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Find and verify inverse functions, decide when an inverse exists, and compose two functions in both orders.
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.
In this lesson you build new functions out of old ones. First you run a function backwards: you find its inverse by undoing its operations in reverse order, you check the inverse by composing the two rules, and you learn why a function that sends two inputs to the same output has no inverse at all. Then you wire two functions in series, feeding the output of one into the other, and you discover that the order you wire them in changes the answer — except for the one pair, a function and its inverse, that cancels out completely.
You already solve $3x + 7 = 22$ by undoing what was done to $x$: take away the $7$, then divide by the $3$. You already read $f(4) = 19$ as "the rule called $f$ turns $4$ into $19$". An inverse is those two habits together: you solve for $x$ but keep the letters, so the answer is a rule rather than a number.
Inverse function, written $f^{-1}$: the rule sending every output of $f$ back to the input it came from. If $f(4) = 19$ then $f^{-1}(19) = 4$.
One-to-one: no two different inputs share an output. Only a one-to-one function has an inverse.
Domain and range: the inputs and the outputs. Inverting swaps them.
Identity function: the rule that leaves every number alone. It is what $f^{-1}(f(x))$ equals.
An inverse function undoes what a function did. If $f$ turns $a$ into $b$, then $f^{-1}$ turns $b$ back into $a$, so $f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x$. That pair of equations is the definition, and the only reliable check.
To find an inverse, write $y = f(x)$, solve for $x$, then rename the letters. For $f(x) = 5x - 2$: from $y = 5x - 2$, add $2$ to get $y + 2 = 5x$, divide to get $x = \frac{y+2}{5}$, so $f^{-1}(x) = \frac{x+2}{5}$. The operations came out in reverse order, each replaced by its opposite — that is what "undo" means in practice.
An inverse exists only when $f$ is one-to-one. $f(x) = x^3$ is, and its inverse is the cube root; $f(x) = x^2$ on all the real numbers is not, because $3$ and $-3$ share the output $9$. On a graph, one-to-one is the horizontal line test: no horizontal line crosses the curve twice.
Another way: picture
The graphs of $f$ and $f^{-1}$ drawn together with the line $y = x$ between them. Each is the other reflected in that line, because swapping input and output swaps the coordinates of every point: $(4, 19)$ becomes $(19, 4)$.
Another way: story
A function is a machine; its inverse is the same machine run backwards. Socks then shoes is undone by shoes then socks — the steps reverse and each one flips. A machine that crushes two different things into one shape cannot be run backwards at all, which is what fails for $x^2$.
"$f^{-1}$ means $\frac{1}{f}$." It does not. If $f(x) = x + 3$ then $f^{-1}(x) = x - 3$, while $\frac{1}{f(x)} = \frac{1}{x+3}$ — a different rule entirely. The $-1$ is borrowed notation, not an exponent.
Undoing in the original order. $f(x) = 5x - 2$ multiplies and then subtracts, so its inverse adds and then divides: $f^{-1}(x) = \frac{x+2}{5}$, not $\frac{x}{5} + 2$. Test one number: $f(3) = 13$, and $\frac{13+2}{5} = 3$.
"Every function has an inverse." $f(x) = x^2$ sends $3$ and $-3$ both to $9$, so nothing decides which one $9$ goes back to. Squaring has an inverse only once the inputs are cut down to $x \ge 0$.
Write $y = 4x + 9$ and solve for $x$.
The inverse is the same relation read the other way round.
$y - 9 = 4x$, so $x = \frac{y - 9}{4}$.
Subtract before dividing: the reverse of the order $f$ used.
Rename: $f^{-1}(x) = \frac{x - 9}{4}$. Check: $f(2) = 17$ and $\frac{17 - 9}{4} = 2$.
One numerical check catches almost every slip.
Compute $g(f(x)) = \frac{(3x - 6) + 6}{3} = \frac{3x}{3} = x$.
Substitute the whole rule, then simplify.
Compute $f(g(x)) = 3 \cdot \frac{x + 6}{3} - 6 = x$.
Both orders must give $x$; one alone is not enough.
Both compositions are the identity, so $g = f^{-1}$.
Write $y = 7x - 1$ and add $1$: $y + 1 = 7x$.
Divide by $7$ and rename: $f^{-1}(x) = \frac{x + 1}{7}$. Check with $f(1) = 6$ and $\frac{6 + 1}{7} = 1$.
$f(x) = 2x - 3$. Write $f^{-1}(x)$.
Answer:
$f(x) = 4x - 5$ and $g(x) = \frac{x + 5}{4}$. What is $f(g(4))$?
Answer:
$f(x) = x^{2}$, defined for every real $x$. Does it have an inverse function?
$f(x) = 4x - 2$. Write $f^{-1}(x)$.
Answer:
You already substitute: given $f(x) = x^2 + 1$, you find $f(3)$ by writing $3$ wherever the rule writes $x$. Composition changes only what you substitute — a whole expression instead of a number. You also know that order matters in arithmetic: $10 - 3$ is not $3 - 10$.
Composition, written $f \circ g$ or $f(g(x))$: apply $g$ first, then apply $f$ to the result. The rule nearest the $x$ runs first.
Inner and outer function: in $f(g(x))$, $g$ is the inner one and $f$ the outer one.
Decompose: split a complicated rule into an inner and an outer piece, the reverse of composing.
Domain of a composition: an input is allowed only if $g$ accepts it and $f$ accepts what $g$ produced.
Composition wires two functions in series: the output of one becomes the input of the next. It is written $f(g(x))$, or $(f \circ g)(x)$, and means apply $g$, then apply $f$. To build the rule, write out $f$ and replace every $x$ in it by the whole of $g$'s rule.
With $f(x) = x^2$ and $g(x) = x + 3$: $$f(g(x)) = (x + 3)^2 = x^2 + 6x + 9 \qquad g(f(x)) = x^2 + 3$$ The order changes the answer: $f \circ g$ and $g \circ f$ are different functions.
To evaluate at a number, work from the inside out and stay numerical: $f(g(2))$ means $g(2) = 5$, then $f(5) = 25$. Building the general rule first gives the same answer, but the inside-out route is faster and harder to get wrong.
One pair always commutes: a function and its inverse, since $f(f^{-1}(x)) = x$. Composition also runs backwards: $\sqrt{3x - 1}$ decomposes into the inner rule $3x - 1$ and the outer rule $\sqrt{\ }$.
Another way: picture
Two boxes wired in a line. A number enters the box labelled $g$, a new number leaves along the wire and enters the box labelled $f$. Swapping the boxes gives a different machine: the same parts, a different output.
Another way: story
Composition is a two-step recipe, and steps do not commute. Doubling a price and then adding tax is not the same as adding tax and then doubling.
Reading left to right. $f(g(x))$ does not run $f$ first. The brackets say $g$ acts on $x$ and $f$ acts on the result, so the inner rule goes first even though it is written second in the name $f \circ g$.
"$f(g(x))$ is the same as $g(f(x))$." Take $f(x) = x^2$ and $g(x) = x + 1$: then $f(g(x)) = (x+1)^2 = x^2 + 2x + 1$, while $g(f(x)) = x^2 + 1$. They agree only for special pairs — inverses, for instance.
"$f(g(x))$ means multiply $f$ by $g$." It does not. $f(x)g(x) = x^2(x+1) = x^3 + x^2$, which matches neither composition. Composition feeds one output into the next; multiplication combines two outputs.
$f(g(x))$: write $g$'s rule wherever $f$ writes $x$, giving $2(x^2) - 1 = 2x^2 - 1$.
Square first, then double and subtract.
$g(f(x))$: write $f$'s rule wherever $g$ writes $x$, giving $(2x - 1)^2 = 4x^2 - 4x + 1$.
Double and subtract first, then square the whole thing.
The two results differ, so $f \circ g \neq g \circ f$. At $x = 3$ they give $17$ and $25$.
A single number is enough to prove two rules different.
$f(g(3))$: inside first, $g(3) = 15$; then $f(15) = 19$.
Stay with numbers; do not build the general rule.
$g(f(3))$: inside first, $f(3) = 7$; then $g(7) = 35$.
Same operations, other order.
$19 \neq 35$: the shift gets multiplied by $5$ in one order and not in the other.
$f(g(5))$: $g(5) = 3$, then $f(3) = 9$.
$g(f(5))$: $f(5) = 25$, then $g(25) = 23$. The two orders disagree.
$f(x) = x^2$ and $g(x) = x + 1$. Write $f(g(x))$.
Answer:
$f(x) = x^2$ and $g(x) = x - 9$. Write $g(f(x))$.
Answer:
$f(x) = 4x$ and $g(x) = x + 2$. What is $f(g(-3))$?
Answer:
$f(x) = x^2$ and $g(x) = x + 7$. Write $f(g(x))$.
Answer:
A photo is $362$ px wide. Filter P doubles the width. Filter Q adds $38$ px to the width. Give the width after P then Q, and the width after Q then P.
P then Q: pq px. Q then P: qp px
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$f(x) = 8x - 6$ and $g(x) = \frac{x + 6}{8}$. What is $f(g(-3))$?
Answer:
$f(x) = x^2$ and $g(x) = x - 8$. Write $g(f(x))$.
Answer:
You can invert a linear function and compose two functions in either order. Find the inverse of $f(x) = 6x - 5$, and for $f(x) = x^2$ and $g(x) = x + 4$ write both $f(g(x))$ and $g(f(x))$ and say why they differ.
8. Your turn: find the inverse of $f(x) = 7x - 1$, step 2
19. Your turn: $f(x) = x^2$, $g(x) = x - 2$. Find $f(g(5))$ and $g(f(5))$, step 2