OMPT drill — Functions

Inverse and composite functions

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

Lesson

Composition chains two machines together: $f (g (x))$ means run $x$ through $g$ first, then feed whatever comes out into $f$ . Read it inside-out, always. The notation is the only hard part, because $f (g (x))$ and $g (f (x))$ look like siblings and are usually completely different functions; order matters in composition the way it matters in getting dressed: socks then shoes is not shoes then socks.

An inverse function runs a machine backwards. Where $f$ takes 3 to 11, the inverse $f^{- 1}$ takes 11 back to 3. Finding it is a two-step ritual; swap the roles of input and output, then solve for the new output, and checking it is even easier: composing a function with its inverse must give you plainly $x$ back. If $f^{- 1} (f (x))$ simplifies to anything else, one of the two is wrong, and the check just paid for itself.

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.