Functions

A relation that assigns each input to exactly one output. Two parts: must act on every domain element, and must be well-defined (no input gets two outputs).

A graph represents a function iff every vertical line meets it in at most one point. Two intersections ⇒ one $x$ would map to two $y$s ⇒ not a function.

Set of allowed inputs. Natural domain = largest subset of $\mathbb{R}$ that gives real outputs.

Set of outputs actually achieved. Read it off the $y$-axis from the graph.

Denominator $\ne 0$ · radicand $\ge 0$ for even roots · argument $> 0$ for $\ln/\log$ · argument in $[-1,1]$ for $\arcsin/\arccos$. Intersect the allowed sets.

$\{x:x\ge 3\}$ ≡ $[3,\infty)$. $\{x:-2\le x<5\}$ ≡ $[-2,5)$. Inverted bracket means open endpoint.

(1) Write $y=f(x)$. (2) Swap $x$ and $y$. (3) Solve for $y$. That's $f^{-1}(x)$.

Graph of $f^{-1}$ = reflection of $f$ across the line $y=x$.

Iff $f$ is one-to-one. If $f$ fails the horizontal line test, restrict the domain to one side of the turning point first.

$D(f^{-1}) = R(f)$ and $R(f^{-1}) = D(f)$.

$x$ must be in $D(g)$ AND $g(x)$ must be in $D(f)$. Compose inner function first; then check the outer's domain restriction kicks in.

Test with $f(x)=2x$, $g(x)=x+1$: $(f\circ g)(1)=4$, $(g\circ f)(1)=3$.

$f^{-1} = f$. Examples: $y=x$, $y=-x$, $y=k/x$ (any $k$), $y=a-x$. Their graphs are symmetric about $y=x$ AND look the same swapped.

If no domain is stated, use the largest set giving real outputs. Always check denominators, radicals, logs.

$f^{-1}$ is the INVERSE function, NOT $1/f(x)$. They mean completely different things.

If you skip the one-to-one check, the 'inverse' you write down won't be a function. Always test (or restrict).

$k>0$ shifts UP by $k$. Intuitive direction.

$h>0$ shifts RIGHT by $h$ (counter-intuitive: SUBTRACTING $h$ moves right). The vector form is $\binom{h}{k}$.

Stretches by factor $a$ from $x$-axis. $a>1$ ⇒ further from x-axis. $0<a<1$ ⇒ closer (compression).

Stretches by factor $1/b$ from $y$-axis (reciprocal!). $b>1$ ⇒ COMPRESSION; $0<b<1$ ⇒ STRETCH.

$a$ vert stretch · $b$ horiz stretch · $h$ horiz shift · $k$ vert shift. Order matters when mixing inside and outside.

$y=|f(x)|$ reflects parts BELOW the $x$-axis ABOVE it. $y=f(|x|)$ replaces the $x<0$ part with a mirror image of the $x>0$ part across the $y$-axis.

Zeros of $f$ ⇒ vertical asymptotes of $1/f$. Where $f$ has a min, $1/f$ has a max (at the same $x$). $1/f$ has the same sign as $f$.

VA at $x=-d/c$; HA at $y=a/c$. Equivalent transformed form $y=\tfrac{A}{B(x-h)}+k$.

Odd, axes are asymptotes, self-inverse. $D = R = \mathbb{R}\setminus\{0\}$.

Domain: $ax+b\ge 0$. Range: $y\ge 0$ (or $\le 0$ if outer minus). Curve starts at boundary and rises with decreasing slope.

Solve both cases and CHECK each against the original equation.

Set denominator $=0$; the values are VAs (unless they cancel with the numerator).

Compare degrees: numerator deg $<$ denominator deg ⇒ $y=0$. Equal degrees ⇒ $y=$ leading-coeff ratio. Higher numerator ⇒ no HA (slant or beyond).

Apply inside-the-bracket transformations first (in reverse algebra order): translate inside ⇒ then stretch inside. Outside transformations follow in natural order.

$f(x-h)$ shifts RIGHT by $h$, $f(x+h)$ shifts LEFT by $h$. Inside the bracket the sign flips its apparent meaning.

If $B<0$ there is no solution. If $B$ contains $x$, always check — squaring or splitting can introduce false roots.

Every output comes from at most one input. Horizontal-line test: ≤ 1 intersection. Required for an inverse.

At least two different inputs share an output. Fails the horizontal-line test.

Every element of the stated co-domain is achieved as an output. Range = co-domain.

Restrict a many-to-one function to ONE side of its turning point to force one-to-one. Example: $x^2$ on $\mathbb{R}$ is many-to-one; on $[0,\infty)$ it's one-to-one with inverse $\sqrt{x}$.

Graph symmetric about the $y$-axis. Polynomials: only even powers + constant. Examples: $x^2,\cos x,|x|$.

Graph has $180^\circ$ rotational symmetry about origin. Polynomials: only odd powers, no constant. Examples: $x^3,\sin x,1/x$.

At least one $x$ with $f(-x) \ne \pm f(x)$. Mixed-degree polynomials like $x^2+x$.

Only $f(x)=0$ satisfies both. $f=f$ and $f=-f$ ⇒ $2f=0$ ⇒ $f=0$.

Set the radicand $\ge 0$ and solve. Don't forget to flip the inequality if $a<0$.

$y=c+k\sqrt{ax+b}$ starts at $(x_0, c)$ where $x_0$ is the boundary. If $k>0$ the curve rises; if $k<0$ it falls below $c$.

Bare $\sqrt{\dots}\ge 0$ ⇒ range $\ge 0$. Outer constant shifts the range floor; outer negative gives range $\le c$.

Proper fraction (deg numerator $<$ deg denominator). Factor denominator into distinct linears: $\dfrac{f(x)}{(x-r_1)(x-r_2)} = \dfrac{A}{x-r_1}+\dfrac{B}{x-r_2}$.

Find $A_i$ by covering up $(x-r_i)$ in the original and substituting $x=r_i$ into what remains. Works because every other term has $(x-r_i)$ as a factor and vanishes.

$(x-r)^{2}$ in denominator ⇒ TWO pieces: $\dfrac{A}{x-r} + \dfrac{B}{(x-r)^{2}}$.

Different rules on different intervals. Domain = union of the intervals. Evaluate by selecting the rule for the input's interval.

Inverses require one-to-one. Integration $\int_{-a}^{a}f$ vanishes if $f$ is odd (saves work!). Symmetry gives quick range arguments and graphs faster sketching.

If $f$ is odd and defined at $0$, then $f(0)=0$ (set $x=0$ in $f(-x)=-f(x)$). If $f(0)\ne 0$, $f$ cannot be odd.

Numerator MUST be lower degree than denominator. If not, polynomial-divide first; the quotient is a polynomial part plus a proper fraction.

Mixed practice over function definition, domain/range, all families (rational/radical/modulus/piecewise), partial fractions, classification (one-to-one, even/odd), composition, inverses, transformations.

18 questions across 4 sections (A: 5, B: 5, C: 4, D: 4 longer). Designed for ~60 min in one sitting; pause and resume any time.

Vertical line test (graph) ⇔ no two ordered pairs share an $x$.

Combine: denominator $\ne 0$, radicand $\ge 0$ (even root), argument $>0$ (log).

Read $y$-values reached. For shifted-vertex parabolas, range is bounded by the $y$ of the vertex.

VA: $x=-d/c$. HA: $y=a/c$.

Domain: $ax+b\ge 0$. Range starts at $y=c$; goes up if $k>0$, down if $k<0$.

For distinct linear factors: cover $(x-r_i)$ and substitute $x=r_i$ to read off $A_i$.

If many-to-one, restrict the domain (e.g. one side of the parabola vertex) before inverting.

Swap $x$ and $y$, solve for $y$. Geometric: reflect graph in $y=x$. Domain/range swap.

Examples: $1/x$, $a-x$, $(x-k)/(x-1)$ for certain $k$.

$a$ vert stretch · $b$ horiz · $h$ right · $k$ up.

Zeros of $f$ ⇒ VAs of $1/f$. Min of $f$ ⇒ max of $1/f$.

$|f(x)|$: flip below-axis up. $f(|x|)$: replace $x<0$ part with mirror of $x>0$ part.

$f(x-h)$ shifts RIGHT by $h$ (subtracting $h$ moves right). Always counter-intuitive.

Modulus equations: split into cases and CHECK in the original. Squaring or expanding can add false roots.

$D(f\circ g)$ = $\{x \in D(g) : g(x) \in D(f)\}$. Don't just take $D(g)$.