Relations and Functions: Additional Practice (Set 2)

The parabola has its vertex at $x = 3$ and is one-to-one on each side of it.

$a, b \ge 3, \; (a-3)^2 = (b-3)^2 \;\Rightarrow\; a = b$

The largest interval that stays on one side and still contains $5$ is $x \ge 3$.

Assume two inputs give the same output, $f(a) = f(b)$, so $a^3 + a = b^3 + b$. Move everything to one side and group the terms.

$(a^3 - b^3) + (a - b) = 0$

Factor $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ and pull out the shared factor $a - b$.

$(a - b)(a^2 + ab + b^2 + 1) = 0$

Complete the square on the second factor: $a^2 + ab + b^2 = \left(a + \tfrac{b}{2}\right)^2 + \tfrac{3}{4}b^2 \ge 0$, so $a^2 + ab + b^2 + 1 \ge 1$ and can never be zero. That forces $a - b = 0$, hence $a = b$: different inputs give different outputs, and $f$ is one-to-one on all reals.

The numerator root needs a nonnegative inside; the denominator root needs a strictly positive inside.

$2 - x \ge 0 \text{ and } x + 1 > 0 \;\Rightarrow\; x \le 2 \text{ and } x > -1$

The overlap is $-1 < x \le 2$; the left end is open because $x = -1$ makes the denominator zero.

The square root increases across the domain, so the endpoints map to the range's endpoints.

$\sqrt{1} = 1 \quad \text{and} \quad \sqrt{9} = 3$

Every value between is achieved, so the range is $1 \le y \le 3$.

Factor and cancel, but only where the cancelled factor is nonzero.

$\frac{(x-3)(x+3)}{x-3} = x + 3, \quad x \ne 3$

The denominator forces $x \ne 3$, so $f$ equals $x + 3$ on its domain but has a hole at $x = 3$. The plain rule $x + 3$ has a larger domain, so the two are different functions.

Look for two inputs that share an output.

$1^4 = 1 = (-1)^4$

The inputs $1$ and $-1$ both give $1$, so $x^4$ is not one-to-one. The linear rules and $x^3$ never repeat an output.

The inside of the root must be nonnegative, and the output runs from the largest root down to zero.

$9 - x^2 \ge 0 \;\Rightarrow\; -3 \le x \le 3, \qquad 0 \le \sqrt{9 - x^2} \le 3$

The root peaks at $3$ when $x = 0$ and drops to $0$ at $x = \pm 3$, so the range is $0 \le y \le 3$.

Solve for the output $y$.

$xy = 12 \;\Rightarrow\; y = \frac{12}{x}, \quad x \ne 0$

Each nonzero $x$ gives one $y$, so it is a function of $x$; the input $x = 0$ has no output at all, so it is excluded from the domain.

At the single input $x = 2$ there are many outputs, but each $y$ corresponds to just $x = 2$.

$x = 2 \text{ for every } y \in [0, 1]$

So reading $x$ as the input fails, while reading $y$ as the input succeeds: it is a function of $y$ but not of $x$.

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

$16 - x^2 > 0 \;\Rightarrow\; -4 < x < 4$

The endpoints $x = \pm 4$ make the denominator zero, so they are excluded and the interval is open.

The domain is fixed before any algebra, so values that make a denominator zero are barred from the start.

$x = 2 \text{ excluded} \;\Rightarrow\; \text{any candidate } x = 2 \text{ is extraneous}$

Clearing denominators can produce such a candidate, but the domain, part of the problem's data, rules it out.

Simplify $g$ but track where it is undefined.

$g(x) = \frac{(x - 3)(x + 3)}{x - 3} = x + 3, \quad x \ne 3$

The function $f$ accepts every real number, while $g$ excludes $3$. Different domains make them different functions, even though the formula simplifies to the same thing.