Optional Function Refresher
Cheat sheet.

Domain restrictions at a glance

Range tactics

• Quadratics: complete the square; vertex gives the extreme.
• $\sqrt{\cdot}$ outputs are $\ge 0$; $a^x$ outputs are $> 0$.
• When stuck: solve $y = f(x)$ for $x$; the $y$ that admit a solution form the range.

Key method

At each boundary, record the left-hand behavior, the right-hand behavior, and the function value separately. They can be three different things. A function may be defined at a boundary without being continuous there.

Example

Let $f(x)=x+1$ for $x<1$ and $f(x)=3-x$ for $x\ge1$. Draw the first line only left of $1$ and the second only on and right of $1$. Both approach $2$, and the second assigns $f(1)=2$, so the graph joins continuously. The left slope is $1$ and right slope is $-1$, so there is a corner.

Avoid this mistake

Continuity does not require the same formula or slope on both sides. Conversely, differentiability requires more than the pieces simply meeting.

Key method

For $g(x)=f(x-h)+k$, translate every point by $(h,k)$. A forbidden input $u$ of $f$ becomes the forbidden input $u+h$ of $g$. Vertical shifts do not change the domain.

Example

Starting from $f(x)=1/x$, the function $g(x)=1/(x-2)+3$ has vertical asymptote $x=2$ and horizontal asymptote $y=3$. Its domain excludes $2$ and its range excludes $3$. The point $(1,1)$ becomes $(3,4)$.

Avoid this mistake

A shift inside a denominator is still an input shift. Solve the denominator-zero equation rather than inferring a sign visually.

Key method

For $x\ne0$, it equals $x/|x|$. At zero, that quotient is undefined, so the separate definition is essential.

Example

The left-hand limit at zero is $-1$ and the right-hand limit is $1$. Consequently, no two-sided limit exists there, even though the function value $0$ is defined. Away from zero, the function is locally constant and its derivative is zero.

Avoid this mistake

Defining a value at a jump cannot make the two one-sided limits equal. The derivative of $|x|$ is signum only for $x\ne0$; $|x|$ is not differentiable at zero.

Key method

Each horizontal segment includes its left endpoint and excludes its right endpoint. The domain is all real numbers and the range is the integers.

Example

Since $-3\le-2.4<-2$, $\lfloor-2.4\rfloor=-3$. At an integer $n$, the left-hand limit is $n-1$ while the right-hand limit and function value are $n$. Thus the function jumps at every integer and is continuous at every noninteger.

Avoid this mistake

Truncating decimal digits agrees with floor for positive inputs but fails for negative nonintegers. Also, floor does not distribute over addition.