Topic 2.8

Inverse Functions

Ridge's model $R(t) = 1200 + 300t$ answers the question “how many subscribers after $t$ months?” But the channel's owner usually asks the opposite: “how many months until we reach 3,000?” That question runs the function backwards, from output to input. A function that runs another one backwards is called its inverse.

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01
$(a, b)$ on $f$ ⇔ $(b, a)$ on $f^{-1}$ . The graph of $f^{-1}$ is the reflection of $f$ across $y = x$ .

Practice 33 problems Next: Chapter 2.9 Logarithmic Expressions

Remember from 2.7: the identity function $I(x) = x$ changes nothing. An inverse is a function that, composed with the original, gives exactly that: it undoes everything the original did.

Undoing a Function

CONCEPT

Inverse Function

If $f$ sends input a to output $b$ , its inverse $f^{-1}$ sends $b$ back to a: $f(a) = b$ means $f^{-1}(b)$ = a.

Composing them in either order returns the original input.

The inputs of $f^{-1}$ are the outputs of $f$ , and the outputs of $f^{-1}$ are the inputs of $f$ . So the domain and range trade places.

f^{-1}\big(f(x)\big) = x \qquad\text{and}\qquad f\big(f^{-1}(x)\big) = x

REAL-LIFE EXAMPLE

Months Until 3,000

Solve $s = 1200 + 300t$ for $t$ . The result is a new function that takes a subscriber count and returns the month:

s = 1200 + 300\,t \quad\Rightarrow\quad t = \frac{s - 1200}{300} = R^{-1}(s)

$R^{-1}(3000) = (3000 - 1200)/300 = 6$ months, and indeed $R(6) = 3000$ . For every $t$ , $R^{-1}(R(t)) = t$ : going forward and then back leaves you where you started.

COMMON MISTAKE

“ $f^{-1}(x)$ means $1/f(x)$ .”

The −1 here is not an exponent. It is notation for the function that undoes $f$ . For $f(x) = 2x + 1$ , the inverse is $f^{-1}(x) = (x - 1)/2$ , so $f^{-1}(7) = 3$ (because $f(3) = 7$ ). The reciprocal $1/f(7) = 1/15$ is a completely different number.

From a Table: Swap the Rows

Nova's subscriber counts from 2.1:

t (month)	0	1	2	3	4
N(t)	256	384	576	864	1296

To get a table for $N^{-1}$ , swap the rows: the subscriber counts become the inputs and the months become the outputs. For example $N^{-1}(864)$ = 3, because $N(3) = 864$ . We can already use $N^{-1}$ from the table, even though we can't yet write its formula; that needs logarithms (2.9 and 2.10).

COMMON MISTAKE

“This table has an inverse too: $x$ : 1, 2, 3, 4, 5 and $g(x)$ : 4, 7, 2, 7, 9.”

The output 7 comes from both $x = 2$ and $x = 4$ . So $g^{-1}(7)$ would have to be 2 AND 4, and a function can't give two outputs for one input. $g$ has no inverse. If an output repeats, there is no inverse.

Worked example

Example (table). $f$ is given by $x$ : 0, 1, 2, 3 and $f(x)$ : 5, 8, 11, 14. Find $f^{-1}(11)$ , $f^{-1}(14)$ , and $f^{-1}(f(2))$ .

01
$f^{-1}(11)$ asks: which INPUT gave the output 11? $f(2) = 11$ , so $f^{-1}(11)$ = 2. Likewise $f^{-1}(14)$ = 3.
02
$f^{-1}(f(2))$ : $f(2) = 11$ , and $f^{-1}(11)$ = 2. Going forward and back returns the input 2, as it always must.

When Does an Inverse Exist?

CONCEPT

One-to-One Functions

A function is one-to-one if different inputs always give different outputs. Only one-to-one functions have inverse functions.

Horizontal line test: if some horizontal line crosses the graph more than once, the function is not one-to-one.

Functions that are always increasing or always decreasing (lines with nonzero slope, and every exponential function from 2.3) are one-to-one.

Left: $y = 5$ crosses the parabola at $x = 0$ and $x = 4$ , so $f(0) = f(4) = 5$ and $f$ has no inverse. Right: keeping only $x \ge 2$ makes $f$ one-to-one.

When a function isn't one-to-one, you can often restrict its domain to a piece that is. Different restrictions give different inverses, so the restriction is part of the answer.

Quick check

A table gives $f(x) = 4, 7, 4, 9$ at $x = 1, 2, 3, 4$ . Is $f$ invertible?

Finding an Inverse Formula

CONCEPT

Three Steps

Step 1. Write $y = f(x)$ .

Step 2. Solve for $x$ in terms of $y$ .

Step 3. Rename: swap $x$ and $y$ to write $f^{-1}(x)$ . Then check $f(f^{-1}(x)) = x$ with a number.

Why swap $x$ and $y$ at the end? After solving, x = (a formula in y) already IS the inverse: it takes an output $y$ and returns the input $x$ . Swapping the letters just rewrites it in the usual form, with $x$ as the input. On a graph, that swap is exactly the reflection across $y = x$ (Section 5).

Worked example

Example 1. Find the inverse of $f(x) = (2x - 5)/3$ .

01
Solve for $x$ , then rename:
$y = \frac{2x - 5}{3} \;\Rightarrow\; 3y = 2x - 5 \;\Rightarrow\; x = \frac{3y + 5}{2} \;\Rightarrow\; f^{-1}(x) = \frac{3x + 5}{2}$
02
Check with a number: $f^{-1}(4) = (12 + 5)/2 = 8.5$ , and $f(8.5) = (17 - 5)/3 = 4$ ✓.
03
Check with algebra: $f(f^{-1}(x)) = (2 \cdot (3x + 5)/2 - 5)/3 = (3x + 5 - 5)/3 = 3x/3 = x$ ✓.

Worked example

Example 2. $f(x) = (x - 2)^{2} + 1$ restricted to $x \ge 2$ . Find $f^{-1}$ and its domain.

01
Solve $y = (x - 2)^{2} + 1$ for $x$ . Taking a square root gives TWO candidates, just as in 2.1 and 2.5A:
$y = (x - 2)^{2} + 1 \;\Rightarrow\; (x - 2)^{2} = y - 1 \;\Rightarrow\; x = 2 \pm \sqrt{y - 1}$
02
Now the restriction decides. We need $x \ge 2$ , and $2 - \sqrt{y - 1}$ is at most 2 (equal to 2 only when $y = 1$ ), so it points to the wrong half of the parabola. Keep the + sign: $f^{-1}(x) = 2 + \sqrt{x - 1}$ .
03
Domain and range swap: $f$ has domain $x \ge 2$ and range $y \ge 1$ , so $f^{-1}$ has domain $x \ge 1$ and range $y \ge 2$ . Check: $f^{-1}(5) = 2 + 2 = 4$ , and $f(4) = 5$ ✓.
If we had restricted to $x \le 2$ instead, the other sign would win: $f^{-1}(x) = 2 - \sqrt{x - 1}$ . Same parabola, different piece, different inverse.

COMMON MISTAKE

“ $x = 2 \pm \sqrt{y - 1}$ , so $f^{-1}(x) = 2$ ± $\sqrt{x - 1}$ .”

A function must give ONE output. The ± shows two candidates; the domain restriction of $f$ picks exactly one. Without a restriction, a parabola has no inverse at all.

Worked example

Example 3. $f(x) = x^{2} + 3$ restricted to $x \ge 0$ . Find $f^{-1}$ .

01
$y = x^{2} + 3$ gives $x^{2} = y - 3$ , so $x = \pm \sqrt{y - 3}$ .
02
The restriction $x \ge 0$ keeps the + sign: $f^{-1}(x) = \sqrt{x - 3}$ , with domain $x \ge 3$ (the range of f).
03
Check: $f(2) = 7$ and $f^{-1}(7) = \sqrt{4} = 2$ ✓.

Quick check

Find the inverse of $f(x) = 2x - 8$ .

The Graph: Reflect Across y = x

Swapping inputs and outputs turns every point $(a, b)$ on $f$ into the point $(b, a)$ on $f^{-1}$ . Geometrically, that swap is a reflection across the line $y = x$ .

$(3, 2)$ on $f$ becomes $(2, 3)$ on $f^{-1}$ ; $(4, 5)$ becomes $(5, 4)$ . The two graphs are mirror images across $y = x$ .

KEY RULE

$(a, b)$ on $f$ ⇔ $(b, a)$ on $f^{-1}$ . The graph of $f^{-1}$ is the reflection of $f$ across $y = x$ .

Worked example

Try it yourself. (a) Find the inverse of $h(x) = 5 - 2x$ and check that $h(h^{-1}(3))$ = 3. (b) If $f(2) = 9$ , what is $f^{-1}(9)$ ? Which point is on the graph of $f^{-1}$ ?

Answers: (a) $y = 5 - 2x$ gives $x = (5 - y)/2$ , so $h^{-1}(x) = (5 - x)/2$ . $h^{-1}(3) = 1$ and $h(1) = 3$ ✓. (b) $f^{-1}$ (9) = 2; the point $(9, 2)$ .

Practice

Worked example

P1. Find the inverse of $f(x) = x^{3} + 4$ and find $f^{-1}(12)$ .

Answer: x = ∛ $(y - 4)$ , so $f^{-1}(x)$ = ∛ $(x - 4)$ . A cube root has one real value, so no restriction is needed. $f^{-1}(12)$ = ∛8 = 2.

Worked example

P2. Use $R^{-1}$ to find how many months it takes Ridge to reach 4,500 subscribers.

Answer: $R^{-1}(4500) = (4500 - 1200)/300 = 11$ months.

Worked example

P3. $f(x) = x^{2} - 6$ restricted to $x \le 0$ . Find $f^{-1}(x)$ , its domain, and $f^{-1}(10)$ .

Answer: $x = \pm \sqrt{y + 6}$ ; the restriction $x \le 0$ picks the − sign: $f^{-1}(x) = -\sqrt{x + 6}$ , domain $x \ge -6$ . $f^{-1}(10)$ = −4, and $f(-4) = 10$ ✓.

Worked example

P4. Find the inverse of $f(x) = (x + 1)/(x - 2)$ .

Answer: $y(x - 2) = x + 1 \;\Rightarrow\; xy - 2y = x + 1 \;\Rightarrow\; xy - x = 2y + 1 \;\Rightarrow\; x(y - 1) = 2y + 1 \;\Rightarrow\; x = (2y + 1)/(y - 1)$ . So $f^{-1}(x) = (2x + 1)/(x - 1)$ . Check: $f(3) = 4$ and $f^{-1}(4) = 9/3 = 3$ ✓.

Common slips

“ $f^{-1}(x)$ means $1/f(x)$ .”

The −1 here is not an exponent. It is notation for the function that undoes $f$ . For $f(x) = 2x + 1$ , the inverse is $f^{-1}(x) = (x - 1)/2$ , so $f^{-1}(7) = 3$ (because $f(3) = 7$ ). The reciprocal $1/f(7) = 1/15$ is a completely different number.
“This table has an inverse too: $x$ : 1, 2, 3, 4, 5 and $g(x)$ : 4, 7, 2, 7, 9.”

The output 7 comes from both $x = 2$ and $x = 4$ . So $g^{-1}(7)$ would have to be 2 and 4, and a function can't give two outputs for one input. $g$ has no inverse. If an output repeats, there is no inverse.
“ $x = 2 \pm \sqrt{y - 1}$ , so $f^{-1}(x) = 2$ ± $\sqrt{x - 1}$ .”

A function must give one output. The ± shows two candidates; the domain restriction of $f$ picks exactly one. Without a restriction, a parabola has no inverse at all.

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5 lines to re-read the night before.

01
$f^{-1}$ undoes $f$ : $f(a) = b$ means $f^{-1}(b)$ = a, and $f^{-1}$ ( $f(x)) = f( f^{-1}$ (x)) = $x$ . $f^{-1}$ is not $1/f$ .
02
Only one-to-one functions have inverses (horizontal line test). Restrict the domain if necessary.
03
To find $f^{-1}$ : solve $y = f(x)$ for $x$ , then swap names. When a square root gives ±, the domain restriction chooses the sign.
04
Domain and range swap. On the graph, $(a, b)$ ↔ $(b, a)$ : a reflection across $y = x$ .
05
Next, in 2.9: the inverse of an exponential, the logarithm. It finally answers “when does Nova reach 10,000?”

Practice 33 problems Next: Chapter 2.9 Logarithmic Expressions

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Inverse Functions

Undoing a Function

From a Table: Swap the Rows

When Does an Inverse Exist?

Finding an Inverse Formula

The Graph: Reflect Across y = x

Practice

Recap card

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