on the graph of $y = f(x)$ .

(a) Input given — enter at $x$ . Find 3 in the $x$ row, drop to the $f(x)$ row: $f(3) = 5$ .

(b) Output given — this is the reverse trip. Now 3 is on the output side, so search the $f(x)$ row for 3 and read the $x$ above it: $x = 1$ . Notice the two questions used the same number, 3, and produced different answers, 5 and 1, because the 3 entered through different doors.

(c) Any column is a point. The column $x = 1$ , $f(x) = 3$ is the statement $f(1) = 3$ , which is the point $(1, 3)$ on the graph of $y = f(x)$ . Inputs are $x$ -coordinates; outputs are $y$ -coordinates.

Check. Both answers survive a re-read: the column headed 3 holds 5, and the only column holding the output 3 is headed 1. Answers: 5; $x = 1$ ; $(1, 3)$ .

Trap watch. Answering (b) with 5 means the reverse question was answered forwards. Answering (a) with 1 means it was answered backwards. And writing the point as $(3, 1)$ reverses the pair — the coordinate that goes first is always the one that went into the function.

Practice

Answer before opening the explanation. Three items hand you a figure — two tables and one graph — and the values you need appear nowhere in the sentence. Two are student-produced response, where a notation slip has no multiple-choice safety net. Every wrong choice below is one specific, named misreading; when you miss one, log the misreading, not the item number.

12 questions — 10 multiple choice, 2 student-produced response. Every wrong choice has its own explanation.

Question 1 Warm-up

The function f is defined by f(x) = x² - 5x. What is the value of f(-2)?

Show the answer Choice D

Why it is right

Replace every x with the input, and write the input inside parentheses so its sign travels with it: f(-2) = (-2)² - 5(-2). The square of a negative number is positive, so (-2)² = 4, and subtracting a negative adds, so -5(-2) = +10. Therefore f(-2) = 4 + 10 = 14.

Why each other choice fails

Choice A: Both signs were lost: the square was computed as -(2²) = -4 and the second term as -5(2) = -10, giving -4 - 10 = -14. Neither copy of x kept the negative sign of the input.
Choice B: The square was handled correctly (4) but the second term was computed as -5(2) = -10 instead of -5(-2) = +10, so the input's negative sign was dropped in the linear term only.
Choice C: The negative sign was applied outside the square, -(2²) = -4, instead of squaring the whole input, (-2)² = 4. The second term, +10, was right, giving -4 + 10 = 6.

Question 2 Standard

For the function h, h(-3) = 8. Which point must lie on the graph of y = h(x) in the xy-plane?

Show the answer Choice B

Why it is right

A statement of the form h(a) = b packs an input and an output into one line: the number inside the parentheses is the input and goes on the horizontal axis, and the number after the equals sign is the output and goes on the vertical axis. So h(-3) = 8 is exactly the ordered pair (-3, 8), and the graph of y = h(x) must pass through it.

Why each other choice fails

Choice A: Moves the input's negative sign onto the output. The -3 sits inside the parentheses, so it is the x-coordinate; the output 8 is positive and stays positive.
Choice C: Drops the negative sign from the input. The point (3, 8) would come from h(3) = 8, a statement about a different input the problem never makes.
Choice D: Reverses the pair, reading 8 as the input and -3 as the output. That would be the statement h(8) = -3, which is not what h(-3) = 8 says.

Question 3 Standard

A film archive is digitizing its reels. The function R defined by R(w) = 480 - 15w gives the number of reels still waiting to be digitized w weeks after the project begins. For what value of w is R(w) = 300?

Show the answer Choice A

Why it is right

R(w) = 300 places 300 on the output side, so w is the unknown and the equation must be solved, not evaluated: 480 - 15w = 300. Subtract 480 from both sides to get -15w = -180, then divide by -15 to get w = 12. Check by evaluating: R(12) = 480 - 15(12) = 480 - 180 = 300 reels still waiting.

Why each other choice fails

Choice B: Divides the given output by the rate, 300 ÷ 15 = 20, ignoring the 480 reels the archive started with. That would be the answer only if the model were R(w) = 15w.
Choice C: Solves R(w) = 0 instead: 480 ÷ 15 = 32 is the week the backlog disappears entirely, not the week 300 reels remain.
Choice D: Flips a sign when moving 480 across the equals sign, producing 15w = 480 + 300 and w = 52. A quick check kills it: R(52) = 480 - 780 = -300, a negative number of reels.

Question 4 Standard

The function f is defined by f(x) = 12 for all values of x. What is the value of f(x + 5)?

Show the answer Choice B

Why it is right

The rule assigns the output 12 to every input, so whatever is written inside the parentheses is irrelevant to the result. Feeding in x + 5 gives f(x + 5) = 12, exactly as f(3), f(-100) and f(x) all equal 12. Changing the input of a constant function changes nothing, because there is no x on the output side for the new input to reach.

Why each other choice fails

Choice A: Subtracts 5 from the output, as if f(x + 5) meant f(x) - 5. Nothing is subtracted here, and the shift direction is invented on top of that.
Choice C: Adds the 5 to the output instead of to the input: this is f(x) + 5, a different expression. The 5 is inside the parentheses, so it goes into the machine, and this machine ignores what it is given.
Choice D: Reads f(x + 5) as multiplication, f times (x + 5), and then multiplies 12 by 5. The notation f( ) names a function evaluated at an input; it is never a product.

Question 5 Standard Student-produced response

The function h is defined by h(x) = (x - 5)². What is the value of h(2)?

Your answer

Show the answer 9

Why it is right

Substitute 2 for x inside the parentheses first, because the parentheses are part of the input to the squaring: h(2) = (2 - 5)² = (-3)². Squaring -3 gives 9, since a negative times a negative is positive. The subtraction happens before the square, which is the whole point of the parentheses in the rule.

Answers students type instead

49: Flips the sign of the 5 and computes (2 + 5)² = 49. The rule subtracts 5 from the input, so the inside is 2 - 5, not 2 + 5.
-1: Squares first and subtracts after, computing 2² - 5 = -1. That evaluates a different function, x² - 5; the parentheses in (x - 5)² say the subtraction comes first.
-9: Gets 2 - 5 = -3 but then squares only the 3 and keeps the minus sign, computing -(3²) = -9. Squaring a negative number always produces a positive result.

Question 6 Harder

At a bakery, the function n defined by n(h) = 24h gives the number of loaves baked during h hours of oven time. The function c defined by c(L) = 1.5L + 20 gives the cost, in dollars, of the flour needed to bake L loaves. Which expression gives the cost, in dollars, of the flour needed for h hours of oven time?

Show the answer Choice C

Why it is right

Hours go into n and come out as loaves; loaves go into c and come out as dollars. So the hours must pass through n first and the result must be fed to c: the cost is c(n(h)) = c(24h) = 1.5(24h) + 20 = 36h + 20. Checking with a number keeps it honest: 2 hours gives n(2) = 48 loaves and c(48) = 1.5(48) + 20 = 92 dollars, and 36(2) + 20 = 92 as well.

Why each other choice fails

Choice A: Adds the two rates, 24 + 1.5 = 25.5, instead of composing the functions. Rates in different units (loaves per hour and dollars per loaf) can be multiplied, never added.
Choice B: Composes the rates correctly but drops the 20-dollar constant from c. The flour cost still includes that fixed 20 dollars no matter how many hours the oven runs.
Choice D: Composes in the wrong order, computing n(c(h)) = 24(1.5h + 20) = 36h + 480. That feeds hours into the cost function and dollars into the loaf-count function, so the units do not line up.

Question 7 Harder

The table gives the values of the functions f and g at four inputs. What is the value of f(g(4))?

x	f(x)	g(x)
1	6	4
2	8	7
3	1	2
4	5	3

Values of the functions f and g at four inputs.

Show the answer Choice A

Why it is right

Work from the inside out. First find g(4): go to the row where x is 4 and read the g(x) column, which gives 3. That output becomes the input of f, so the expression is now f(3). Go back to the x column, find the row where x is 3, and read the f(x) column: 1. Therefore f(g(4)) = f(3) = 1.

Why each other choice fails

Choice B: Stops after the inner step. 3 is the value of g(4), which is the input to f, not the final output.
Choice C: Reads f(4) = 5, applying the outer function to the original input and skipping g entirely. The 4 belongs to g; f never sees it.
Choice D: Reads the table backwards, searching the g(x) column for the value 4, landing on the row x = 1, and then reporting f(1) = 6. Table lookups go from the x column outward, never from an output column back to x.

Question 8 Harder

The graph of y = f(x) is shown in the xy-plane, and the function g is defined by g(x) = x + 3. What is the value of f(g(-1))?

The graph of y = f(x), with its two x-intercepts marked.

Show the answer Choice A

Why it is right

The inner function is evaluated first: g(-1) = -1 + 3 = 2. That output becomes the input of f, so read the graph at x = 2. The curve's lowest point sits at (2, -1), so f(2) = -1, and therefore f(g(-1)) = -1. Only the second step uses the graph; the first is pure substitution.

Why each other choice fails

Choice B: Stops at the inner value g(-1) = 2 and reports it. That number is the input handed to f, not the value of the composition.
Choice C: Reads the graph at x = -1, giving f(-1) = 8, which applies f to the original input and ignores g completely.
Choice D: Composes in the wrong order: g(f(-1)) = g(8) = 8 + 3 = 11. In f(g(-1)) the function written closest to the number acts first.

Question 9 Harder

The function f is defined by f(x) = 3x² - 5. Which expression is equivalent to f(2x)?

Show the answer Choice C

Why it is right

Substitution replaces every x in the rule with the entire new input, wrapped in parentheses: f(2x) = 3(2x)² - 5. The parentheses matter because the square applies to the whole input: (2x)² = 4x². So f(2x) = 3(4x²) - 5 = 12x² - 5. A numeric check confirms it: at x = 1, f(2) = 3(4) - 5 = 7, and 12(1)² - 5 = 7.

Why each other choice fails

Choice A: Squares only the x and multiplies the coefficient by 2, computing 3(2x²) = 6x². The 2 is inside the parentheses being squared, so it contributes a factor of 4, not 2.
Choice B: Doubles the whole output instead of the input, computing 2f(x) = 2(3x² - 5) = 6x² - 10. Doubling what goes in is not the same as doubling what comes out.
Choice D: Lets the square reach the coefficient 3 as well, computing (3 · 2x)² = 36x². The 3 sits outside the parentheses in 3(2x)², so it is multiplied after the squaring, never squared.

Question 10 Hardest

The function f is defined by f(x) = 3x - 5, and the function g is defined by g(x) = f(x + 2). Which equation defines g?

Show the answer Choice C

Why it is right

g feeds x + 2 into f, so every x in f's rule is replaced by the whole quantity x + 2: g(x) = 3(x + 2) - 5. Distribute before simplifying: 3x + 6 - 5 = 3x + 1. Check at one input: g(4) should equal f(6) = 13, and 3(4) + 1 = 13.

Why each other choice fails

Choice A: Shifts the input the other way, computing f(x - 2) = 3(x - 2) - 5 = 3x - 11. The rule adds 2 to the input, so the substituted quantity is x + 2.
Choice B: Adds 2 to the output instead of the input, computing f(x) + 2 = 3x - 5 + 2 = 3x - 3. The 2 is inside the parentheses, so it is processed by f and gets multiplied by 3 on the way through.
Choice D: Adds the 2 to the coefficient, producing 5x - 5. Substitution replaces x; it never edits the numbers already written in the rule.

Question 11 Hardest

The table gives the values of the function f at four inputs. The function g is defined by g(x) = f(x + 1). What is the value of g(2)?

x	f(x)
1	5
2	9
3	6
4	11

Selected values of the function f.

Show the answer Choice B

Why it is right

Substitute 2 for x in the definition of g before touching the table: g(2) = f(2 + 1) = f(3). Now the table is used once, at x = 3, where f(x) is 6. So g(2) = 6. The +1 is inside the parentheses, so it changes which row of the table is read, not the number read from it.

Why each other choice fails

Choice A: Reads f(1) = 5, shifting the input in the wrong direction. g(x) = f(x + 1) sends the input one row further down the table, not one row back.
Choice C: Reads f(2) = 9, ignoring the +1 and treating g as if it were f. The two functions agree nowhere on this table.
Choice D: Computes f(2) + 1 = 9 + 1 = 10, adding 1 to the output rather than to the input. That would be the function f(x) + 1, which is a different rule.

Question 12 Hardest Student-produced response

The function f is defined by f(x) = x² - 2x. If f(k) = 24 and k > 0, what is the value of k?

Your answer

Show the answer 6

Why it is right

f(k) = 24 puts 24 on the output side, so k is an unknown input and the rule must be solved: k² - 2k = 24. Move everything to one side to get k² - 2k - 24 = 0, which factors as (k - 6)(k + 4) = 0, so k = 6 or k = -4. The condition k > 0 keeps only k = 6. Check: f(6) = 36 - 12 = 24.

Answers students type instead

12: Reads the rule as if it were 2k, giving 2k = 24 and k = 12. Check it: f(12) = 144 - 24 = 120, not 24.
24: Repeats the output. 24 is the value f produces, not the input that produces it; f(24) = 576 - 48 = 528.
-4: The other root of the quadratic. It does satisfy f(-4) = 16 + 8 = 24, but the problem restricts k to positive values, so it must be rejected.

Common mistakes

Composing in the wrong order — computing $g(f(4))$ when the stem wrote $f(g(4))$ . The function printed closest to the number goes first, always.
Stopping at the inner value — reporting $g(4)$ and forgetting that it was only the input to $f$ . Writing the intermediate number on paper, then rewriting the expression, makes this nearly impossible.
Reading $f(x+1)$ as $f(x)+1$ — moving a shift from the input side to the output side. Inside the parentheses changes what goes in; outside changes what comes out.
Substituting without parentheses — turning $f(2x)$ into $3 \cdot 2x^2 - 5$ instead of $3(2x)^2 - 5$ , so the exponent reaches only part of the input. The same slip drops the sign in $f(-2)$ .
Replacing only the first $x$ — a rule containing the variable twice needs both copies replaced, every time.
Running a table lookup backwards — searching an output column for the input value. Tables are entered at the $x$ column; anything else answers a question that was not asked.
Evaluating when the question asked you to solve — computing $f(24)$ when the stem said $f(k) = 24$ . A given output means an equation, not a substitution.
Reversing a point — writing $f(a) = b$ as $(b, a)$ . The input is the first coordinate because it is the input.
Treating $f(x)$ as a product — multiplying $f$ by $x$ , or “cancelling” the $f$ . The parentheses here announce an input, not multiplication.
Keeping an inadmissible solution — solving $f(k) = 24$ into two roots and reporting the one the stem excluded with a condition such as $k > 0$ . Read the restriction before choosing.

FAQ

Does $f(x)$ mean $f$ times $x$ ? No, and this is worth over-learning. $f$ is the name of a rule and $x$ is what you hand it; the parentheses are punctuation, not multiplication. You can never divide both sides by $f$ , and $f(2x)$ is not $2f(x)$ — for $f(x) = 3x^2 - 5$ they are $12x^2 - 5$ and $6x^2 - 10$ , and setting those equal gives $6x^2 = -5$ , so they agree at no real value of $x$ at all.

Is $f(g(x))$ the same as $g(f(x))$ ? Almost never. Composition is not commutative: order changes the answer, and the test builds a distractor from exactly that. With $f(x) = x^2 + 4x$ and $g(x) = 2x - 7$ , $f(g(5)) = 21$ while $g(f(5)) = 83$ .

What does $(f \circ g)(x)$ mean if it appears? It is another way to write $f(g(x))$ — $g$ first, then $f$ . Digital SAT stems overwhelmingly use the nested form, but the circle notation is not exotic and reads the same way: right to left.

Can I just plug the choices in? On a “find the input” item, yes, and it is often the fastest safe route: feed each choice into the rule and keep the one that returns the stated output. It has the side benefit of forcing you to treat the choices as inputs, which is the direction the question intended. On “which expression is equivalent” items, do the same thing with one convenient number — substitute $x = 1$ into the stem expression and into each choice, and discard everything that disagrees.

How do I enter an answer here on a grid-in? Type the exact value. Negative outputs are legal on this skill and appear often; a fraction such as 7/3 should be entered as a fraction rather than rounded. If a question restricts the input (“ $k > 0$ ”), enter only the solution that obeys the restriction — a grid-in accepts one answer, so the restriction is doing real work.

A table shows two functions. How do I keep them straight? Read the column headings out loud before the numbers, and physically point at the column you are using. Nearly every composition item built on a two-function table has a wrong choice that comes from reading the correct row in the wrong column, and no amount of arithmetic care catches a column error.

Do I need to know function transformations for this skill? Not for the notation itself, but the two overlap. Recognizing that $g(x) = f(x+2)$ replaces $x$ with $x+2$ is pure substitution — the same move as $f(2x)$ — and it is also the algebra behind a horizontal shift. Doing it as substitution is the reliable route: rewrite the rule, then, if the question asks about the graph, interpret what changed.

Coming next in this domain: interpreting nonlinear graphs.

Editorial team · last reviewed 2026-08-06

Function notation and evaluation

On the test

What this page owns

Pattern recognition

Method

Worked example 1 — a composition, inside-out

Worked example 2 — f(x+1)f(x+1)f(x+1) is not f(x)+1f(x)+1f(x)+1

Worked example 3 — one table, read in both directions

Practice

Question 1 Warm-up

Why it is right

Why each other choice fails

Question 2 Standard

Why it is right

Why each other choice fails

Question 3 Standard

Why it is right

Why each other choice fails

Question 4 Standard

Why it is right

Why each other choice fails

Question 5 Standard Student-produced response

Why it is right

Answers students type instead

Question 6 Harder

Why it is right

Why each other choice fails

Question 7 Harder

Why it is right

Why each other choice fails

Question 8 Harder

Why it is right

Why each other choice fails

Question 9 Harder

Why it is right

Why each other choice fails

Question 10 Hardest

Why it is right

Why each other choice fails

Question 11 Hardest

Why it is right

Why each other choice fails

Question 12 Hardest Student-produced response

Why it is right

Answers students type instead

Common mistakes

FAQ

Related skills

Worked example 2 — $f(x+1)$ is not $f(x)+1$