Relations and Functions: Additional Practice (Set 1)

The inside of the root must be nonnegative, so factor it and find where the product is at least zero.

$(x - 2)(x - 3) \ge 0 \;\Rightarrow\; x \le 2 \text{ or } x \ge 3$

A product of two factors is nonnegative outside the interval between its roots, giving the two-piece domain.

The denominator is at least $1$, so the fraction is positive and at most $1$.

$x^2 + 1 \ge 1 \;\Rightarrow\; 0 < \frac{1}{x^2 + 1} \le 1$

The value $1$ is reached at $x = 0$, and the outputs approach $0$ without ever equalling it, so the range is $0 < y \le 1$.

Each of the three inputs must have exactly one output, and the only output available is $7$.

$\{(1,7),(2,7),(3,7)\}$

This constant function is perfectly valid: repeating an output is allowed, since each input still has a single output.

Isolate the $y$-root and square once.

$\sqrt{y} = 4 - \sqrt{x} \;\Rightarrow\; y = (4 - \sqrt{x})^2, \quad 0 \le x \le 16$

Each allowed $x$ produces exactly one $y$, so it is a function of $x$ on $0 \le x \le 16$.

On this domain the square is smallest at $x = 0$ and largest at the far endpoint $x = -3$.

$0^2 = 0 \quad \text{and} \quad (-3)^2 = 9$

Since $x = 0$ lies in the domain, the minimum is $0$, and the maximum is $9$, so the range is $0 \le y \le 9$.

Swapping the coordinates turns each output into a first coordinate, so the swap is a function when no output was repeated.

$(a, c) \text{ and } (b, c) \text{ with } a \ne b \;\Rightarrow\; \text{swap repeats the input } c$

No repeated output means one-to-one, so the swap is a function exactly when the original is one-to-one.

The root is nonnegative on the domain $x \ge 3$, so it only adds to $2$.

$\sqrt{x - 3} \ge 0 \;\Rightarrow\; 2 + \sqrt{x - 3} \ge 2$

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

The rule must give one output for every real input and never repeat an output.

$a^3 + 1 = b^3 + 1 \;\Rightarrow\; a = b$

The shifted cube is defined for all $x$ and strictly increasing, so it is one-to-one. The squares and absolute value repeat outputs, and $\dfrac{1}{x}$ is undefined at $x = 0$.

Each input $x$ has the single output $5$, but the input $y = 5$ corresponds to every $x$.

$x \mapsto 5 \text{ for all } x$

Reading $x$ as input gives a function; reading $y$ as input fails because $y = 5$ has infinitely many outputs. So it is a function of $x$ but not of $y$.

Solve $y^2 = 0$ for the outputs.

$y^2 = 0 \;\Rightarrow\; y = 0$

The input $0$ has a single output. The relation still fails to be a function of $x$ because other inputs (like $x = 4$) have two outputs; only one bad input is needed.

Each input gets exactly one of the two outputs, so it is a function; but many inputs share an output.

$f(1) = 1 = f(2)$

Two different inputs give the same output, so it is not one-to-one, though it is a perfectly valid function with range $\{-1, 1\}$.

A function is a relation with the extra condition of one output per input.

$\text{functions} \subseteq \text{relations}$

So every function is a relation, while a relation such as a circle need not be a function.