Inverse Functions
Introduced| English | Meaning |
|---|---|
| inverse/ɪnˈvɜːs/ | A function that reverses another function's mapping on a suitable one-to-one domain. |
| domain/dəˈmeɪn/ | The set of input values for which a function is defined. |
| range/reɪndʒ/ | The set of output values actually attained by a function on its stated domain. |
| one-to-one/wʌn tuː wʌn/ | Describing a function in which different inputs always have different outputs. |
| reflection/rɪˈflekʃn/ | A transformation that maps each point to its mirror image across a line or plane. |
Recover the input from the output
- A bike hire costs 12 yuan plus 3 yuan per hour, for up to 4 hours. The cost function is $C(t)=12+3t$, with $0\le t\le4$. If the cost is 21 yuan, solving $12+3t=21$ recovers a hire time of 3 hours.
- An inverse 反函数 reverses a function's input-output pairs. A function needs a suitable domain 定义域 before this reverse mapping is also a function.
Exchange inputs and outputs
- If $f(2)=5$, then $f^{-1}(5)=2$, provided an inverse function exists on the stated domain. The original output becomes the inverse's input.
- The domain of $f$ becomes the range 值域 of $f^{-1}$, and the range of $f$ becomes the domain of $f^{-1}$. Keep both sets with the inverse rule.
The inverse of a function…
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If $f$ sends $2 \to 5$, then $f^{-1}$ sends $5 \to 2$: the inverse undoes what the function did.
Check that the function is one-to-one
- A function is one-to-one 一一对应 if each output in its range comes from just one input. The horizontal-line test checks this: no horizontal line may meet the graph more than once.
- On all real inputs, $f(x)=x^2$ fails: both $f(-2)$ and $f(2)$ equal 4. Reversing that pair would require one input, 4, to give two outputs. The reverse relation exists, but it is not an inverse function.
A function has an inverse function only if it is ____ (each output comes from exactly one input).
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If two inputs share an output, the reverse map is ambiguous — so only a one-to-one function is invertible.
Restrict the domain before choosing a branch
- For $f(x)=x^2$ restricted to $x\ge0$, the range is $y\ge0$ and $f^{-1}(x)=\sqrt x$. Both the inverse's domain and range are $[0,\infty)$.
- A different restriction, $x\le0$, gives $f^{-1}(x)=-\sqrt x$, with domain $[0,\infty)$ and range $(-\infty,0]$. At inverse input 9, these branches return 3 and $-3$, respectively. The original domain decides which answer is valid.
The symbol $\sqrt x$ means the nonnegative square root. Writing $\pm\sqrt x$ gives two values for positive $x$, so it does not define a single inverse function. State the original restriction first.
To find the rule for $f^{-1}$, you…
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Swapping $x$ and $y$ reverses the roles of input and output; solving for $y$ gives the inverse rule.
Which restriction makes $f(x)=x^2$ have inverse $f^{-1}(x)=\sqrt x$?
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The nonnegative branch maps nonnegative inputs one-to-one onto nonnegative outputs.
For $f(x)=x^2$ restricted to $x\le0$, find $f^{-1}(9)$.
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The original input must be nonpositive. Its square is 9, so choose −3, not 3.
Find an inverse rule
- For $f(x)=2x+1$ on all real inputs, write $y=2x+1$, swap the variables to get $x=2y+1$, then solve: $f^{-1}(x)=\dfrac{x-1}{2}$. Both functions have all real numbers as their domain and range.
- This also reverses the operations: $f$ doubles, then adds 1; its inverse subtracts 1, then halves. Reverse both the operations and their order.
For $f(x)=2x+1$ on all real inputs, which is its inverse rule?
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Undo the final +1 first, then undo multiplication by 2.
Verify both composition identities
- For this line, $f^{-1}(f(x))=\dfrac{(2x+1)-1}{2}=x$ and $f(f^{-1}(x))=2\left(\dfrac{x-1}{2}\right)+1=x$. These identities hold for every real input, so they check the whole rules rather than one pair.
- In general, $f^{-1}(f(x))=x$ uses $x$ in the domain of $f$, while $f(f^{-1}(y))=y$ uses $y$ in its range. Restrictions matter: $\sqrt{x^2}=|x|$, which equals $x$ only when $x\ge0$.
Select all true statements about inverse functions.
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Not every function is invertible — it must be one-to-one first. The other three are correct.
The identity $\sqrt{x^2}=x$ holds for every real $x$.
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For x=−2, the left side is 2, not −2. In general √(x²)=|x|; restrict x≥0 for equality to x.
Reflect across the identity line
- Swapping $(a,b)$ to $(b,a)$ gives a reflection 反射 across $y=x$. For $f(x)=2x+1$, the point $(2,5)$ becomes $(5,2)$ on its inverse. Use equal scales on the two axes to see the reflection accurately.
- The swap also exchanges intercepts and domains with ranges. For a restricted square, reflect only the chosen branch of the parabola; reflecting the full parabola gives a relation that fails the vertical-line test.
A function and its inverse exchange coordinates
Start with f(2)=5 and f⁻¹(5)=2. Compare the line with the two different restricted square branches.
The graph of $f^{-1}$ is the reflection of the graph of $f$ across the x-axis.
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False. Swapping every $(a, b)$ to $(b, a)$ reflects the graph across the line $y = x$, not the x-axis.
Distinguish inverse from reciprocal
- The superscript in $f^{-1}$ names the inverse function. It does not instruct you to divide 1 by the value of $f$.
- For $f(x)=2x+1$, the inverse at 5 is $f^{-1}(5)=2$, whereas the reciprocal at 5 is $\dfrac{1}{f(5)}=\dfrac{1}{11}$. Check which operation the notation requests.
Read an inverse from a table or graph
- Suppose a one-to-one function has the pairs $(1,5)$, $(4,11)$ and $(7,17)$. Its inverse has $(5,1)$, $(11,4)$ and $(17,7)$, so $f^{-1}(11)=4$. On a graph, locate output 11 and read its corresponding input.
- A few different outputs in a table do not prove that an unstated continuous function is one-to-one everywhere. Use the given domain or graph to justify that a full inverse function exists.
A one-to-one function has $f(1)=5$, $f(4)=11$ and $f(7)=17$. Find $f^{-1}(11)$.
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Reverse the given pair (4,11) to get (11,4).
Keep the context with the inverse
- For the bike hire, $C^{-1}(p)=\dfrac{p-12}{3}$ returns hours from a price $p$ in yuan. Since $0\le t\le4$, the cost ranges from 12 to 24 yuan: the inverse domain is $[12,24]$ and its range is $[0,4]$.
- The formula gives $-1$ at price 9, but that is outside this context. This model assumes time can vary continuously within 4 hours; a charge rounded up to whole hours would need a different rule and may not be one-to-one.
The model $C(t)=12+3t$ is defined for $0\le t\le4$. What is the domain of its inverse?
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The original range, from C(0)=12 to C(4)=24, becomes the inverse domain.
For the bike-hire model with $0\le t\le4$, substituting price 9 into $(p-12)/3$ gives a valid hire time.
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The result −1 hour lies outside the original domain. Price 9 is outside the inverse domain [12,24].
Exponentials and logarithms undo each other
- The function $f(x)=2^x$ is one-to-one on all real inputs and has range $(0,\infty)$. Its inverse is $\log_2 x$, whose domain is $(0,\infty)$ and range is all real numbers.
- For example, $2^3=8$ becomes $\log_2 8=3$. The animation below shows an exponential and its matching logarithm reflected across $y=x$.
Check that the function is one-to-one on its stated domain. Reverse input-output pairs, exchange domain and range, and check both compositions on the appropriate sets. A domain restriction determines the inverse branch and any context limits.
For $f(x)=2^x$ on all real inputs, what is the domain of $f^{-1}(x)=\log_2 x$?
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An exponential has strictly positive outputs. Its inverse therefore accepts positive inputs only, excluding zero.
Explain the recovered input
- Let $f(x)=3x-6$ on all real inputs. Find $f^{-1}(9)$ and explain what the result represents.
- Derive the inverse rule and verify it by composing in both orders. Explain why $f(9)$ and $1/f(9)$ answer different questions.
For $f(x)=3x-6$ on all real inputs, find $f^{-1}(9)$, derive the inverse rule and check both compositions. Distinguish this from $f(9)$ and $1/f(9)$.
15 divided by 3 is 5.
For $f(x)=3x-6$ on all real inputs, what is $f^{-1}(9)$?
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Solving for the input gives 5.