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When a function can be undone, how to undo it, and what restricting the domain buys.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to test a function for injectivity either by deriving equal inputs from equal outputs or by producing one counterexample, say which of injectivity and surjectivity each half of the inverse's definition needs, find an inverse by solving and check it by composing, restrict a domain to make a function invertible and say which piece you kept, and argue that an inverse exists for a strictly monotone function without writing one down.
$f : A \to B$ is
A function has an inverse $f^{-1} : B \to A$ exactly when it is bijective, and then $f^{-1}(f(x)) = x$ and $f(f^{-1}(y)) = y$.
The two conditions do different jobs, and it is worth being able to say which is which. Injectivity is what makes $f^{-1}$ well defined — without it, $f^{-1}(y)$ would have to name two different inputs at once. Surjectivity is what makes it defined everywhere on $B$ — without it, $f^{-1}$ has nothing to say about the parts of $B$ that $f$ never reaches.
And surjectivity is the cheap one: shrinking the codomain to the image makes any function surjective, and costs nothing anyone cares about. Injectivity is the real condition, which is why every technique in this lesson is about producing it.
$f^{-1}$ is not $1/f$. The superscript is notation for the inverse function. $\sin^{-1}$ is not $\csc$. This collision is universal and there is nothing to do about it but notice it.
Another way: picture
The graph of $f^{-1}$ is the graph of $f$ reflected in the line $y = x$: swapping input and output swaps the coordinates. Injectivity is the horizontal line test — no horizontal line meets the graph twice — which becomes the vertical line test after the reflection, which is the condition for the reflection to be a graph at all.
Another way: steps
To find an inverse:
Chasing $f(a) = f(b) \Rightarrow a = b$ through algebra works for simple formulas and stalls quickly. The test that scales is strict monotonicity: a function that is strictly increasing (or strictly decreasing) on an interval is injective there, immediately, because $a < b$ gives $f(a) < f(b)$ and in particular $f(a) \ne f(b)$.
That matters because monotonicity is something the derivative can decide, three units from now: $f' > 0$ on an interval makes $f$ strictly increasing there, so $f$ is injective there, so it has an inverse there. $f(x) = x^3 + 5x$ has $f'(x) = 3x^2 + 5 > 0$ everywhere, so it is invertible on all of $\mathbb{R}$ — and it has no elementary inverse formula whatsoever.
That combination is the normal case, not a curiosity. Most invertible functions cannot have their inverses written down, and the value of the existence argument is that it does not need to.
| Function | Restricted to | So that $f^{-1}$ lands in |
|---|---|---|
| $x^2$ | $[0, \infty)$ | $[0, \infty)$ — hence $\sqrt{4} = 2$ and not $-2$ |
| $\sin x$ | $[-\pi/2, \pi/2]$ | $[-\pi/2, \pi/2]$ |
| $\cos x$ | $[0, \pi]$ | $[0, \pi]$ |
| $\tan x$ | $(-\pi/2, \pi/2)$ | $(-\pi/2, \pi/2)$ |
Every entry in that table is a convention: somebody chose a piece on which the function is one-to-one, and the choice is what makes $\sqrt{\ }$ and $\arcsin$ single-valued. A different choice would give a different, equally valid inverse.
This is why $\sqrt{x^2} = |x|$ rather than $x$, and why $\arcsin(\sin 3)$ is not $3$: the composite returns the representative in the chosen piece, and $3$ is not in $[-\pi/2, \pi/2]$.
Reading $f^{-1}$ as a reciprocal. $f^{-1}(4)$ is the input that maps to $4$; $1/f(4)$ is a different number. Nothing in the notation prevents the confusion, so it has to be prevented by hand.
Deciding injectivity from the formula alone. $x^2$ is injective on $[0, \infty)$ and not on $\mathbb{R}$. The domain is half the question.
Fixing injectivity by shrinking the codomain. Shrinking the codomain fixes surjectivity. Injectivity is about the domain, and only restricting the domain can repair it.
Restricting without saying where. "Restrict $x^2$ so it has an inverse" has two answers, $[0, \infty)$ and $(-\infty, 0]$, giving $\sqrt{y}$ and $-\sqrt{y}$. Both are right; not naming which one you took is what is wrong.
The two questions — does an inverse exist and can I write it down — are separate, and only the first one is a mathematical fact about the function. $f(x) = x^3 + 5x$ is strictly increasing, hence injective, hence invertible; solving $y = x^3 + 5x$ for $x$ needs the cubic formula and produces something unusable. $f(x) = x + \sin x$ is invertible and has no closed-form inverse at all.
This matters practically as well as philosophically: when the mean value theorem or the inverse function theorem hands you an inverse, it hands you existence and a derivative, never a formula — and existence with a derivative turns out to be everything you need.
$f(x) = 2x + 3$ from $\mathbb{R}$ to $\mathbb{R}$: if $2a + 3 = 2b + 3$ then $2a = 2b$ then $a = b$. Injective.
Equal outputs forced equal inputs.
For any $y$, $x = (y - 3)/2$ is a real number with $f(x) = y$. Surjective.
Every target is hit.
So $f$ is bijective, with $f^{-1}(y) = (y - 3)/2$. Compare $g(x) = 2x + 3$ from $\mathbb{R}$ to $[0, \infty)$: same formula, still injective, and now not even a function — $g(-5) = -7$ is not in the codomain.
The sets are part of the claim.
$f(x) = \dfrac{3x - 1}{x + 2}$ on $x \ne -2$. Set $y = \dfrac{3x - 1}{x + 2}$ and clear the fraction: $y(x + 2) = 3x - 1$.
No cancelling yet — collect the unknown.
Gather the $x$ terms: $yx + 2y = 3x - 1$, so $yx - 3x = -1 - 2y$, so $x(y - 3) = -(1 + 2y)$.
All the $x$s on one side, factored.
$x = \dfrac{-(1 + 2y)}{y - 3} = \dfrac{1 + 2y}{3 - y}$, so $f^{-1}(y) = \dfrac{1 + 2y}{3 - y}$, defined for $y \ne 3$.
And $y = 3$ is excluded because $f$ never takes the value $3$ — the image of $f$ is exactly the domain of $f^{-1}$.
Try to factor: $x^3 - x = x(x-1)(x+1)$, which vanishes at three inputs.
Look for repeated outputs.
$g(0) = g(1) = g(-1) = 0$, and those inputs differ.
One pair is already enough.
So $g$ is not injective. On $[1, \infty)$, where $g' = 3x^2 - 1 > 0$, it is.
Match each inverse statement to its correct interpretation.
| $f^{-1}(8)=3$ | Becomes the range of $f^{-1}$ | Becomes the domain of $f^{-1}$ | Means inverse function, not $1/f(x)$ | |
|---|---|---|---|---|
| $f(3)=8$ | ||||
| Domain of $f$ | ||||
| Range of $f$ | ||||
| $f^{-1}(x)$ |
Match each inverse check to the conclusion it supports.
| It equals 4 | $(b,a)$ lies on $y=f^{-1}(x)$ | The principal root $\sqrt{x}$ | The negative root $-\sqrt{x}$ | |
|---|---|---|---|---|
| $f^{-1}(f(4))$ when $4$ is in the domain | ||||
| $(a,b)$ lies on $y=f(x)$ | ||||
| Inverse of $x^2$ on $[0,\infty)$ | ||||
| Inverse of $x^2$ on $(-\infty,0]$ |
Match each function-and-domain pair to the correct inverse conclusion.
| No inverse function: $2$ and $-2$ share an output | Inverse is $f^{-1}(y)=\sqrt y$ for $y\geq0$ | Inverse is $f^{-1}(y)=\sqrt[3]{y}$ | The point $(7,2)$ lies on $f^{-1}$ | |
|---|---|---|---|---|
| $f(x)=x^2$ on $\mathbb R$ | ||||
| $f(x)=x^2$ on $[0,\infty)$ | ||||
| $f(x)=x^3$ on $\mathbb R$ | ||||
| The point $(2,7)$ lies on $f$ |
Is $f(x) = x^4 - 4x^2$, taken on $\mathbb{R}$, injective?
$f(x) = 2x + 3$ on $\mathbb{R}$. Write $f^{-1}(y)$.
Answer:
$f(x) = x^3 + 4x$ is strictly increasing on $\mathbb{R}$, so it has an inverse — though no elementary formula for it. What is $f^{-1}(80)$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$f(t) = 3t + 6$. What is $f^{-1}(30)$?
Answer:
You can decide injectivity, invert a formula and check it, and restrict a domain to produce an inverse. Say in your own words why shrinking the codomain cannot fix injectivity. Next: the limit of a function, where the value at the point stops mattering.
9. Your turn: is $g(x) = x^3 - x$ on $\mathbb{R}$ injective?, step 3