Functions, Relations & Transformations

A relation is any set of ordered pairs $(x,y)$. A function is a special relation in which every input $x$ is paired with exactly one output $y$. In other words: no $x$-value is allowed to repeat with two different $y$-values.

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• The set of all allowed inputs is the domain.
• The set of all resulting outputs is the range.

A graph represents a function if and only if no vertical line crosses it more than once. A vertical line is a set of points with the same $x$; hitting the graph twice would mean one $x$ has two $y$'s.

We write $f(x)$ (read “$f$ of $x$”) for the output of function $f$ at input $x$. The letter $x$ is a placeholder: to evaluate, substitute the given value everywhere $x$ appears and simplify.

A discrete function is defined only at separated points (draw dots), e.g. “number of students” vs. “number of pizzas.” A continuous function is defined on an unbroken interval (draw a solid curve), e.g. “temperature” vs. “time.”

Each family of functions is built from a simple “parent.” Memorize their shapes, domains, and ranges.

Starting from a parent $f(x)$, the graph of $y=a\,f(x-h)+k$ is transformed as follows:

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• $k$: vertical shift --- up if $k>0$, down if $k<0$.
• $h$: horizontal shift --- right if $h>0$, left if $h<0$ (it moves opposite the sign inside).
• $a$: vertical stretch if $|a|>1$, vertical compression if $0<|a|<1$.
• $a<0$: reflection over the $x$-axis. Replacing $x$ with $-x$ reflects over the $y$-axis.

For two functions $f$ and $g$:

The domain of the result is where both pieces are defined, and for the quotient we also exclude any $x$ with $g(x)=0$.

The composition $(f\circ g)(x)=f\bigl(g(x)\bigr)$ means: run $g$ first, then feed its output into $f$. Composition is usually not commutative: $(f\circ g)(x)\ne (g\circ f)(x)$ in general.

The inverse $f^{-1}$ undoes $f$: it swaps inputs and outputs, so if $f(a)=b$ then $f^{-1}(b)=a$. To find it: replace $f(x)$ with $y$, swap $x$ and $y$, then solve for $y$.

A piecewise function uses different formulas on different parts of the domain. To evaluate, first decide which interval the input falls in, then use that piece.

A functional equation constrains an unknown function $f$ instead of an unknown number. The classic trick: substitute a cleverly chosen input to generate a second equation, then treat $f(x)$ and $f(\text{other})$ as two unknowns and eliminate. For equations mixing $f(x)$ with $f(1/x)$, the magic substitution is $x\mapsto 1/x$, which swaps the two terms.

Amplitude is the height; period is one full cycle.

Composing $f$ with itself is written $f^{[n]}=\underbrace{f\circ f\circ\cdots\circ f}_{n\text{ copies}}$, so $f^{[2]}(x)=f(f(x))$. Beware: this is iteration, not the power $\bigl(f(x)\bigr)^n$ and not the inverse $f^{-1}$.

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• If $f^{[n]}=\mathrm{id}$ (the identity $x\mapsto x$) for some $n$, then $f$ is periodic under iteration with period dividing $n$, and $f^{[m]}$ depends only on $m \bmod n$.
• An involution is the special case $n=2$: $f(f(x))=x$, so $f$ is its own inverse, $f^{-1}=f$. Its graph is symmetric across $y=x$.

A function is invertible exactly when it is one-to-one (passes the horizontal line test). You do not need an explicit formula for $f^{-1}$ to know it exists:

For example, $f(x)=x^5+x+1$ is a sum of the strictly increasing pieces $x^5$, $x$, and a constant, so it is strictly increasing and therefore invertible on all of $\mathbb{R}$ --- even though solving $y=x^5+x+1$ for $x$ by hand is hopeless. Monotonicity is also how you restrict a non-injective parent (like $y=x^2$) to make it invertible: choose a branch, e.g. $x\ge 0$.

A function is even if $f(-x)=f(x)$ (graph symmetric across the $y$-axis) and odd if $f(-x)=-f(x)$ (symmetric through the origin). Symmetry halves your work: knowing $f$ on $x\ge 0$ determines it everywhere. Two facts worth memorizing:

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• Products behave like parity signs: $\text{even}\times\text{even}=\text{even}$, $\text{odd}\times\text{odd}=\text{even}$, $\text{even}\times\text{odd}=\text{odd}$.
• Every function splits uniquely into an even part plus an odd part: @@BLOCK0@@

A chain of transformations can always be collapsed into a single rule $y=a\,f\bigl(b(x-h)\bigr)+k$. Track what happens to a generic point, applying inner (horizontal) changes to $x$ and outer (vertical) changes to the whole expression. Order matters, because a horizontal shift done before a horizontal stretch is not the same as after.

Example. Starting from $f$, reflect over the $y$-axis, then shift left $2$, then stretch vertically by $3$, then shift up $1$. Building inside-out, the horizontal reflect-and-shift turns the input into $-(x+2)$, and the outer stretch-and-shift wraps it:

A single expression now encodes all four moves --- feed in one $x$, get the fully transformed $y$.