Single-Variable-Calculus · Unit 5 · Video 4 · Interactive Practice

Inverse Functions: Swap x and y, Reflect Across y = x

IKey Formulas

Formula	Name	What it says
$y = f(x), \quad g(y) = x \ \Longrightarrow\ g = f^{-1}$	Definition of the inverse	Each $y$ must come from just one $x$
$f^{-1}\big(f(x)\big) = x, \qquad f\big(f^{-1}(x)\big) = x$	Undoing, both ways	$\big(\sqrt{x}\,\big)^{2} = x$ for every $x \ge 0$
$(a, b) \text{ on } f \iff (b, a) \text{ on } f^{-1}$	Reflection across $y = x$	The graph of $f^{-1}$ is the mirror image
$\dfrac{d}{dy}\big(f^{-1}(y)\big) = \dfrac{1}{\;dy/dx\;}$	Derivative of the inverse	At the matching point $y = f(x)$ , where $dy/dx \ne 0$

Key Insight: The $-1$ in $f^{-1}$ names the inverse, never a power: for $f(x) = \sqrt{x}$ the inverse is $x^{2}$ while the reciprocal $1/f(x)$ is $1/\sqrt{x}$ . And an inverse exists only on a piece where $f$ takes each value once — the right half of the parabola, or the tangent's branch $-\tfrac{\pi}{2} < x < \tfrac{\pi}{2}$ .

IIEvery Point Has a Partner

The point $(4, 2)$ on $y = \sqrt{x}$ has a twin $(2, 4)$ on $y = x^{2}$ .

Point on the curve: x 4

IIIWhy Only Half the Parabola

Squaring sends $2$ and $-2$ to the same $4$ — which half of it can be undone?

Height c 4 Domain

💡 The tangent needs the same surgery: keep the branch $-\tfrac{\pi}{2} < x < \tfrac{\pi}{2}$ , where it takes every value exactly once.

IVThe Tangent Becomes the Arctangent

Reflection turns the tangent's vertical asymptotes into the arctangent's horizontal ones.

Reflection 0.00

💡 The same equation $\tan y = x$ also yields the slope: implicit differentiation turns it into $\dfrac{d}{dx}\arctan x = \dfrac{1}{1 + x^{2}}$ , the subject of the next video.

VQuiz Questions

Problem 1 · Reflecting a Point

Given: the point $(9, 3)$ on the graph of $f(x) = \sqrt{x}$ — find the point that must lie on the graph of $f^{-1}$ .

✅ Correct! Reflecting across

y = x

trades

(a, b)

for

(b, a)

, and the partner curve is

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

with

3^{2} = 9

❌ That is the reciprocal, not the inverse.

1/f(9) = 1/3

, but

f^{-1}

undoes

f

: it must send

3

back to

9

❌ That reflects through the origin. Reflecting across the diagonal swaps the two coordinates and never changes their signs — and

\sqrt{x}

has no negative outputs to reflect.

❌ That reflects across the $x$ -axis. The mirror line here is the diagonal

y = x

, so the first coordinate becomes the second.

Show solution

The point $(9, 3)$ on $y = \sqrt{x}$ records $f(9) = 3$ . The inverse reverses that arrow:

f(9) = 3 \quad \Longleftrightarrow \quad f^{-1}(3) = 9

and $f^{-1}(3) = 9$ is exactly the statement that $(3, 9)$ lies on the graph of $f^{-1}$ . Geometrically the two points are mirror images across $y = x$ : their midpoint $\left(6, 6\right)$ sits on the diagonal, and the segment joining them has slope $-1$ , perpendicular to it.

Here $f^{-1}(x) = x^{2}$ , taken on $x \ge 0$ — the range of $\sqrt{x}$ — and indeed $3^{2} = 9$ .

Problem 2 · One Output, Two Inputs

Given: $f(x) = x^{2}$ with domain all of $\mathbb{R}$ — choose the true statement.

✅ Correct! Reflecting the whole parabola gives the sideways parabola

y^{2} = x

, with two heights

\pm 2

above

x = 4

— not a graph of a function. Restrict to

x \ge 0

first, and the inverse is

\sqrt{x}

❌ $\pm\sqrt{x}$ is not a function. It assigns two outputs to each

x > 0

; its graph is the whole sideways parabola

y^{2} = x

, which fails the vertical line test.

❌ It fails on the negative half.

\sqrt{(-2)^{2}} = \sqrt{4} = 2 \ne -2

: squaring is undone by

\sqrt{\;}

only after the domain is cut down to

x \ge 0

❌ That is the reciprocal $1/f(x)$ . The superscript

-1

names the inverse function, not a power.

Show solution

An inverse exists only where the function takes each value once. Squaring does not:

2^{2} = 4, \qquad (-2)^{2} = 4

so a function undoing it would have to send $4$ to both $2$ and $-2$ , and a function gives one output per input. The picture says the same thing: reflecting $y = x^{2}$ across $y = x$ produces

y^{2} = x \quad \Longrightarrow \quad y = \pm\sqrt{x}

a parabola lying on its side, cut twice by the vertical line $x = 4$ — at $(4, 2)$ and $(4, -2)$ .

The repair is to restrict squaring to $x \ge 0$ , the half the square root reaches. On that half $x^{2}$ takes each value once, its reflection is the single curve $y = \sqrt{x}$ , and

\sqrt{x^{2}} = x \ \ (x \ge 0), \qquad \big(\sqrt{x}\,\big)^{2} = x \ \ (x \ge 0)

Problem 3 · The Branch You Keep

Given: $\arctan x$ is the angle $y$ with $\tan y = x$ and $-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}$ — evaluate $\arctan\!\left(\tan\tfrac{3\pi}{4}\right)$ , and then $\lim\limits_{x \to \infty} \arctan x$ .

The composite value

The limit at infinity

✅ Correct!

\tan\tfrac{3\pi}{4} = -1

and

\arctan(-1) = -\tfrac{\pi}{4}

; every value of

\arctan

stays strictly inside

\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)

, which is why the graph levels off at the height

\tfrac{\pi}{2}

❌ The arctangent does not return the angle you started with.

\tfrac{3\pi}{4}

lies outside the kept branch, so the composite cannot give it back:

\tan\tfrac{3\pi}{4} = -1

, and the angle in

\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)

with tangent

-1

-\tfrac{\pi}{4}

❌ Work from the inside out. First

\tan\tfrac{3\pi}{4} = -1

; then find the angle in

\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)

whose tangent is

-1

— a negative angle, since the tangent there is negative only to the left of

0

❌ Read the asymptote off the mirror. The tangent's vertical asymptote

x = \tfrac{\pi}{2}

reflects into the horizontal line

y = \tfrac{\pi}{2}

, and the arctangent climbs toward that height without ever reaching it.

Show solution

The composite. The angle $\tfrac{3\pi}{4}$ is not on the branch we kept, so evaluate the inside first:

\tan\frac{3\pi}{4} = -1

Now $\arctan(-1)$ asks for the angle $y$ with $\tan y = -1$ and $-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}$ . That angle is $-\tfrac{\pi}{4}$ , since $\tan\left(-\tfrac{\pi}{4}\right) = -1$ :

\arctan\!\left(\tan\frac{3\pi}{4}\right) = \arctan(-1) = -\frac{\pi}{4}

So $\arctan(\tan y) = y$ only for $y$ on the kept branch — the same caution as $\tan\pi = 0$ with $\arctan 0 = 0 \ne \pi$ .

The limit. On the kept branch the tangent climbs to $+\infty$ as $x \to \tfrac{\pi}{2}^{-}$ , so its graph has the vertical asymptote $x = \tfrac{\pi}{2}$ . Reflection across $y = x$ turns that vertical line into the horizontal line $y = \tfrac{\pi}{2}$ , giving

\lim_{x \to \infty} \arctan x = \frac{\pi}{2}, \qquad \lim_{x \to -\infty} \arctan x = -\frac{\pi}{2}

The arctangent is defined for every $x$ , and its values stay strictly between $-\tfrac{\pi}{2}$ and $\tfrac{\pi}{2}$ — it approaches these heights and never reaches them.

Problem 4 · Asymptotes in the Mirror

Given: $f$ is increasing on all of $\mathbb{R}$ and $f(x) \to 3$ as $x \to \infty$ , so its graph has the horizontal asymptote $y = 3$ — find the matching feature of the graph of $f^{-1}$ .

✅ Correct! Reflection across

y = x

sends the line

y = 3

to the line

x = 3

— the same trade that turned the tangent's asymptotes

x = \pm\tfrac{\pi}{2}

into the arctangent's

y = \pm\tfrac{\pi}{2}

, read backwards.

❌ The asymptote reflects too. Swapping coordinates swaps the two kinds of line: the mirror image of

y = 3

x = 3

, not

y = 3

❌ The inverse is not the reciprocal. Nothing in the reflection turns a

3

into a

\tfrac{1}{3}

; the coordinates are exchanged, not inverted.

❌ Reflection keeps every feature, in mirrored form. The line

y = 3

has a mirror image, namely

x = 3

, and the reflected graph runs along it.

Show solution

Each point of the graph of $f$ reflects by the rule $(a, b) \mapsto (b, a)$ , so take a point far out to the right:

\big(x, f(x)\big) \ \longmapsto \ \big(f(x), x\big)

As $x \to \infty$ the height $f(x)$ climbs toward $3$ without reaching it. In the reflected picture that height becomes the horizontal position: the reflected point sits just to the left of $3$ while its height $x$ runs off to $+\infty$ . That is a vertical asymptote at $x = 3$ .

Equivalently, $f$ is increasing with values approaching $3$ , so $f^{-1}(t) \to \infty$ as $t \to 3^{-}$ :

\lim_{t \to 3^{-}} f^{-1}(t) = \infty

This is the arctangent story in reverse. There $\tan x$ had the vertical asymptote $x = \tfrac{\pi}{2}$ and its inverse gained the horizontal asymptote $y = \tfrac{\pi}{2}$ ; reflection always trades one kind for the other, because it trades the two coordinates.

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