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Lesson 5 — Composition and inverse functions

Composition f∘g as sequential operations. Inverse f⁻¹ undoing the operation. Conditions for invertibility: bijection or domain restriction.

Used in: 1.º ano do EM (15 anos) · Math I japonês cap. 3 · Klasse 10 alemã — Funktionen

(f \circ g)(x) = f(g(x)), \quad f^{-1}(f(x)) = x

Composition chains two functions — the output of $g$ becomes the input of $f$ . Inverse undoes the operation: $f^{-1}$ exists if and only if f is bijective (or the domain is restricted). The identity $f^{-1}(f(x)) = x$ is the universal verification.

Choose your door

Rigorous notation, full derivation, hypotheses

Rigorous definition

Composition of functions

"When we combine functions such that the output of one function becomes the input of another, we create a composition of functions. The resulting function is called a composite function." — OpenStax College Algebra 2e, §3.4

Composition: each solid arrow is a function; the dashed arrow below is the composite f ∘ g — a shortcut that "skips" the intermediate set B.

Inverse function

"In order for a function to have an inverse function, it needs to be a one-to-one function. A function is one-to-one if each output value corresponds to exactly one input value." — Stitz–Zeager Precalculus, §5.2

f and its inverse are symmetric with respect to the line y = x. Reflecting the graph of f across that diagonal gives the graph of f⁻¹.

Worked examples

Example— 1· Computing (f∘g)(x) — noting that order changes the result

Problem: Given $f(x) = 3x + 1$ and $g(x) = x^2$ , compute $(f \circ g)(x)$ and $(g \circ f)(x)$ and compare.

Strategy: Apply the definition $(f \circ g)(x) = f(g(x))$ — first evaluate $g(x)$ , then substitute the result into $f$ . Repeat in the opposite order for $(g \circ f)$ .

Solution:

$(f \circ g)(x) = f(g(x)) = f(x^2) = 3x^2 + 1$ .
$(g \circ f)(x) = g(f(x)) = g(3x + 1) = (3x + 1)^2 = 9x^2 + 6x + 1$ .
Compare: $3x^2 + 1 \neq 9x^2 + 6x + 1$ . The order changed everything.

Verification: At $x = 2$ : $(f \circ g)(2) = 3 \cdot 4 + 1 = 13$ and $(g \circ f)(2) = 7^2 = 49$ . Distinct, confirming non-commutativity.

Source. OpenStax College Algebra 2e §3.4 — CC-BY 4.0.

Example— 2· Finding the inverse of a linear function and verifying

Problem: Determine $f^{-1}(x)$ for $f(x) = 4x - 7$ . Verify that $f \circ f^{-1} = f^{-1} \circ f = \operatorname{id}$ .

Strategy: Standard algorithm: write $y = f(x)$ , swap $x \leftrightarrow y$ , solve for $y$ . Then validate.

Solution:

$y = 4x - 7$ .
Swap: $x = 4y - 7$ .
Isolate: $4y = x + 7 \Rightarrow y = (x + 7)/4$ .
Therefore $f^{-1}(x) = (x + 7)/4$ .

Verification:

$f(f^{-1}(x)) = 4 \cdot \dfrac{x+7}{4} - 7 = (x + 7) - 7 = x$ . ✓
$f^{-1}(f(x)) = \dfrac{(4x - 7) + 7}{4} = \dfrac{4x}{4} = x$ . ✓

Source. OpenStax College Algebra 2e §5.7 — CC-BY 4.0.

Example— 3· Domain of a composite with a square root — beware of restrictions

Problem: Let $f(x) = \sqrt{x}$ and $g(x) = x^2 - 4$ . Determine $(f \circ g)(x)$ and its maximum domain.

Strategy: The composite exists where $g(x)$ falls in the domain of $f$ , i.e., where $g(x) \geq 0$ .

Solution:

$(f \circ g)(x) = f(g(x)) = \sqrt{x^2 - 4}$ .
For it to exist: $x^2 - 4 \geq 0 \Rightarrow x^2 \geq 4 \Rightarrow |x| \geq 2$ .
Domain: $(-\infty, -2] \cup [2, +\infty)$ .

Verification: At $x = 3$ : $g(3) = 5$ ; $f(5) = \sqrt{5}$ . And $\sqrt{9-4} = \sqrt{5}$ . ✓. At $x = 1$ : $g(1) = -3 < 0$ , outside the domain of $f$ — correct.

Source. Stitz–Zeager Precalculus §5.1 — CC-BY-NC-SA.

Example— 4· Inverse of a quadratic function with domain restriction

Problem: $f(x) = x^2 - 4x + 7$ is not invertible on $\mathbb{R}$ . Restrict the domain so that it becomes invertible and determine $f^{-1}$ .

Strategy: Complete the square to identify the vertex. Restrict to one branch (convention: right branch) and invert.

Solution:

$f(x) = (x-2)^2 + 3$ . Vertex $(2, 3)$ , parabola increasing for $x \geq 2$ .
Restrict to $A = [2, +\infty)$ : $f$ is strictly increasing, hence injective.
$y = (x-2)^2 + 3 \Rightarrow (x-2)^2 = y - 3$ . Since $x \geq 2$ , we have $x - 2 = \sqrt{y-3}$ , so $x = 2 + \sqrt{y-3}$ .
$f^{-1}: [3, +\infty) \to [2, +\infty)$ , $f^{-1}(y) = 2 + \sqrt{y-3}$ .

Verification: $f^{-1}(7) = 2 + \sqrt{4} = 4$ . $f(4) = (4-2)^2 + 3 = 7$ . ✓

Source. OpenStax College Algebra 2e §5.7 — CC-BY 4.0.

Example— 5· Chain composition: currency conversion modeled as inverse of a composite

Problem: Dollar-to-real conversion via $f(d) = 5.00 \cdot d$ (simplified rate). Real-to-loyalty-points conversion via $g(r) = r/10$ . (a) Model the composite function dollar to points. (b) Determine the inverse points to dollar. (c) How many dollars are 150 points worth?

Strategy: Direct composition is $g \circ f$ . The inverse is $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$ — note the reversed order.

Solution:

$(g \circ f)(d) = g(f(d)) = g(5d) = 5d/10 = d/2$ .
$f^{-1}(r) = r/5$ and $g^{-1}(p) = 10p$ .
$(g \circ f)^{-1}(p) = f^{-1}(g^{-1}(p)) = f^{-1}(10p) = 10p/5 = 2p$ .
For $p = 150$ : $2 \cdot 150 = 300$ dollars.

Verification: 300 USD to 1,500 BRL to 150 points. ✓

Source. Yoshiwara — Modeling, Functions, and Graphs ch. 4 — free.

Exercise list

40 exercises · 10 with worked solution (25%)

Application 18Understanding 9Modeling 6Challenge 4Proof 3

Ex. 5.1Understanding
How do you determine the domain of the quotient of two functions, $f/g$ ?
Select the correct option
All $x$ where $f$ and $g$ are defined and $g(x) \neq 0$ .All $x$ where $f(x) \neq 0$ .All $x$ where $f$ and $g$ are defined, ignoring $g(x)=0$ .Only the $x$ where $f(x)=g(x)$ .
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 1
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The domain of $f/g$ includes all values where both functions are defined, excluding the points where the denominator $g(x)$ is zero.
Show step-by-step (with the why)
1. 1. Determine the domain of $f$ and the domain of $g$ separately.
2. 2. Take the intersection of those two domains.
3. 3. Exclude from the intersection all $x$ such that $g(x) = 0$ .
4. 4. The resulting set is the domain of $f/g$ .
Ex. 5.2UnderstandingAnswer key
What is the composition of two functions, $f \circ g$ ?
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The output of $g$ becomes the input of $f$ : $(f\circ g)(x) = f(g(x))$ .The output of $f$ becomes the input of $g$ : $(f\circ g)(x) = g(f(x))$ .It is the sum $f(x) + g(x)$ .It is the product $f(x) \cdot g(x)$ .
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 2
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By definition, $(f \circ g)(x) = f(g(x))$ : evaluate $g$ first, and pass the result to $f$ .
Show step-by-step (with the why)
1. 1. Evaluate $g$ at the point $x$ .
2. 2. Use that result as the argument of $f$ .
3. 3. The final value is $(f \circ g)(x)$ .
Ex. 5.3Understanding
If the order is reversed when composing two functions, can the result be the same as in the original order? If so, give an example.
Select the correct option
Yes, for example $f(x)=x$ and $g(x)=x$ satisfy $f \circ g = g \circ f$ .No, never: composition never commutes.Yes, always: $f \circ g = g \circ f$ for any functions.Yes, but only when $f(x) = -g(x)$ .
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 3
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In general $f \circ g \neq g \circ f$ , but for equal functions the compositions coincide. Example: $f(x) = 2x$ and $g(x) = 2x$ give $(f \circ g)(x) = 4x = (g \circ f)(x)$ .
Ex. 5.4UnderstandingAnswer key
How do you determine the domain of the composition $f \circ g$ ?
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The set of $x$ in the domain of $g$ such that $g(x)$ belongs to the domain of $f$ .The set of $x$ where $f(x)$ is defined.The set of $x$ where $g(x)$ is defined.The simple intersection of the domains of $f$ and $g$ , with no additional condition.
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 4
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For $f(g(x))$ to make sense, $x$ must be in the domain of $g$ and additionally $g(x)$ must fall within the domain of $f$ .
Show step-by-step (with the why)
1. 1. Write the domain of $g$ .
2. 2. For each $x$ in that domain, compute $g(x)$ and check whether it belongs to the domain of $f$ .
3. 3. Keep only the $x$ that satisfy both conditions.
Ex. 5.5Application
With $f(x) = 2x^2 + 1$ and $g(x) = 3x + 5$ , compute $f(g(2))$ and find $f(g(x))$ .
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$f(g(2)) = 243$ and $f(g(x)) = 2(3x+5)^2+1$ $f(g(2)) = 125$ and $f(g(x)) = 6x^2+11$ $f(g(2)) = 99$ and $f(g(x)) = (3x+5)^2+2$ $f(g(2)) = 243$ and $f(g(x)) = 9x^2+30x+26$
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 76–77
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$g(2) = 3(2)+5 = 11$ ; $f(11) = 2(121)+1 = 243$ . For the general formula: $f(g(x)) = 2(3x+5)^2+1$ .
Show step-by-step (with the why)
1. 1. Compute $g(2) = 3 \cdot 2 + 5 = 11$ .
2. 2. Compute $f(11) = 2 \cdot 11^2 + 1 = 2 \cdot 121 + 1 = 243$ .
3. 3. For the general formula: substitute $g(x) = 3x+5$ into $f$ : $f(g(x)) = 2(3x+5)^2 + 1$ .
Ex. 5.6Application
With $f(x) = 2x^2 + 1$ and $g(x) = 3x + 5$ , compute $g(f(-3))$ .
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$g(f(-3)) = 62$ $g(f(-3)) = 9$ $g(f(-3)) = -13$ $g(f(-3)) = 54$
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 78
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$f(-3) = 2(-3)^2+1 = 2 \cdot 9+1 = 19$ ; then $g(19) = 3(19)+5 = 62$ .
Show step-by-step (with the why)
1. 1. Compute $f(-3) = 2(-3)^2+1 = 18+1 = 19$ .
2. 2. Compute $g(19) = 3 \cdot 19+5 = 57+5 = 62$ .
Ex. 5.7Application
With $g(x) = 3x + 5$ , find $(g \circ g)(x)$ .
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$(g \circ g)(x) = 9x + 20$ $(g \circ g)(x) = 6x + 5$ $(g \circ g)(x) = 9x + 10$ $(g \circ g)(x) = 3x + 10$
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 79
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$(g \circ g)(x) = g(g(x)) = g(3x+5) = 3(3x+5)+5 = 9x+15+5 = 9x+20$ .
Ex. 5.8Application
With $f(x) = 4x + 8$ and $g(x) = 7 - x^2$ , compute $f(g(0))$ and $g(f(0))$ .
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$f(g(0)) = 36$ and $g(f(0)) = -57$ $f(g(0)) = 8$ and $g(f(0)) = 7$ $f(g(0)) = 36$ and $g(f(0)) = 57$ $f(g(0)) = 28$ and $g(f(0)) = -57$
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 72
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$g(0) = 7$ , so $f(g(0)) = 4(7)+8 = 36$ . On the other hand, $f(0) = 8$ , so $g(f(0)) = 7-8^2 = 7-64 = -57$ .
Show step-by-step (with the why)
1. 1. Compute $g(0) = 7 - 0^2 = 7$ .
2. 2. Compute $f(7) = 4(7)+8 = 36$ .
3. 3. Compute $f(0) = 4(0)+8 = 8$ .
4. 4. Compute $g(8) = 7 - 64 = -57$ .
Ex. 5.9Application
With $f(x) = 5x + 7$ and $g(x) = 4 - 2x^2$ , compute $f(g(0))$ and $g(f(0))$ . (Ans: 27 and $-94$ )
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$f(g(0)) = 27$ and $g(f(0)) = -4$ $f(g(0)) = 7$ and $g(f(0)) = 4$ $f(g(0)) = 4$ and $g(f(0)) = -28$ $f(g(0)) = 27$ and $g(f(0)) = 4$
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 73
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$g(0) = 4$ , so $f(4) = 5(4)+7 = 27$ . Also $f(0) = 7$ and $g(7) = 4 - 2(49) = 4-98 = -94$ . (Ans: $f(g(0))=27$ , $g(f(0))=-94$ )
Ex. 5.10Application
With $f(x) = x + 4$ and $g(x) = 12 - x^3$ , compute $f(g(0))$ and $g(f(0))$ . (Ans: 16 and $-52$ )
Select the correct option
$f(g(0)) = 16$ and $g(f(0)) = 12$ $f(g(0)) = 12$ and $g(f(0)) = 16$ $f(g(0)) = 4$ and $g(f(0)) = 12$ $f(g(0)) = 16$ and $g(f(0)) = -12$
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 74
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$g(0) = 12 - 0 = 12$ , so $f(12) = 12+4 = 16$ . Also $f(0) = 4$ and $g(4) = 12 - 4^3 = 12-64 = -52$ . (Ans: 16 and $-52$ )
Ex. 5.11Application
With $f(x) = \frac{1}{x+2}$ and $g(x) = 4x + 3$ , compute $f(g(0))$ and $g(f(0))$ . (Ans: $\frac{1}{5}$ and $5$ )
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$\frac{1}{2}$ and $3$ $2$ and $\frac{1}{3}$ $\frac{1}{2}$ and $\frac{1}{3}$ $3$ and $2$
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 75
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$g(0) = 4(0)+3 = 3$ and $f(3) = \frac{1}{3+2} = \frac{1}{5}$ . (Ans: $f(g(0)) = 1/5$ )
Ex. 5.12UnderstandingAnswer key
With $f(x) = x^3 + 1$ and $g(x) = \sqrt[3]{x-1}$ , compute $(f \circ g)(x)$ and $(g \circ f)(x)$ and compare the results.
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$(f \circ g)(x) = x = (g \circ f)(x)$ , both compositions equal the identity. $(f \circ g)(x) = x^3 + 1$ and $(g \circ f)(x) = x - 1$ . $(f \circ g)(x) = x + 1$ and $(g \circ f)(x) = x - 1$ .The compositions differ: $(f \circ g)(x) \neq (g \circ f)(x)$ .
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 80
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With $f(x) = x^3+1$ and $g(x) = \sqrt[3]{x-1}$ : $(f \circ g)(x) = (\sqrt[3]{x-1})^3+1 = x$ and $(g \circ f)(x) = \sqrt[3]{x^3+1-1} = x$ . Both are the identity, confirming that $g = f^{-1}$ .
Show step-by-step (with the why)
1. 1. Compute $f(g(x)) = f(\sqrt[3]{x-1}) = (\sqrt[3]{x-1})^3+1 = x-1+1 = x$ .
2. 2. Compute $g(f(x)) = g(x^3+1) = \sqrt[3]{x^3+1-1} = \sqrt[3]{x^3} = x$ .
3. 3. Conclude: both compositions give $x$ , so $f$ and $g$ are inverses of each other.
Ex. 5.13Application
With $f(x) = x^3 + 1$ and $g(x) = \sqrt[3]{x-1}$ , compute $(f \circ g)(2)$ and $(g \circ f)(2)$ .
Select the correct option
$(f \circ g)(2) = 2$ and $(g \circ f)(2) = 2$ $(f \circ g)(2) = 9$ and $(g \circ f)(2) = 2$ $(f \circ g)(2) = 2$ and $(g \circ f)(2) = 9$ $(f \circ g)(2) = 3$ and $(g \circ f)(2) = 3$
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 81
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$g(2) = \sqrt[3]{2-1} = 1$ , so $f(1) = 1+1 = 2$ . Also $f(2) = 9$ and $g(9) = \sqrt[3]{8} = 2$ . Both yield 2.
Ex. 5.14Understanding
With $f(x) = x^3 + 1$ and $g(x) = \sqrt[3]{x-1}$ , what is the domain of $(g \circ f)(x)$ ?
Select the correct option
All of $\mathbb{R}$ , since $x^3+1$ is defined for all $x$ and the cube root accepts any real.Only $x \geq 0$ , since the cube root requires a non-negative argument.Only $x \geq 1$ . $x \neq 0$ .
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 82
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$(g \circ f)(x) = \sqrt[3]{x^3}$ , and the cube root is defined for all reals. Therefore the domain is $\mathbb{R}$ .
Ex. 5.15Challenge
Let $F(x) = (x+1)^5$ , $f(x) = x^5$ and $g(x) = x+1$ . True or false: $(f \circ g)(x) = F(x)$ ?
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True: $(f \circ g)(x) = (x+1)^5 = F(x)$ .False: $(f \circ g)(x) = x^5+1 \neq F(x)$ .False: $(f \circ g)(x) = x^6$ .True, but only for $x \geq 0$ .
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 86
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$(f \circ g)(x) = f(g(x)) = f(x+1) = (x+1)^5 = F(x)$ . True.
Ex. 5.16Challenge
With $f(x) = x^2 + 2$ (for $x \geq 0$ ) and $g(x) = \sqrt{x-2}$ , compute $(f \circ g)(6)$ and $(g \circ f)(6)$ .
Select the correct option
$(f \circ g)(6) = 6$ and $(g \circ f)(6) = 6$ $(f \circ g)(6) = 38$ and $(g \circ f)(6) = 6$ $(f \circ g)(6) = 6$ and $(g \circ f)(6) = 36$ $(f \circ g)(6) = 38$ and $(g \circ f)(6) = 36$
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 87
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For $f(x) = x^2+2$ ( $x \geq 0$ ) and $g(x) = \sqrt{x-2}$ : $g(6) = 2$ , $f(2) = 6$ ; and $f(6) = 38$ , $g(38) = 6$ . Both yield 6.
Show step-by-step (with the why)
1. 1. $g(6) = \sqrt{6-2} = \sqrt{4} = 2$ .
2. 2. $f(2) = 4+2 = 6$ . Therefore $(f \circ g)(6) = 6$ .
3. 3. $f(6) = 36+2 = 38$ .
4. 4. $g(38) = \sqrt{38-2} = \sqrt{36} = 6$ . Therefore $(g \circ f)(6) = 6$ .
Ex. 5.17Modeling
The function $D(p)$ gives the number of items demanded when the price is $p$ . The function $C(x)$ is the cost of producing $x$ items. To determine the production cost when the price is 6, which composition would you evaluate?
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$C(D(6))$ : first find the quantity demanded at price 6, then compute production cost. $D(C(6))$ : apply the demand function to the cost. $D(C(x)) = 6$ : solve the equation. $C(D(p)) = 6$ : solve the equation for p.
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 90
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To find the production cost when the price is 6, first determine how many items will be demanded — $D(6)$ — then compute the cost of producing that quantity: $C(D(6))$ .
Ex. 5.18Modeling
The function $A(d)$ gives the pain level (0–10) with $d$ mg of analgesic. The amount of drug after $t$ minutes is $m(t)$ . What would you do to find when the patient will have pain level 4?
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Solve $A(m(t)) = 4$ : find the instant $t$ when pain reaches level 4.Evaluate $A(m(4))$ .Evaluate $m(A(4))$ .Solve $m(A(d)) = 4$ .
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 91
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Pain is a function of dose, which is a function of time: $A(m(t))$ gives the pain level as a function of time. To find when pain reaches 4, solve $A(m(t)) = 4$ .
Ex. 5.19Modeling
A store offers 30% off the price $x$ of selected items, then an additional 15% off at checkout. Write the final price function $P(x)$ using composition.
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$P(x) = 0.595x$ $P(x) = 0.45x$ $P(x) = 0.55x$ $P(x) = 0.85x$
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 92
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First apply the 30% discount: $d(x) = 0.7x$ . Then the additional 15% discount at checkout: $P(x) = 0.85 \cdot d(x) = 0.85 \cdot 0.7x = 0.595x$ .
Show step-by-step (with the why)
1. 1. 30% discount: price becomes $d(x) = 0.7x$ .
2. 2. Additional 15% discount: $P(x) = 0.85 \cdot 0.7x$ .
3. 3. Multiply: $0.85 \times 0.7 = 0.595$ .
Ex. 5.20ModelingAnswer key
A raindrop creates a circular ripple. The radius (in inches) grows with time (in minutes) as $r(t) = 25\sqrt{t}+2$ . Express the area of the ripple as a function of time and evaluate it at $t=2$ .
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$A(t) = \pi(25\sqrt{t}+2)^2$ and $A(2) = \pi(25\sqrt{2}+2)^2 \approx 5765$ sq in $A(t) = 25\pi t + 4$ and $A(2) = 54\pi$ $A(t) = \pi(25t+2)^2$ and $A(2) = \pi(52)^2$ $A(t) = 25\pi\sqrt{t}$ and $A(2) \approx 111$ sq in
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 93
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The radius grows as $r(t) = 25\sqrt{t}+2$ . The area is $A = \pi r^2$ , so the composition gives $A(t) = \pi(25\sqrt{t}+2)^2$ . At $t=2$ : $r = 25\sqrt{2}+2 \approx 37.36$ , $A \approx \pi(37.36)^2 \approx 4384$ sq in.
Ex. 5.21ChallengeAnswer key
The number of bacteria in a food is $N(T) = 23T^2 - 56T + 1$ (with $3 < T < 33$ °C). The temperature after $t$ hours out of the refrigerator is $T(t) = 5t + 1.5$ . Find the composite function $N(T(t))$ .
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$N(T(t)) = 23(5t+1.5)^2 - 56(5t+1.5)+1$ $N(T(t)) = 23t^2 - 56t + 1.5$ $N(T(t)) = 115t^2 - 280t + 1$ $N(T(t)) = 23(5t)^2 - 56(5t)$
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 97
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Substitute $T(t) = 5t+1.5$ into $N(T) = 23T^2 - 56T+1$ : $N(T(t)) = 23(5t+1.5)^2 - 56(5t+1.5)+1$ .
Ex. 5.22Understanding
Why is the horizontal line test effective for determining whether a function is injective (one-to-one)?
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The test checks whether any horizontal line crosses the graph more than once: if so, the function is not injective.The test checks whether the graph passes through the origin.The test checks whether any vertical line crosses the graph more than once.The test checks whether the function is surjective.
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 1
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A function is injective (one-to-one) if and only if no horizontal line crosses its graph more than once. The horizontal line test checks precisely this condition visually.
Ex. 5.23Understanding
Why do we restrict the domain of $f(x) = x^2$ to find its inverse function?
Select the correct option
Because $f(x) = x^2$ is not injective on $\mathbb{R}$ : $f(2) = f(-2) = 4$ , preventing a global inverse.Because $f(x) = x^2$ is not surjective.Because the inverse of $x^2$ would be a constant function.Because $x^2$ grows too fast.
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 2
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For a global inverse to exist, $f$ must be bijective. Since $f(2) = f(-2) = 4$ , the function is not injective on $\mathbb{R}$ . Restricting to $[0, +\infty)$ , it becomes injective and invertible, with $f^{-1}(x) = \sqrt{x}$ .
Ex. 5.24ProofAnswer key
Can a function be its own inverse? Explain and give an example.
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Yes, for example $f(x) = x$ (identity) satisfies $f^{-1} = f$ .No, a function can never be its own inverse.Yes, but only constant functions are their own inverses.Yes, but only exponential functions.
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 3
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A function is its own inverse if $f(f(x)) = x$ for all $x$ . Examples: $f(x) = x$ (identity), $f(x) = 1/x$ (for $x \neq 0$ ), and $f(x) = a - x$ for any constant $a$ .
Ex. 5.25Understanding
How do you find the inverse of a function algebraically?
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Swap $x$ and $y$ , then isolate $y$ : write $y = f(x)$ , swap to $x = f(y)$ , solve for $y$ .Take the reciprocal of $f(x)$ : $f^{-1}(x) = 1/f(x)$ .Differentiate $f(x)$ and invert the result.Reflect the graph across the $y$ -axis.
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 5
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Standard algorithm: (1) write $y = f(x)$ ; (2) swap $x \leftrightarrow y$ obtaining $x = f(y)$ ; (3) isolate $y$ — the result is $f^{-1}(x)$ . Always verify with $f(f^{-1}(x)) = x$ .
Ex. 5.26Application
Find $f^{-1}(x)$ for $f(x) = x + 3$ .
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$f^{-1}(x) = x - 3$ $f^{-1}(x) = x + 3$ $f^{-1}(x) = \frac{1}{x+3}$ $f^{-1}(x) = 3 - x$
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 7
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Swap $y = x+3$ to $x = y+3$ , isolate: $y = x-3$ . Therefore $f^{-1}(x) = x-3$ .
Ex. 5.27Application
Find $f^{-1}(x)$ for $f(x) = 2 - x$ .
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$f^{-1}(x) = 2 - x$ $f^{-1}(x) = x - 2$ $f^{-1}(x) = \frac{1}{2-x}$ $f^{-1}(x) = x + 2$
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 9
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Swap $y = 2-x$ to $x = 2-y$ , isolate: $y = 2-x$ . Therefore $f^{-1}(x) = 2-x$ — the function is its own inverse!
Ex. 5.28Application
Find $f^{-1}(x)$ for $f(x) = \frac{x}{x+2}$ .
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$f^{-1}(x) = \frac{2x}{1-x}$ $f^{-1}(x) = \frac{x}{x+2}$ $f^{-1}(x) = \frac{1-x}{2x}$ $f^{-1}(x) = \frac{x+2}{x}$
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 11
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Swap: $x = \frac{y}{y+2}$ . Solve: $x(y+2) = y$ , $xy+2x = y$ , $2x = y - xy = y(1-x)$ , therefore $y = \frac{2x}{1-x}$ .
Show step-by-step (with the why)
1. 1. Write $y = \frac{x}{x+2}$ and swap $x \leftrightarrow y$ : $x = \frac{y}{y+2}$ .
2. 2. Multiply both sides by $(y+2)$ : $x(y+2) = y$ .
3. 3. Expand: $xy + 2x = y$ .
4. 4. Isolate $y$ : $2x = y(1-x)$ , therefore $f^{-1}(x) = \frac{2x}{1-x}$ .
Ex. 5.29Application
Find $f^{-1}(x)$ for $f(x) = \frac{2x+3}{5x+4}$ .
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$f^{-1}(x) = \frac{4x-3}{5-2x}$ $f^{-1}(x) = \frac{5x+4}{2x+3}$ $f^{-1}(x) = \frac{3-4x}{2x-5}$ $f^{-1}(x) = \frac{2x+3}{5x+4}$
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 12
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Swap: $x = \frac{2y+3}{5y+4}$ . Solve: $x(5y+4) = 2y+3$ , $5xy+4x = 2y+3$ , $5xy-2y = 3-4x$ , $y(5x-2) = 3-4x$ . Therefore $f^{-1}(x) = \frac{3-4x}{5x-2} = \frac{4x-3}{2-5x}$ .
Ex. 5.30Application
Find a domain where $f(x) = (x+7)^2$ is injective and non-decreasing. Then find $f^{-1}$ restricted to that domain.
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Domain $[-7, +\infty)$ and $f^{-1}(x) = \sqrt{x} - 7$ Domain $(-\infty, 0]$ and $f^{-1}(x) = \sqrt{x+7}$ Domain $[-7, +\infty)$ and $f^{-1}(x) = \sqrt{x+7}$ Domain $[0, +\infty)$ and $f^{-1}(x) = \sqrt{x} - 7$
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 13
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Restrict to $[-7, +\infty)$ where $f$ is increasing. Swap: $x = (y+7)^2$ , isolate: $y = \sqrt{x}-7$ . Therefore $f^{-1}(x) = \sqrt{x}-7$ , domain $[0,+\infty)$ .
Ex. 5.31ApplicationAnswer key
Find a domain where $f(x) = x^2 - 5$ is injective and non-decreasing, then determine $f^{-1}$ on that domain.
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Domain $[0, +\infty)$ and $f^{-1}(x) = \sqrt{x+5}$ Domain $(-\infty, 0]$ and $f^{-1}(x) = \sqrt{x+5}$ Domain $[0, +\infty)$ and $f^{-1}(x) = \sqrt{x} + 5$ Domain $[-5, +\infty)$ and $f^{-1}(x) = \sqrt{x}-5$
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 15
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On $[0, +\infty)$ , $f(x) = x^2-5$ is increasing. Swap: $x = y^2-5$ , isolate: $y = \sqrt{x+5}$ . Therefore $f^{-1}(x) = \sqrt{x+5}$ , domain $[-5, +\infty)$ .
Show step-by-step (with the why)
1. 1. Restrict to $[0, +\infty)$ (the branch where $f$ is increasing).
2. 2. Write $y = x^2-5$ and swap: $x = y^2-5$ .
3. 3. Isolate $y$ : $y^2 = x+5$ , $y = \sqrt{x+5}$ (positive since $y \geq 0$ ).
Ex. 5.32Proof
Use function composition to verify that $f(x) = \sqrt[3]{x-1}$ and $g(x) = x^3 + 1$ are inverses of each other.
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Yes: $f(g(x)) = x$ and $g(f(x)) = x$ , therefore they are inverses.No: the compositions do not yield the identity.Yes, but only for $x \geq 0$ .No: $f(g(x)) = x$ but $g(f(x)) \neq x$ .
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 17
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$f(g(x)) = f(\sqrt[3]{x-1}) = (\sqrt[3]{x-1})^3+1 = x$ . And $g(f(x)) = g(x^3+1) = \sqrt[3]{x^3} = x$ . Both compositions are the identity, confirming they are inverses.
Ex. 5.33ProofAnswer key
Use composition to verify that $f(x) = -3x+5$ and $g(x) = \frac{x-5}{-3}$ are inverses of each other.
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Yes: both compositions yield $x$ , confirming they are inverses.No: $f(g(x)) \neq x$ .No: $g(f(x)) \neq x$ .No: neither composition gives the identity.
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 18
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$f(g(x)) = f\!\left(\frac{x-5}{-3}\right) = -3 \cdot \frac{x-5}{-3}+5 = (x-5)+5 = x$ . And $g(f(x)) = g(-3x+5) = \frac{-3x+5-5}{-3} = \frac{-3x}{-3} = x$ .
Ex. 5.34Application
The Celsius-to-Fahrenheit conversion is $f(x) = \frac{9}{5}x + 32$ . Find the inverse function and explain its meaning.
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$f^{-1}(x) = \frac{5}{9}(x-32)$ — converts Fahrenheit to Celsius $f^{-1}(x) = \frac{9}{5}(x+32)$ $f^{-1}(x) = \frac{5}{9}x - 32$ $f^{-1}(x) = \frac{9}{5}x - 32$
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 45
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Swap: $x = \frac{9}{5}y+32$ . Isolate: $x-32 = \frac{9}{5}y$ , therefore $y = \frac{5}{9}(x-32)$ . The inverse converts Fahrenheit to Celsius.
Show step-by-step (with the why)
1. 1. Start from $y = \frac{9}{5}x+32$ and swap $x \leftrightarrow y$ : $x = \frac{9}{5}y+32$ .
2. 2. Subtract 32: $x-32 = \frac{9}{5}y$ .
3. 3. Multiply by $\frac{5}{9}$ : $y = \frac{5}{9}(x-32)$ .
4. 4. Interpret: given a value in °F, the formula returns °C.
Ex. 5.35ModelingAnswer key
The circumference of a circle is $C(r) = 2\pi r$ . Express the radius as a function of the circumference, call it $r(C)$ , and compute $r(36\pi)$ .
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$r(C) = \frac{C}{2\pi}$ and $r(36\pi) = 18$ (radius as a function of circumference) $r(C) = 2\pi C$ and $r(36\pi) = 72\pi^2$ $r(C) = \frac{2\pi}{C}$ and $r(36\pi) = \frac{1}{18}$ $r(C) = \frac{C}{\pi}$ and $r(36\pi) = 36$
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 46
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From the formula $C = 2\pi r$ , isolate $r = C/(2\pi)$ . Therefore $r(C) = C/(2\pi)$ . For $C = 36\pi$ : $r = 36\pi/(2\pi) = 18$ .
Ex. 5.36ModelingAnswer key
A car at 50 mph travels $d(t) = 50t$ miles in $t$ hours. Express the time as a function of distance, call it $t(d)$ , and compute $t(180)$ .
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$t(d) = \frac{d}{50}$ and $t(180) = 3.6$ hours $t(d) = 50d$ and $t(180) = 9000$ hours $t(d) = \frac{50}{d}$ and $t(180) \approx 0.28$ hours $t(d) = d - 50$ and $t(180) = 130$ hours
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 47
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From $d = 50t$ , isolate $t = d/50$ . For $d = 180$ : $t = 180/50 = 3.6$ hours. Interpretation: it takes 3.6 hours to travel 180 miles at 50 mph.
Ex. 5.37Application
With $f(x) = \sqrt{x}$ and $g(x) = x^2 + 1$ , determine $(f \circ g)(x)$ and its domain.
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$(f \circ g)(x) = f(g(x)) = f(x^2 + 1) = \sqrt{x^2+1}$ , with domain $\mathbb{R}$ $(f \circ g)(x) = \sqrt{x^2}+1 = |x|+1$ $(f \circ g)(x) = x+1$ for $x \geq 0$ $(f \circ g)(x)$ is not defined for $x < 0$
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Solve online OpenStax College Algebra 2e · §3.4 · ex. 88
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Substituting $g(x) = x^2+1$ into $f(u) = \sqrt{u}$ : $(f \circ g)(x) = \sqrt{x^2+1}$ . Since $x^2+1 \geq 1 > 0$ for all $x$ , the domain is $\mathbb{R}$ .
Ex. 5.38Application
Find $f^{-1}(x)$ for $f(x) = x - 5$ .
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$f^{-1}(x) = x + 5$ $f^{-1}(x) = x - 5$ $f^{-1}(x) = 5 - x$ $f^{-1}(x) = \frac{1}{x+5}$
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 8
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Swap: $x = y-5$ gives $y = x+5$ . Therefore $f^{-1}(x) = x+5$ . Verification: $f(f^{-1}(x)) = (x+5)-5 = x$ .
Ex. 5.39Application
Find $f^{-1}(x)$ for $f(x) = 3 - x$ and verify that $f$ is its own inverse.
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$f^{-1}(x) = 3 - x$ $f^{-1}(x) = x - 3$ $f^{-1}(x) = x + 3$ $f^{-1}(x) = \frac{1}{3-x}$
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 10
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Swap: $x = 3-y$ , isolate: $y = 3-x$ . Therefore $f^{-1}(x) = 3-x$ , which is the function itself (self-inverse).
Ex. 5.40Challenge
Show that $f(x) = a - x$ is its own inverse for every real number $a$ .
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$f(x) = a - x$ is its own inverse for any real $a$ : $f(f(x)) = a-(a-x) = x$ . $f(x) = a - x$ is its own inverse only for $a = 0$ . $f(x) = a - x$ is never its own inverse, since $f(f(x)) = 2a - x \neq x$ . $f(x) = a - x$ is its own inverse only for $a = 1$ .
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Solve online OpenStax College Algebra 2e · §3.7 · ex. 6
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Compute $f(f(x)) = f(a-x) = a-(a-x) = a-a+x = x$ . Since $f(f(x)) = x$ for any $a$ , the function is its own inverse for all real $a$ .
Show step-by-step (with the why)
1. 1. Apply $f$ once: $f(x) = a-x$ .
2. 2. Apply $f$ again to the result: $f(a-x) = a-(a-x)$ .
3. 3. Simplify: $a-a+x = x$ . Therefore $f \circ f = \mathrm{id}$ for any $a$ .

Sources

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OpenStax College Algebra 2e — Jay Abramson et al. · 2022, 2nd ed · EN · CC-BY 4.0 · §3.4 (composition) and §5.7 (inverse). Primary source for blocks A and C.
Stitz–Zeager Precalculus — Carl Stitz, Jeff Zeager · 2013, v3 · EN · CC-BY-NC-SA · §5.1 (composition and domain) and §5.2 (inverses). Primary source for block B.
Yoshiwara — Modeling, Functions, and Graphs — Katherine Yoshiwara · 2020 · EN · free · ch. 4 (inverse in unit modeling). Primary source for block D.
Hammack — Book of Proof (3rd ed) — Richard Hammack · 2018 · EN · free · ch. 12 (composition, inverse, bijection, proofs). Primary source for block E.
Active Calculus 2.0 — Matt Boelkins · 2024 · EN · CC-BY-NC-SA · §1.5 (composition as anticipation of the chain rule).