Relations and Functions: Practice

The root is in a denominator, so its inside must be strictly positive.

$x^2 - 4 > 0 \;\Rightarrow\; x^2 > 4 \;\Rightarrow\; x < -2 \text{ or } x > 2$

Both $x = \pm 2$ give $\sqrt{0} = 0$ in the denominator and are excluded, so the domain splits into two pieces.

Test an input by solving for the outputs.

$x = 0 \;\Rightarrow\; |y| = 1 \;\Rightarrow\; y = 1 \text{ or } y = -1$

One input has two outputs, so the diamond is not a function of $x$.

Complete the square to expose the lowest output of the upward parabola.

$x^2 - 4x + 1 = (x - 2)^2 - 3 \ge -3$

The minimum $-3$ is reached at $x = 2$ and the outputs rise without bound, so the range is $y \ge -3$.

Being a function of $x$ is the vertical line test; being a function of $y$ is the horizontal line test.

$\text{vertical line test and horizontal line test both pass}$

So every vertical line and every horizontal line meets $R$ at most once. It need not be a line or a single point.

The original is a function of $x$ when its inputs are distinct; its swap fails when an output repeats. The set $\{(1,5),(2,5),(3,6)\}$ has distinct inputs but the output $5$ twice.

$\text{swap} = \{(5,1),(5,2),(6,3)\} \text{ repeats the first coordinate } 5$

So it is a function whose swap is not. The set with a repeated input $1$ is not even a function of $x$; the other two swap to functions.

The output $\dfrac{1}{x}$ can be any nonzero value, but never $0$.

$\frac{1}{x} = y \ne 0 \;\Rightarrow\; x = \frac{1}{y}$

Every nonzero $y$ is reached, and no input makes $\dfrac{1}{x} = 0$, so the range is all real numbers except $0$.

A nonconstant linear function is one-to-one because it never repeats an output.

$2a - 7 = 2b - 7 \;\Rightarrow\; a = b$

Each of the others repeats an output (for instance both $c$ and a partner value give the same result), so only $2x - 7$ is one-to-one on all reals.

A formula names a relation, but only the one-output-per-input test decides whether it is a function.

$x = 3 \;\Rightarrow\; y = \pm 4$

One input gives two outputs, so the circle fails the vertical line test and is not a function of $x$, formula or not.

The definition of a function asks only for one output per input, not for a formula.

$\text{rational} \mapsto 1, \qquad \text{irrational} \mapsto 0$

Every real input gets exactly one of the two outputs, so it is a perfectly good function, even though no analytic expression of the kind Euler had in mind can describe it.

On $x \le 0$ the square is strictly decreasing, so it never repeats an output.

$a, b \le 0, \; a^2 = b^2 \;\Rightarrow\; a = b$

Thus $g$ is one-to-one while $f$ (which has $f(2) = f(-2)$) is not. Same formula, different domain, different function.

The horizontal line test is the vertical line test applied to $y$ as the input.

$\text{swap } (x, y) \to (y, x)$

Passing it means the swapped relation is itself a function, which is exactly the condition an inverse rule requires. The inverse-functions lesson develops this fully.

The absolute value is never negative, so subtracting it can only lower the output from $3$.

$-|x - 1| \le 0 \;\Rightarrow\; 3 - |x - 1| \le 3$

The maximum $3$ is reached at $x = 1$, and the outputs fall without bound, so the range is $y \le 3$.