Worked examples — Inverse functions — finding f⁻¹(x), horizontal line test

This is the "throw everything at it" page for the parent topic. We are going to build a matrix of every kind of situation the topic can put in front of you, then grind through worked examples until every single box is ticked. No box left blank means no surprise on the exam.

If a word here is new — "one-to-one", "domain", "swap method" — the parent note builds those first. Here we use them, on hard cases.

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The scenario matrix

Think of a function-inverse problem as a machine you must reverse. The trouble always comes from the type of machine. Here is the full list of machine-types this topic hides, and the one worry each one carries.

Cell	Case class	The worry it hides	Example that hits it
A	Straight-line machine ($f(x)=mx+c$)	slope sign flips the graph, but inverse still exists	Ex 1
B	Negative slope / decreasing	does "decreasing" break invertibility? (no)	Ex 2
C	Even power on all of $\mathbb{R}$	fails the line test — must restrict	Ex 3
D	Even power, but only the negative branch	inverse picks the negative root, sign trap	Ex 4
E	Fraction machine (rational)	domain hole moves when roles swap	Ex 5
F	Root machine with a shift	hidden domain/range restrictions	Ex 6
G	Exponential ↔ log pair	the "undo" is a different letter entirely	Ex 7
H	Degenerate / constant machine	zero information out — no inverse at all	Ex 8
I	Real-world word problem (units!)	inverse must make physical sense	Ex 9
J	Exam twist: self-inverse ($f=f^{-1}$)	graph is its own reflection	Ex 10

We now walk every cell. Watch the figure for the geometric ones — the reflection across the line $y=x$ is the whole story.

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Cell A — Straight-line machine

Forecast: guess before reading — will the inverse also be a straight line? Steeper or shallower?

1. Write $y=4x-9$. Why this step? We name the output $y$ so we have two letters to swap.
2. Swap $x\leftrightarrow y$: $x=4y-9$. Why this step? Swapping literally exchanges the input and output roles — the definition of "undo".
3. Solve for $y$: $x+9=4y\Rightarrow y=\frac{x+9}{4}$. Why this step? We isolate the new output to read off the machine.
4. $f^{-1}(x)=\frac{x+9}{4}$.

Verify: $f(f^{-1}(x))=4\cdot \frac{x+9}{4}-9=(x+9)-9=x$ ✓. The slope became $\frac{1}{4}$ — reciprocal of $4$ — so yes, the inverse is a shallower line. Steepness of $f$ becomes flatness of $f^{-1}$ because the reflection swaps rise and run.

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Cell B — Negative slope (decreasing)

Forecast: a decreasing line — one-to-one or not?

1. Line test check: every horizontal line hits a straight line exactly once (unless the line is horizontal). Slope $-\frac{1}{2}\ne 0$, so it passes. One-to-one confirmed. Why this step? Mistake 3 in the parent: never invert before checking.
2. Write $y=-\frac{1}{2}x+6$.
3. Swap: $x=-\frac{1}{2}y+6$.
4. Solve: $x-6=-\frac{1}{2}y\Rightarrow y=-2(x-6)=-2x+12$. Why this step? Multiply by $-2$ to undo the $-\frac{1}{2}$.
5. $f^{-1}(x)=-2x+12$.

Verify: $f(f^{-1}(x))=-\frac{1}{2}(-2x+12)+6=(x-6)+6=x$ ✓. Both slopes are negative — decreasing functions are perfectly invertible; the reflection of a downhill line is still a single-valued line.

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Cell C — Even power on all of $\mathbb{R}$

Forecast: will one horizontal line hit it twice?

1. Line test: set $y=1$. Then $x^2-3=1\Rightarrow x^2=4\Rightarrow x=2$ and $x=-2$. Two hits → fails. Why this step? Two inputs, one output = not one-to-one = no inverse.
2. Fix — restrict domain to $x\ge 0$ (keep only the right half of the parabola). Why this step? Chopping off the left branch removes the duplicate.
3. On $x\ge 0$: $y=x^2-3\Rightarrow x=y^2-3\Rightarrow y^2=x+3\Rightarrow y=\sqrt{x+3}$ (positive root, since range $y\ge 0$).
4. $f^{-1}(x)=\sqrt{x+3}$, with domain $x\ge -3$ (the original range).

Verify: take $x=2$ in $f$: $f(2)=4-3=1$. Then $f^{-1}(1)=\sqrt{1+3}=2$ ✓. See Domain and Range: original range $[-3,\infty )$ becomes inverse domain $[-3,\infty )$.

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Cell D — Even power, the negative branch

Forecast: same formula as Ex 3, or does a sign flip?

1. The right branch is gone; we keep $x\le 0$. Line test now passes (left half alone hits each level once).
2. $y=x^2-3\Rightarrow y^2=x+3$. Solving $y=\pm \sqrt{x+3}$.
3. Choose the sign from the range. Here $x\le 0$, so the output $y$ of $f^{-1}$ must be $\le 0$. Pick the negative root. Why this step? $f^{-1}$'s outputs must land back in the original domain $x\le 0$.
4. $f^{-1}(x)=-\sqrt{x+3}$, domain $x\ge -3$.

Verify: take $x=-2$: $f(-2)=4-3=1$; $f^{-1}(1)=-\sqrt{4}=-2$ ✓. Same parabola, opposite branch → opposite sign in the inverse. This is exactly the trap: the algebra gives $\pm$, the domain picks which.

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Cell E — Fraction machine (moving hole)

Forecast: where will the new forbidden value be?

1. $y=\frac{x+4}{x-1}$.
2. Swap: $x=\frac{y+4}{y-1}$.
3. Clear the fraction: $x(y-1)=y+4\Rightarrow xy-x=y+4$. Why this step? Multiply out so $y$ leaves the denominator.
4. Collect $y$: $xy-y=x+4\Rightarrow y(x-1)=x+4\Rightarrow y=\frac{x+4}{x-1}$. Why this step? Factor $y$ so we can isolate it.
5. $f^{-1}(x)=\frac{x+4}{x-1}$, $x\ne 1$.

Verify: interestingly $f^{-1}=f$ here (see Cell J too!). Check $f(2)=\frac{6}{1}=6$, then $f^{-1}(6)=\frac{10}{5}=2$ ✓. The horizontal asymptote of $f$ is $y=1$ (never output $1$), which becomes the inverse's forbidden input $x=1$. The hole moved from domain to range to domain of inverse — as Mistake 2 warns.

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Cell F — Root machine with hidden restrictions

Forecast: the answer will be a parabola — but on which half?

1. Original domain/range: need $x-2\ge 0\Rightarrow x\ge 2$. Since $\sqrt{\cdot }\ge 0$, outputs satisfy $y\ge 5$. So range is $[5,\infty )$. Why this step? These become the inverse's range/domain — write them down first.
2. $y=\sqrt{x-2}+5\Rightarrow x=\sqrt{y-2}+5$.
3. Isolate the root: $x-5=\sqrt{y-2}$, valid only when $x-5\ge 0$.
4. Square: $(x-5{)}^2=y-2\Rightarrow y=(x-5{)}^2+2$. Why this step? Squaring undoes the square root — but only the branch where $x-5\ge 0$.
5. $f^{-1}(x)=(x-5{)}^2+2$, domain $x\ge 5$.

Verify: $f(6)=\sqrt{4}+5=7$; $f^{-1}(7)=(2{)}^2+2=6$ ✓. Without the "$x\ge 5$" the parabola would fail the line test — the restriction is not optional.

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Cell G — Exponential ↔ log pair

Forecast: what operation undoes "$3$ to the power"?

1. $y=2\cdot 3^x-5$.
2. Swap: $x=2\cdot 3^y-5$.
3. Isolate the power: $x+5=2\cdot 3^y\Rightarrow 3^y=\frac{x+5}{2}$.
4. Take ${\log }_3$ of both sides — this is the tool that answers "to what power must I raise 3?": $y={\log }_3\left(\frac{x+5}{2}\right).$ Why this tool? A logarithm is defined as the inverse of an exponential; nothing algebraic can pull $y$ out of an exponent, only a log can. See Exponential and Logarithmic Functions.
5. $f^{-1}(x)={\log }_3\left(\frac{x+5}{2}\right)$, valid when $\frac{x+5}{2}>0\Rightarrow x>-5$.

Verify: $f(0)=2\cdot 1-5=-3$; $f^{-1}(-3)={\log }_3\left(\frac{2}{2}\right)={\log }_{3}1=0$ ✓. Domain $x>-5$ matches the horizontal asymptote $y=-5$ of the original exponential.

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Cell H — Degenerate / constant machine

Forecast: trust your gut — reversible or hopeless?

1. Line test: the graph is the horizontal line $y=7$. The horizontal line $y=7$ overlaps it at infinitely many points. Massive failure. Why this step? Every input maps to $7$, so $f^{-1}(7)$ would have to be every real number at once.
2. Conclusion: no inverse exists, and no domain restriction can save it — restricting the domain still leaves every point mapping to $7$.

Verify: try $f(1)=7$ and $f(2)=7$. Different inputs, same output ⇒ not one-to-one ⇒ not a bijection ⇒ no inverse. This is the degenerate corner of the matrix: zero information out means zero recovery.

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Cell I — Real-world word problem (units matter)

Forecast: what will the inverse mean physically?

1. $y=\frac{9}{5}c+32$ (here $y$ is Fahrenheit, $c$ Celsius).
2. Swap: $c=\frac{9}{5}y+32$ — treat $c$ now as the known Fahrenheit reading.
3. Solve: $c-32=\frac{9}{5}y\Rightarrow y=\frac{5}{9}(c-32)$. Why this step? Multiply by $\frac{5}{9}$ to undo the $\frac{9}{5}$.
4. $F^{-1}(c)=\frac{5}{9}(c-32)$: input Fahrenheit, output Celsius.

Verify: $F(100)=\frac{9}{5}(100)+32=212$°F (boiling water ✓). Then $F^{-1}(212)=\frac{5}{9}(212-32)=\frac{5}{9}(180)=100$°C ✓. Units confirm the inverse is a genuine "reverse converter", not the reciprocal $\frac{1}{F(c)}$ (Mistake 1).

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Cell J — Exam twist: self-inverse

Forecast: what would the graph look like if a function equals its own inverse?

1. $y=\frac{6}{x}$.
2. Swap: $x=\frac{6}{y}$.
3. Solve: $xy=6\Rightarrow y=\frac{6}{x}$. Why this step? Same equation returns — that is the definition of self-inverse.
4. So $f^{-1}(x)=\frac{6}{x}=f(x)$.

Verify (composition): $f(f(x))=\frac{6}{6/x}=\frac{6x}{6}=x$ ✓ for all $x\ne 0$. Geometry: the graph is symmetric across $y=x$, so reflecting it leaves it unchanged — that is precisely what "own inverse" looks like. Concrete check: $f(3)=2$ and $f(2)=3$, the pair $(3,2)$ and $(2,3)$ mirror across the diagonal.

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Recall Quick self-test on the matrix

Which cell forces you to pick between $+\sqrt{}$ and $-\sqrt{}$ using the domain? ::: Cells C and D (even powers on a restricted branch). Which case has NO inverse no matter what you do? ::: Cell H, the constant function. What tool is the only way to free a variable from an exponent? ::: The logarithm (Cell G). When $f=f^{-1}$, what does the graph do under reflection across $y=x$? ::: Nothing — it maps onto itself.