Inverse functions — finding f⁻¹(x), horizontal line test: Question bank

This page is a rapid-fire drill of the ideas behind inverses — no heavy algebra, just the traps your brain naturally falls into. Each line is a question ::: answer reveal. Cover the right side, commit to an answer out loud, then check. If you get one wrong, follow the wikilink and re-read.

Parent: Inverse functions (Hinglish version).

True or false — justify

The claim

f^{- 1} (x) = \frac{1}{f ( x )}

is true.

False.

f^{- 1}

is the reverse machine, defined by

f (f^{- 1} (x)) = x

; the reciprocal

\frac{1}{f ( x )}

instead satisfies

f (x) \cdot \frac{1}{f ( x )} = 1

. Different operations entirely — see Function Composition.

Every function has an inverse.

False. Only one-to-one functions do. If two inputs share an output (like

f (2) = f (- 2) = 4

), the reverse machine cannot decide which input to return.

f

passes the vertical line test it must have an inverse.

False. The vertical line test only confirms

f

is a function. Invertibility needs the horizontal line test (each output used at most once).

The graph of

f^{- 1}

is the reflection of

f

across the line

y = x

True. A point

(a, b)

f

means

f (a) = b

, so

(b, a)

lies on

f^{- 1}

; swapping coordinates is exactly reflection over

y = x

f (x) = x^{2}

has no inverse, and nothing can be done about it.

False on the second half. On all of

R

it fails the horizontal line test, but restricting the domain to

x \geq 0

makes it one-to-one, giving

f^{- 1} (x) = x

f

is one-to-one on its domain, then

f^{- 1}

is also one-to-one.

True. The reflection over

y = x

preserves the "at most one hit per horizontal line" property (horizontal and vertical lines swap under reflection), so the inverse is itself invertible — its inverse is

f

The domain of

f^{- 1}

equals the domain of

f

False. The domain of

f^{- 1}

equals the range of

f

; input and output roles swap. See Domain and Range.

Every strictly increasing function is one-to-one.

True. If

x_{1} < x_{2}

forces

f (x_{1}) < f (x_{2})

, then unequal inputs always give unequal outputs, so no horizontal line is hit twice.

A function that has an inverse must be a bijection between its domain and range.

True. It must be one-to-one (injective) so the reverse is unambiguous, and onto its range (surjective onto that range) so every output has a source — that's a bijection.

e^{x}

and

ln x

are inverses of each other on their natural domains.

True.

ln (e^{x}) = x

for all real

x

and

e^{l n x} = x

for

x > 0

; this is the classic pair in Exponential and Logarithmic Functions.

Spot the error

f (x) = x - 2

inverts to

f^{- 1} (x) = x^{2} + 2

for all real

x

The algebra is right but the domain is missing. Since

f

outputs only

y \geq 0

, the inverse's domain is

x \geq 0

; so

f^{- 1} (x) = x^{2} + 2

only for $x \geq 0$ .

f (x) = x^{2}

inverts to

f^{- 1} (x) = \pm x

The

\pm

gives two outputs for one input, so it is not a function at all. The real fix is to restrict

f

's domain first, then take a single branch.

"Swap

x

and

y

y = 2 x + 3

: I get

y = 2 x - 3

That's not swapping, that's negating a constant. Swapping means writing

x = 2 y + 3

, then solving to get

y = \frac{x - 3}{2}

"For

f (x) = \frac{2 x + 1}{x - 3}

the inverse keeps the restriction

x \neq = 3

The restriction moves. The original range excludes

2

(horizontal asymptote), so the inverse's domain excludes

2

f^{- 1} (x) = \frac{3 x + 1}{x - 2}, x \neq = 2

"I found

g (x)

with

f (g (x)) = x

, so

g

is the inverse of

f

Not yet — you must also check

g (f (x)) = x

. A one-sided identity can hold on a restricted set without

g

being the full two-sided inverse.

sin x

has an inverse, so

arcsin (sin (4)) = 4

4

is outside the restricted range

[- \frac{π}{2}, \frac{π}{2}]

where

sin

was made one-to-one, so

arcsin (sin 4) \neq = 4

. Inverse trig only rewinds inputs inside the restricted domain — see Trigonometric Functions.

"To invert

f (x) = 3 x - 7

I take

\frac{1}{3 x - 7}

That is the reciprocal, giving

1

under multiplication, not

x

under composition. The true inverse is

f^{- 1} (x) = \frac{x + 7}{3}

Why questions

Why does the horizontal line test detect invertibility, rather than the vertical line test?

A horizontal line

y = c

collects all inputs that share the output

c

; two intersections mean two inputs map to one output, breaking one-to-one and making reversal ambiguous.

Why do we bother restricting the domain instead of just accepting

\pm x

Because a function must return exactly one output per input. Restricting removes the duplicate branch so the reverse machine gives a single, unambiguous answer.

Why is the swap method (

x \leftrightarrow y

) legitimate and not just a trick?

(a, b)

lies on

f

then

(b, a)

lies on

f^{- 1}

; swapping the letters in the equation literally exchanges the input and output roles, so solving the swapped equation is solving

f (y) = x

Why must the domain and range trade places for

f^{- 1}

The inverse takes outputs of

f

as its inputs and returns the original inputs, so whatever set

f

produced (its range) becomes what

f^{- 1}

consumes (its domain), and vice versa.

Why does verifying with composition catch algebra slips that just "reversing steps" might miss?

Composition tests the definition directly —

f (f^{- 1} (x))

must collapse to exactly

x

. Any sign error or dropped term leaves a residue that won't simplify to

x

, exposing the mistake.

Why is a strictly monotonic function guaranteed invertible on its domain?

Strict monotonicity means the output never repeats a value as the input moves, so every horizontal line meets the graph at most once — the exact condition for one-to-one.

Edge cases

Does the constant function

f (x) = 5

have an inverse?

No. Every input maps to

5

, so infinitely many inputs share one output — the line

y = 5

hits the graph everywhere, failing the horizontal line test completely.

f (x) = x

(the identity) its own inverse?

Yes. It sends every input to itself, so reversing does nothing;

f (f (x)) = x

trivially, and its reflection over

y = x

is itself.

Can a function equal its own inverse without being the identity?

Yes — such functions are called self-inverse (involutions). Example:

f (x) = \frac{1}{x}

satisfies

f (f (x)) = x

, yet it is not the identity.

What happens to

f^{- 1}

at an

x

-value that was excluded from

f

's range, like

x = 2

for

f (x) = \frac{2 x + 1}{x - 3}

It is undefined there. The excluded output of

f

becomes an excluded input of

f^{- 1}

, showing up as a denominator zero in

f^{- 1} (x) = \frac{3 x + 1}{x - 2}

f (x) = 3 x + 5 - 2

invertible without any domain restriction?

Yes. Cube root is defined and strictly increasing on all of

R

, so it passes the horizontal line test everywhere; both

f

and

f^{- 1} (x) = (x + 2)^{3} - 5

have domain and range

R

f

is only defined on a single point, say domain

{3}

with

f (3) = 8

, does it have an inverse?

Yes, a trivial one:

f^{- 1}

has domain

{8}

with

f^{- 1} (8) = 3

. One-to-one is satisfied vacuously since there's only one input.

Recall Quick self-test

Statement: "Passing the vertical line test guarantees an inverse." ::: False — that only guarantees $f$ is a function; you need the horizontal line test for invertibility. Statement: " $f^{- 1} (x)$ and $1/ f (x)$ are the same." ::: False — the first undoes $f$ by composition, the second is a reciprocal under multiplication. Statement: "Domain of $f$ = range of $f^{- 1}$ ." ::: True — input/output roles swap, so $f$ 's domain becomes $f^{- 1}$ 's range.