OMPT drill — Functions

OMPT inverse and composite functions practice: lesson and 10 problems

Chaining functions and undoing them. The mechanical part is easy; the traps are order of composition and domain after inversion. Attempt each problem before opening the solution; the notation only becomes automatic by writing it yourself.

Tested inOMPT-A OMPT-B OMPT-D OMPT-F

OMPT inverse and composite functions: the lesson

What the OMPT tests in inverse and composite functions

Composition and inverses sit in the Functions chapter of OMPT-A, B, D and F. The sub-skills are: evaluating a composition at a number, $f (g (2))$ ; writing the formula of $f (g (x))$ or $g (f (x))$ and simplifying it; finding the inverse of a linear, rational, root or exponential function; stating the domain of that inverse; and reading a fact about $f^{- 1}$ as a fact about $f$ , such as " $f^{- 1} (5) = 2$ , so $f (2) = 5$ ". The heavy families are the inverse formula and the composition in the correct order. A typical question reads "Let $f (x) = 2 x + 3$ and $g (x) = x^{2}$ . Find $g (f (x))$ " or "Find $f^{- 1} (x)$ and state its domain".

On OMPT-A and OMPT-B I expect one or two items among the four to six Functions questions. On OMPT-D and OMPT-F the composition is usually a step inside a longer question, for example the chain rule in differentiation, which is composition wearing a different hat, or a two-stage pricing model that has to be run backwards.

Notation and answer format the OMPT expects

$f (g (x))$ and $(f \circ g) (x)$ mean the same thing: apply $g$ first, then $f$ . Read the small circle as "after".
$f^{- 1} (x)$ is the inverse function, not $\frac{1}{f ( x )}$ . The $- 1$ is a label, not an exponent.
Type an inverse as an expression in $x$ , for example (x + 1)/3 or sqrt(x - 2) or log_2(x). Exact forms only; no decimals unless the question says "round to".
When a domain is asked for, type it as a condition such as x >= 0 or x > -5. When a composition is evaluated at a number, the answer is that number, exact.

Worked example 1: the order experiment

Let $f (x) = 2 x + 3$ and $g (x) = x^{2}$ . Compute $f (g (2))$ and $g (f (2))$ .
For $f (g (2))$ , inner first: $g (2) = 4$ . Then outer: $f (4) = 8 + 3 = 11$ .
For $g (f (2))$ , the inner is now $f$ : $f (2) = 7$ . Then $g (7) = 49$ .
As formulas: $f (g (x)) = 2 x^{2} + 3$ , while $g (f (x)) = (2 x + 3)^{2} = 4 x^{2} + 12 x + 9$ . Substituting $x = 2$ confirms 11 and 49.
11 versus 49 from the same two machines and the same input. Composition does not commute, and the OMPT checks whether you respect that by asking for the order students do not expect.

Worked example 2: a linear inverse, with check

Find the inverse of $f (x) = 3 x - 1$ .
Write $y = 3 x - 1$ and solve for $x$ : $y + 1 = 3 x$ , so $x = \frac{y + 1}{3}$ .
Swap the letters so the inverse reads as a function of $x$ : $f^{- 1} (x) = \frac{x + 1}{3}$ .
Check by composing: $f^{- 1} (f (x)) = \frac{( 3 x - 1 ) + 1}{3} = \frac{3 x}{3} = x$ . Clean $x$ , so the inverse is right. A numeric check works too: $f (2) = 5$ and $f^{- 1} (5) = 2$ .
The slip here is undoing the operations in the wrong order, dividing by 3 first and then adding 1, which gives $\frac{x}{3} + 1$ . The inverse undoes the last operation first.

The full lesson covers the domain of a composition, the one-to-one condition and the horizontal line test, inverses of rational, root and exponential functions with their domains, solving an equation of the form $f (g (x)) = c$ , a parameter question about $f (f (x))$ and a two-stage pricing problem run backwards. It ends with the edge cases the OMPT likes and the slips I mark most often.

How each OMPT variant tests inverse and composite functions

The chapter weights are the official ones from the omptest.org learning-assessment pages. The question counts are my estimates of how much of that chapter this one topic takes up. The 21-question papers (OMPT-D and OMPT-F) ask long multi-part questions, so a topic there usually shows up as a step inside a bigger problem.

OMPT-A
52 questions
Testedroughly 1 to 2 of 52 questions, inside the 4 to 6 for the Functions chapter
Official chapter: Quadratic equations and functions, 22% of the paper.
Evaluate f(g(2)) and g(f(2)), write the formula of a composition, and find the inverse of a linear or simple rational function. Occasionally the domain of the inverse or a fact read from f inverse (5) = 2.
OMPT-B
64 questions
Testedroughly 1 to 2 of 64 questions
Official chapter: Quadratic equations and functions, 10% of the paper.
Same shapes as OMPT-A, and the decomposition of a function into inner and outer parts returns as the chain rule in the differentiation chapter.
OMPT-C
30 questions
Not tested
OMPT-C has no Functions chapter; nothing is composed or inverted beyond rearranging a linear formula. Spend the time on linear equations and geometry instead.
OMPT-D
21 questions
Tested1 to 2 of the 21 long questions, usually as a step inside a bigger problem
Official chapter: Functions, equations and inequalities, 30% of the paper.
Inverses of exponential and root functions with their domains, then the chain rule and inverse-function reasoning inside long differentiation questions.
OMPT-E
40 questions
Not tested
OMPT-E has a Functions chapter aimed at models for the statistics chapters; composition and inverses are not named skills there. Study function basics and exponential functions instead.
OMPT-F
21 questions
Tested1 to 2 of the 21 long questions, usually as a step inside a bigger problem
Official chapter: Functions, 15% of the paper.
Business processes composed in order (a charge, then a percentage), then run backwards to recover an input, and composition inside the chain rule for a cost or revenue derivative.
OMPT-G
26 questions
Not tested
OMPT-G uses functions as physics and chemistry models; composition and inverses are not named. Put the time into trigonometry, geometry and sequences.

OMPT inverse and composite functions practice problems

10 original problems in the formats the OMPT uses: typed numeric answers, multiple choice, and spot-the-error. The last few are harder than the exam average on purpose. Solutions and the common mistake sit behind each problem.

Problem 1Numeric answer

Let $f (x) = 3 x - 4$ and $g (x) = x^{2}$ . Compute $f (g (2))$ .

Show the worked solution

Answer: 8

I work from the inside out: $g (2) = 4$ , then $f (4) = 3 \cdot 4 - 4 = 8$ . The inner function always fires first — the notation $f (g (2))$ hands $g$ 's output to $f$ .

Problem 2Multiple choice

What is the inverse of $f (x) = \frac{x + 7}{2}$ ?

A $f^{- 1} (x) = 2 x - 7$
B $f^{- 1} (x) = \frac{2}{x + 7}$
C $f^{- 1} (x) = 2 x + 7$
D $f^{- 1} (x) = \frac{x - 7}{2}$

Show the worked solution

Answer: A — $f^{- 1} (x) = 2 x - 7$

I write $y = (x + 7) /2$ , swap the roles of $x$ and $y$ , and solve: $x = (y + 7) /2$ gives $2 x = y + 7$ , so $y = 2 x - 7$ . The inverse undoes the original in reverse order: first undo the division (multiply by 2), then undo the $+ 7$ (subtract 7). Check: $f (3) = 5$ and $f^{- 1} (5) = 3$ .

Problem 3Numeric answer

Let $f (x) = 2 x + 1$ and $g (x) = x^{2} - 3$ . Compute $g (f (1))$ .

Show the worked solution

Answer: 6

Inside out again: $f (1) = 3$ , then $g (3) = 9 - 3 = 6$ . Compare with the other order: $f (g (1)) = f (- 2) = - 3$ . Same two functions, different answer — order is everything in composition.

Problem 4Multiple choice

Suppose $f$ has an inverse and $f^{- 1} (5) = 2$ . Which statement must be true?

A $f (2) = 5$
B $f (5) = 2$
C $f (5) = \frac{1}{2}$
D $f (2) = \frac{1}{5}$

Show the worked solution

Answer: A — $f (2) = 5$

$f^{- 1} (5) = 2$ says: the input that $f$ sends to 5 is 2. In other words, $f (2) = 5$ . Inverse functions swap input and output; each fact about $f^{- 1}$ is the mirrored fact about $f$ .

Problem 5Spot the error

A student finds the inverse of $y = 2 x + 6$ . Step 1: swap variables, $x = 2 y + 6$ . Step 2: solve for $y$ : $y = \frac{x}{2} - 6$ . Step 3: so $f^{- 1} (x) = \frac{x}{2} - 6$ . Which step contains the error?

Show the worked solution

Answer: Step 2

The swap in Step 1 is correct. In Step 2 the student divided by 2 first and then subtracted the full 6, but the 6 must be removed before dividing: $x - 6 = 2 y$ gives $y = \frac{x - 6}{2} = \frac{x}{2} - 3$ . A composition check exposes the slip: with the student's formula, $f (1) = 8$ but $8/2 - 6 = - 2 \neq = 1$ . With the correct inverse, $(8 - 6) /2 = 1$ .

Problem 6Numeric answer

Let $f (x) = \frac{x}{x - 1}$ for $x \neq = 1$ . Compute $f (f (7))$ .

Show the worked solution

Answer: 7

First $f (7) = \frac{7}{6}$ . Then $f (\frac{7}{6}) = \frac{7/6}{7/6 - 1} = \frac{7/6}{1/6} = 7$ . We are back where we started — this function is its own inverse, so applying it twice returns every input unchanged.

Problem 7Numeric answer

Let $f (x) = x^{2} + 1$ and $g (x) = 3 x - 2$ . Compute $f (g (2)) - g (f (2))$ .

Show the worked solution

Answer: 4

Inside out, twice. For $f (g (2))$ : $g (2) = 4$ , then $f (4) = 17$ . For $g (f (2))$ : $f (2) = 5$ , then $g (5) = 13$ . The difference is $17 - 13 = 4$ . As formulas, $f (g (x)) = (3 x - 2)^{2} + 1$ and $g (f (x)) = 3 x^{2} + 1$ ; at $x = 2$ those give 17 and 13 again.

Problem 8Multiple choice

What is the inverse of $f (x) = \frac{x}{x + 2}$ , $x \neq = - 2$ ?

A $f^{- 1} (x) = \frac{2 x}{1 - x}$
B $f^{- 1} (x) = \frac{x + 2}{x}$
C $f^{- 1} (x) = \frac{x}{x - 2}$
D $f^{- 1} (x) = \frac{1 - x}{2 x}$

Show the worked solution

Answer: A — $f^{- 1} (x) = \frac{2 x}{1 - x}$

I set $y = \frac{x}{x + 2}$ and clear the fraction: $y (x + 2) = x$ , so $x y + 2 y = x$ . Every term with $x$ goes to one side: $x y - x = - 2 y$ , so $x (y - 1) = - 2 y$ and $x = \frac{- 2 y}{y - 1} = \frac{2 y}{1 - y}$ . Swapping letters, $f^{- 1} (x) = \frac{2 x}{1 - x}$ , with $x \neq = 1$ . Check: $f (2) = \frac{2}{4} = \frac{1}{2}$ , and $f^{- 1} (\frac{1}{2}) = \frac{1}{1/2} = 2$ .

Problem 9Numeric answer

Let $f (x) = 2 x - 3$ and $g (x) = x^{2}$ . The equation $g (f (x)) = 25$ has two solutions. Give their sum.

Show the worked solution

Answer: 3

Inside out: $g (f (x)) = (2 x - 3)^{2}$ , so the equation is $(2 x - 3)^{2} = 25$ . Square-rooting with both signs: $2 x - 3 = 5$ gives $x = 4$ , and $2 x - 3 = - 5$ gives $x = - 1$ . Check: $f (4) = 5$ , $g (5) = 25$ ; $f (- 1) = - 5$ , $g (- 5) = 25$ . The sum is $4 + (- 1) = 3$ .

Problem 10Numeric answer

A taxi charges a fixed 4 euros plus 2.50 euros per kilometre, so a ride of $d$ km costs $c (d) = 4 + 2.5 d$ . A promotion then takes 20% off the whole fare, $s (x) = 0.8 x$ . A passenger paid 23.20 euros with the promotion. How many kilometres was the ride?

Show the worked solution

Answer: 10

The passenger paid the discount applied to the fare, which is $s (c (d)) = 0.8 (4 + 2.5 d) = 3.2 + 2 d$ ; the order matters, since the discount acts on the whole fare including the 4 euros. Running it backwards: $3.2 + 2 d = 23.2$ , so $2 d = 20$ and $d = 10$ km. Equivalently, undo the discount first, $23.2/0.8 = 29$ , then undo the fare, $(29 - 4) /2.5 = 10$ . Check forwards: $4 + 25 = 29$ , and $0.8 \cdot 29 = 23.2$ .

OMPT inverse and composite functions: questions students ask

Are inverse and composite functions on the OMPT-A?

Yes, inside the Functions chapter. OMPT-A asks you to evaluate a composition at a number, write the formula of f(g(x)) or g(f(x)), find the inverse of a linear or simple rational function, and sometimes state the domain of that inverse. Expect one or two such items among the four to six Functions questions. OMPT-B, D and F test the same skills; OMPT-C, E and G do not name them.

How do I find the inverse of a function for the OMPT?

Write y equals f(x), solve that equation for x in terms of y, then swap the letters so the answer reads as a function of x. For 3x minus 1 that gives (x plus 1)/3. Check by composing: f inverse of f(x) must simplify to plain x. If the original involved a root or a square, add the domain: the domain of the inverse is the range of the original.

What is the difference between f(g(x)) and g(f(x)) on the OMPT?

The order. f(g(x)) applies g first and then f; g(f(x)) applies f first and then g. With f(x) = 2x + 3 and g(x) = x squared, f(g(2)) is 11 and g(f(2)) is 49. Read a composition from the inside bracket outwards, and when the notation (f after g)(x) with a small circle appears, it means f(g(x)): the function nearest the input goes first.

Is f to the power minus 1 the same as 1 over f on the OMPT?

No. f inverse, written with a superscript minus 1, is the function that undoes f, and 1 over f(x) is the reciprocal of its output. For f(x) = 3x minus 1 the inverse is (x plus 1)/3, while the reciprocal is 1/(3x minus 1). The OMPT plants the reciprocal among the multiple-choice options in most inverse questions.

Why does the inverse of a square root function need a domain on the OMPT?

Because a square root never outputs a negative number, so its inverse must never accept one. The inverse of the square root of (x minus 2) is x squared plus 2, but only for x greater than or equal to 0; without that restriction the formula would claim that a root equals minus 1. The domain of the inverse is the range of the original function, and the OMPT marks the formula alone as incomplete.

Test hubs that cover inverse and composite functions:OMPT-A OMPT-B OMPT-D OMPT-F All 28 topics

OMPT inverse and composite functions practice: lesson and 10 problems

OMPT inverse and composite functions: the lesson

What the OMPT tests in inverse and composite functions

Notation and answer format the OMPT expects

Worked example 1: the order experiment

Worked example 2: a linear inverse, with check

How each OMPT variant tests inverse and composite functions

OMPT inverse and composite functions practice problems

OMPT inverse and composite functions: questions students ask

Related OMPT topics and test hubs