What Is a Function?: Free Response

Substitute each number for $x$ and simplify.

$f(4) = 4^2 - 2(4) = 16 - 8 = 8.$

$f(-1) = (-1)^2 - 2(-1) = 1 + 2 = 3.$

Replace every $x$ in $x^2-2x$ with the block $(k+1)$, keeping it inside parentheses until it is expanded.

$f(k+1) = (k+1)^2 - 2(k+1).$

Expand each piece separately and combine.

$(k+1)^2 - 2(k+1) = \left(k^2+2k+1\right) - \left(2k+2\right) = k^2 - 1.$

Compute the second expression to compare against.

$f(k) + 1 = \left(k^2-2k\right)+1 = k^2-2k+1.$

Against part B's $f(k+1)=k^2-1$, the two agree only where $k^2-2k+1=k^2-1$, that is where $-2k+1=-1$, or $k=1$; for every other $k$ they differ. The difference is procedural, not just numerical: $f(k+1)$ replaces $x$ with $k+1$ everywhere the rule uses it, including inside the squared term, while $f(k)+1$ leaves the rule's own output alone and tacks on a separate $+1$ afterward.

The flat fee is paid regardless of time, and the hourly charge multiplies the rate by the number of hours.

$C(h) = 40 + 25h.$

Substitute $h=3$:

$C(3) = 40 + 25(3) = 40 + 75 = 115.$

A $3$-hour rental costs $115$ dollars.

As an algebraic expression, $40+25h$ accepts any real number for $h$: negative hours, zero hours, fractions of an hour all produce a value with no error. But the situation being modeled restricts $h$ further. A rental duration cannot be negative, and a genuine rental involves some positive amount of time, so the sensible domain is

$h > 0.$

The formula's own domain, all real numbers, is wider than what the situation allows; the model's domain is the situation's restriction applied on top of the formula's.

Substitute directly.

$C(0) = 40 + 25(0) = 40 + 0 = 40.$

The flat fee term does not depend on $h$ at all, so it survives no matter what is substituted, including $0$. The formula has no way to know that $h=0$ represents a rental that never took place; it simply computes.

That is exactly the gap part B pointed at. The formula's own domain includes $h=0$, since nothing about the expression breaks there. But the contextual domain from part B excluded it. So $h=0$ is a case where the formula produces a perfectly good number while the situation says that input should not be used at all.

Choose a rule where two different inputs land on the same result, then check the definition on the input side, not the output side.

Let $g(x) = (x-3)^2$.

$g(1) = (1-3)^2 = (-2)^2 = 4, \qquad g(5) = (5-3)^2 = 2^2 = 4.$

Both inputs give the output $4$. The definition of a function only asks whether any single input has two different outputs. Here $1$ has exactly one output, and $5$ has exactly one output; no input is paired with two different values, so the definition holds regardless of the repeated output.

State the definition carefully: a relation is a function when every input is paired with exactly one output. Read closely, this constrains inputs, not outputs. Nothing in that sentence says an output value must be used only once.

So a repeated output, as in part A, changes nothing: two different inputs may perfectly well share the same output. What WOULD break the definition is the opposite pairing, one input tied to two different outputs, such as a relation containing both

$(2,7) \quad \text{and} \quad (2,11).$

There the input $2$ has two outputs, and that single input is what the definition rules out.

Check the inputs first, since that is what the definition constrains.

$-3 \mapsto 10, \qquad 0 \mapsto 4, \qquad 3 \mapsto 10.$

The three inputs are $-3$, $0$, and $3$, all different from one another, and each appears exactly once in the set. No input is paired with two outputs, so by the definition this set is a function.

Now check the outputs separately, since that is a different question. The outputs are $10$, $4$, $10$: the value $10$ appears twice, coming from the two different inputs $-3$ and $3$. So yes, an output repeats. But that fact belongs to a different question than functionhood, and by part B's reasoning it changes nothing about the first conclusion.

A square root needs a nonnegative quantity underneath it, so set up and solve that inequality.

$2x - 6 \ge 0 \quad \Longrightarrow \quad 2x \ge 6 \quad \Longrightarrow \quad x \ge 3.$

Two separate restrictions apply here, and both must be satisfied.

The square root needs $10-x \ge 0$:

$10 - x \ge 0 \quad \Longrightarrow \quad x \le 10.$

The denominator must not be zero:

$x + 4 \ne 0 \quad \Longrightarrow \quad x \ne -4.$

Both restrictions apply at once, so the domain is $x \le 10$ and $x \ne -4$.

The squared term $(x+1)^2$ can never be negative, for any real $x$:

$(x+1)^2 \ge 0.$

Subtracting a nonnegative quantity from $9$ can only decrease $9$ or leave it unchanged, so

$r(x) = 9 - (x+1)^2 \le 9.$

The greatest possible output, $9$, happens exactly when the squared term is $0$, at $x=-1$. So the range is every real number $y \le 9$.

Test $x=15$ against each restriction separately.

$15 \ne -4 \ \text{(passes denominator)}, \qquad 10-15 = -5 < 0 \ \text{(fails square root)}.$

A domain check that looked only at the denominator would wrongly admit it.

Test $x=-4$ the other way.

$-4 \le 10 \ \text{(passes square root)}, \qquad -4+4 = 0 \ \text{(fails denominator)}.$

A domain check that looked only at the square root would wrongly admit it. Each restriction catches inputs the other one misses, so a domain built from only one of them is too generous in exactly the direction the other one guards against. Both restrictions have to be applied and their results combined for the domain to be correct.

Substitute $x=4$ into the rule exactly as written, before simplifying anything.

$f(4) = \frac{4^2-9}{4-3} = \frac{16-9}{4-3} = \frac{7}{1} = 7.$

This matches the value given at the start of the question.

Substitute $x=10$ into the original rule.

$f(10) = \frac{10^2-9}{10-3} = \frac{100-9}{7} = \frac{91}{7} = 13.$

Factor the numerator to see where the simplification comes from.

$f(x) = \frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3}.$

Canceling the factor of $x-3$ is only legal where that factor is not zero, that is where $x \ne 3$; dividing by zero was never allowed, and the cancellation step does not repeal that. So the simplified rule $x+3$ agrees with $f$ everywhere $f$ is defined, but it says nothing about $x=3$, because that value was excluded before the cancellation ever happened.

At $x=3$ the original rule reads $\dfrac{0}{0}$, undefined, while the simplified form $x+3$ would happily return $6$. That gap is the point: a simplified rule can look defined at a value the original rule never accepted, so the domain has to be read from the ORIGINAL expression, before any canceling. The domain of $f$ is all real numbers except $x=3$.