Functions Review
Cheat sheet.

Key method

The notation $f:A\to B$ specifies the domain $A$ and codomain $B$. The rule $f(x)$ tells how to assign outputs. The actual outputs form the range, which is a subset of $B$.

Example

Let $f:\mathbb R\to\mathbb R$ have rule $f(x)=x^2$. Then $f(-3)=9$ and $f(3)=9$. These equal outputs do not violate the function definition: each input still has only one square. The equation $y^2=x$ describes a relation with two possible $y$ values for positive $x$, unless a branch is selected.

Avoid this mistake

$f(x)$ is the output, not multiplication of $f$ by $x$. The input symbol is a placeholder: $f(t)=t^2$ describes the same rule.

Key method

For $f(x)=x^2-3x$, input $u$ gives $f(u)=u^2-3u$. If $u$ is an expression, expand only after substitution. Check domain restrictions first.

Example

Compute $f(a+h)$:

Subtracting $f(a)$ leaves $2ah+h^2-3h$. If $h\ne0$, dividing by $h$ gives $2a+h-3$, the difference quotient used in differentiation.

Avoid this mistake

$f(a+h)$ is usually not $f(a)+f(h)$. A nonlinear function does not distribute over addition.

Key method

For $f:A\to B$, the range is $f(A)=\{f(x):x\in A\}\subseteq B$. Restricting the domain can shrink the range. Enlarging the codomain alone does not change any computed value.

Example

For $f:[-2,3]\to\mathbb R$, $f(x)=x^2$, the minimum is $0$ at $x=0$ and the maximum is $9$ at $x=3$. The range is $[0,9]$, not $[4,9]$: squaring only the endpoints misses the interior minimum. With domain $[1,3]$, the same rule has range $[1,9]$.

Avoid this mistake

The codomain cannot be inferred uniquely from the formula. State it explicitly when asking whether a function is onto.

Key method

A vertical line $x=a$ may meet the graph at at most one point. A formula that fails at an input can define a function on a smaller domain, but cannot silently retain the forbidden input.

Example

The circle $x^2+y^2=1$ fails the vertical-line test: at $x=0$ it has $y=1$ and $y=-1$. Its upper half $y=\sqrt{1-x^2}$ is a function on $[-1,1]$. The rule $1/x$ is a function on $\mathbb R\setminus\{0\}$, but not on all of $\mathbb R$.

Avoid this mistake

The horizontal-line test checks one-to-one behavior, not whether the relation is a function in the first place.

Key method

Useful methods include completing the square, monotonicity on the stated domain, and solving $y=f(x)$ for an allowed $x$. Endpoint inclusion follows from whether an input actually attains the candidate output.

Example

For $f(x)=(x-2)^2+3$ on $\mathbb R$, the square gives $f(x)\ge3$. Conversely, for any $y\ge3$, choose $x=2+\sqrt{y-3}$, which produces $f(x)=y$. The range is exactly $[3,\infty)$. For $1/x$, output zero is impossible, while every nonzero $y$ is attained by $x=1/y$.

Avoid this mistake

A graph window can suggest a range but cannot prove behavior outside its visible portion.

Key method

Use brackets for included endpoints and parentheses for excluded endpoints. A missing point removes an output from the range only if no other point on the graph has that same height.

Example

Consider the line segment $y=2x+1$ for $-1<x\le2$. Its horizontal projection is $(-1,2]$. Since the line increases, its vertical projection is $(-1,5]$. If a graph has a hole at $(0,1)$ but also passes through $(2,1)$, the output $1$ still belongs to its range.

Avoid this mistake

Do not interpret the edge of a plotted window as a mathematical endpoint. Look for an explicit domain, endpoint marker or continuation arrow.

Key method

To find the range, solve $y=p(x)/q(x)$ for $x$, then check for impossible targets and forbidden solutions. Removable holes can exclude an otherwise attainable output.

Example

For $f(x)=(x+1)/(x-2)$, the domain excludes $2$. Solve $y(x-2)=x+1$:

This expression never equals $2$, so every $y\ne1$ is attained. At $y=1$, the equation would say $-2=1$, impossible. The range excludes $1$.

Avoid this mistake

For $(x^2-1)/(x-1)$, canceling gives $x+1$ only when $x\ne1$. The hole means the range also excludes $2$.

Key method

For distinct inputs,

Positive $m$ means increasing, negative $m$ decreasing, and zero $m$ constant. A vertical line is not the graph of a function $y=f(x)$.

Example

A line through $(1,3)$ and $(4,9)$ has slope $(9-3)/(4-1)=2$. Substituting $(1,3)$ into $y=2x+b$ gives $b=1$. Hence $f(x)=2x+1$ and increasing $x$ by $0.5$ increases the output by $1$.

Avoid this mistake

The intercept is the output at $x=0$, not the input at which the graph crosses the horizontal axis.

Key method

To evaluate, first decide which condition the input satisfies, then use only that formula. At a boundary, inspect strict and non-strict inequality signs to determine endpoint inclusion.

Example

Let $f(x)=x+2$ for $x<0$ and $f(x)=x^2$ for $x\ge0$. Then $f(-1)=1$, $f(0)=0$, and $f(2)=4$. The left branch approaches $2$ near zero, while the value at zero is determined by the second branch. This is a jump, despite both formulas being individually continuous.

Avoid this mistake

Using both formulas at a boundary without checking the case conditions can assign two incompatible outputs.

Key method

For an expression $u(x)$, solve $u(x)=0$ to locate the breakpoints before splitting into intervals.

Example

For $|2x-6|$, the breakpoint is $x=3$. If $x<3$, then $2x-6<0$ and the value is $6-2x$. If $x\ge3$, the value is $2x-6$. Both branches meet at zero when $x=3$, producing a corner rather than a discontinuity.

Avoid this mistake

$|a+b|$ is not generally $|a|+|b|$. For $a=1,b=-1$, the two sides are $0$ and $2$.

Key method

Test the whole formula, not merely its visible exponents. Absolute value and cosine are even; many expressions containing odd powers are neither even nor odd.

Example

For $f(x)=x^4-3x^2+2$, substitution gives $f(-x)=(-x)^4-3(-x)^2+2=f(x)$. For $g(x)=x^2+x$, $g(-x)=x^2-x$ is generally different. Restricting $x^2$ to $[0,\infty)$ removes the symmetric domain, so the restricted function is not even under the standard definition.

Avoid this mistake

Matching $f(1)$ and $f(-1)$ alone does not prove evenness; the equality must hold throughout the domain.

Key method

If zero belongs to the domain, oddness forces $f(0)=0$. That necessary condition alone does not prove a function is odd.

Example

For $f(x)=x^3-2x$, $f(-x)=-x^3+2x=-(x^3-2x)$. The reciprocal $1/x$ is also odd on its symmetric domain excluding zero. In contrast, $x^3+1$ fails because the constant does not change sign.

Avoid this mistake

The zero function is both even and odd on any symmetric domain. Most functions are neither.

Key method

Even + even is even; odd + odd is odd. Products obey: even × even is even, odd × odd is even, and even × odd is odd. A sum of a nonzero even part and a nonzero odd part is generally neither.

Example

The function $x\sin x$ is even: both $x$ and $\sin x$ are odd, so their sign changes cancel. The function $x\cos x$ is odd. For $f(x)=x^2+x$, its even part is $(f(x)+f(-x))/2=x^2$ and its odd part is $(f(x)-f(-x))/2=x$.

Avoid this mistake

An odd power in a numerator does not determine the parity of an entire rational expression; inspect the denominator too.

Key method

$-f(x)$ reflects across the horizontal axis. $f(-x)$ reflects across the vertical axis. $-f(-x)$ reflects through the origin. Reflection across $y=x$ exchanges coordinates and describes an inverse relation.

Example

If $(2,5)$ lies on $f$, then $(2,-5)$ lies on $-f$, $(-2,5)$ lies on $f(-x)$, and $(-2,-5)$ lies on $-f(-x)$. For the even function $x^2$, horizontal input reflection leaves the graph unchanged, but output reflection gives a downward-opening parabola.

Avoid this mistake

The graph $y=-x^2$ is not a right-to-left reflection of $y=x^2$; it is an up-to-down reflection.

Key method

A point $(u,f(u))$ becomes $(u+h,f(u)+k)$. Positive $h$ shifts right, while positive $k$ shifts up.

Example

The vertex of $f(x)=x^2$ is $(0,0)$. For $g(x)=(x-2)^2+3$, that vertex moves to $(2,3)$. The original points $(1,1)$ and $(-1,1)$ move to $(3,4)$ and $(1,4)$, preserving symmetry about the new vertical line $x=2$.

Avoid this mistake

The sign inside looks reversed because the old input $0$ now occurs at $x=h$. Derive the shift from that equation instead of memorizing a slogan.

Key method

For $g(x)=a f(bx)$ with nonzero $a,b$, $(u,v)$ becomes $(u/b,av)$. Thus $|a|>1$ stretches vertically, while $|b|>1$ compresses horizontally. Negative factors also reflect.

Example

For $f(x)=x^2$, $g(x)=3f(x)$ triples every output. The point $(2,4)$ becomes $(2,12)$. For $h(x)=f(2x)$, the same old input $2$ is reached at $x=1$, so its point becomes $(1,4)$. Although $f(2x)=4x^2$ here, horizontal and vertical scaling are different constructions.

Avoid this mistake

A multiplier of zero collapses a graph and is not an invertible stretch. Track this as its own case.

Key method

Its domain contains exactly the inputs in the domain of $g$ whose outputs lie in the domain of $f$. Both stages must be legal.

Example

Let $f(u)=\sqrt u$ and $g(x)=x-3$. Then $(f\circ g)(x)=\sqrt{x-3}$ with domain $x\ge3$. Reversing the order gives $(g\circ f)(x)=\sqrt x-3$ with domain $x\ge0$. The different order changes both the rule and its domain.

Avoid this mistake

Composition is not multiplication. $f(g(x))$ generally differs from $f(x)g(x)$.

Key method

Both sides mean $f(g(h(x)))$. With the identity function $I(x)=x$, $f\circ I=I\circ f=f$ on the appropriate domain.

Example

For $f(x)=x+1$ and $g(x)=x^2$, $(f\circ g)(x)=x^2+1$, while $(g\circ f)(x)=(x+1)^2$. At $x=1$, the outputs are $2$ and $4$. Thus associativity does not imply commutativity. An inverse is special: composing it with the original function returns the identity on the corresponding set.

Avoid this mistake

Canceling $f$ from $f(a)=f(b)$ requires injectivity. It is not a general algebraic cancellation rule.

Key method

Strictly increasing or strictly decreasing functions on an interval are injective there. Restricting the domain can make a noninjective rule injective.

Example

For $f(x)=5x+1$, equality $5a+1=5b+1$ forces $a=b$. For $g(x)=x^2$ on $\mathbb R$, $g(-2)=g(2)$, so it is not injective. On $[0,\infty)$, $a^2=b^2$ and nonnegativity force $a=b$, making the restriction injective.

Avoid this mistake

A graph passing the vertical-line test can still fail the horizontal-line test. It is a function, but may not have an inverse function on the full domain.

Key method

To disprove onto behavior, exhibit a single element of the codomain with no preimage. You must know the codomain before making this judgment.

Example

For $f:\mathbb R\to\mathbb R$ with $f(x)=x^2$, the target $-1$ cannot be reached by a real input. Thus $f$ is not onto. If instead the declared codomain is $[0,\infty)$, every target $y$ has preimage $\sqrt y$, so the same formula and domain become onto that codomain.

Avoid this mistake

A function can be one-to-one without being onto, or onto without being one-to-one. These are independent properties.

Key method

For $f:A\to B$, surjectivity means: for every $y\in B$, there exists $x\in A$ with $f(x)=y$. Verify both that the constructed input belongs to $A$ and that substitution returns $y$.

Example

Take $f:\mathbb R\to\mathbb R$, $f(x)=3x-2$. Given any real $y$, set $x=(y+2)/3$. This is real and $f(x)=3(y+2)/3-2=y$. Therefore the function is onto. On domain $[0,\infty)$ with the same codomain, it is not onto: its range starts at $-2$.

Avoid this mistake

Checking a few targets is evidence, not a proof for every target. Use a symbolic arbitrary $y$.

Key method

If $f:A\to B$ is bijective, then $f^{-1}:B\to A$ satisfies

Surjectivity guarantees existence of the reverse assignment; injectivity guarantees uniqueness.

Example

The function $f:[0,\infty)\to[0,\infty)$, $f(x)=x^2$, is bijective. Its inverse is $\sqrt x$. If its domain expands to all real numbers, uniqueness is lost. If its codomain expands to all real numbers, some targets have no preimage.

Avoid this mistake

An inverse is not a reciprocal: $f^{-1}(x)$ usually differs from $1/f(x)$.

Key method

Write $y=f(x)$, solve for $x$ in terms of $y$, then rename the input of the inverse. Domain and range exchange roles. Verify by composition on the appropriate domains.

Example

For $f(x)=3x-5$, solve $y=3x-5$ to get $x=(y+5)/3$. Hence $f^{-1}(x)=(x+5)/3$. For $g(x)=x^2$ on $x\ge0$, take the nonnegative branch: $g^{-1}(x)=\sqrt x$, also defined for $x\ge0$.

Avoid this mistake

Writing both $\pm\sqrt x$ does not define a single inverse function; the original domain decides which branch to use.

Key method

If $f=g\circ h$ and both maps are bijective between their specified sets, then

The last operation performed is the first operation undone.

Example

For $f(x)=2(x-3)^3+5$, first subtract $3$, cube, multiply by $2$, then add $5$. Reverse these operations: subtract $5$, divide by $2$, take a cube root, add $3$.

Cube roots are single-valued over the reals, so no branch restriction is needed.

Avoid this mistake

Undoing in the same order is wrong. Squaring also requires a domain restriction before a square root can undo it.

Key method

If $(a,b)$ lies on a bijective function's graph, $(b,a)$ lies on its inverse graph. Vertical and horizontal features exchange: domain becomes range, and a vertical asymptote becomes a horizontal one where applicable.

Example

The exponential $y=2^x$ passes through $(0,1)$ and $(1,2)$. Its inverse $y=\log_2 x$ passes through $(1,0)$ and $(2,1)$. The exponential's horizontal asymptote $y=0$ corresponds to the logarithm's vertical asymptote $x=0$.

Avoid this mistake

A graph and its inverse need not meet on the displayed window. Reflection does not mean reflection across the vertical axis.

Key method

Strict increase means $x_1<x_2\Rightarrow f(x_1)<f(x_2)$ for every pair in the interval. Strict decrease reverses the output inequality. Later, derivative signs provide an efficient test.

Example

For $f(x)=(x-1)^2$, the graph decreases toward its vertex on $(-\infty,1]$ and increases away from it on $[1,\infty)$. The values on both intervals are nonnegative, illustrating why positive output does not mean increasing.

Avoid this mistake

A flat tangent at one point does not automatically mean a change in monotonicity. For example, $x^3$ increases through zero.

Key method

An absolute minimum at $a$ satisfies $f(a)\le f(x)$ for every allowed $x$. A local minimum needs this only in a neighborhood of $a$ within the domain. Maximum reverses the inequality.

Example

For $f(x)=(x-1)^2$ on $[0,3]$, the absolute minimum is $0$ at $x=1$. Endpoint values are $1$ and $4$, so the absolute maximum is $4$ at $x=3$. On the open interval $(0,3)$, values approach $4$ but never attain it, so no absolute maximum exists.

Avoid this mistake

A bound need not be attained. The words “maximum” and “minimum” require actual inputs that produce those values.