Function Notation and Evaluation: Free Response

Follow the description in the order it is given. The input is added to $3$ first, and only the result of that is squared. Writing the input as $x$,

$f(x) = (x + 3)^2.$

To evaluate at $-10$, put $-10$ in the slot, wrapped in parentheses so the sign survives:

$f(-10) = (-10 + 3)^2 = (-7)^2 = 49.$

To evaluate at $2w$, copy the whole expression $2w$ into the slot. The slot squares everything it holds, so the entire sum $2w + 3$ is squared:

$f(2w) = (2w + 3)^2 = 4w^2 + 12w + 9.$

The middle term $12w$ is the piece lost by squaring $2w$ and $3$ separately, which is what happens when the input is not kept inside parentheses.

Because $f(x)$ is a number, the statement $f(x) = 81$ is an ordinary equation in the input:

$(x + 3)^2 = 81.$

Undoing the square keeps both signs, so $x + 3 = 9$ or $x + 3 = -9$:

$x = 6 \qquad \text{or} \qquad x = -12.$

Check both against the rule: $(6 + 3)^2 = 81$ and $(-12 + 3)^2 = 81$. Both numbers are inputs, not outputs. Each is a number the rule sends to $81$, and there is nothing odd about two different inputs sharing one output.

Three different objects hide behind one letter.

$f$ is the function itself: the rule together with the set of inputs it accepts. It is not a single output number. That does not make it untouchable: expressions like $2f$ and $f + 5$ do have a meaning, the pointwise one, and wherever that reading is defined what they name is another function, the rule sending $x$ to $2f(x)$ or to $f(x) + 5$. What they never name is a number. $f(x)$ is the other kind of object. It is a number, the output the rule produces at the input $x$. And $f(-10)$ is one particular such number, namely $49$. The parentheses mark an input; they never mean multiplication, so $f(-10)$ is not $-10$ times $f$.

Now the letter. Picture the rule with a box where the input goes:

$f(\,\square\,) = (\square + 3)^2.$

The box does not care what letter is written inside it, only what is put in. Writing $f(s) = (s + 3)^2$ accepts the same inputs, sends them into the same codomain, and at any input $c$ returns

$f(c) = (c + 3)^2$

either way. Same domain, same codomain, same value at every input, so it is the same function. Only the writing changed, which is why the input letter is safe to rename whenever another letter is more convenient.

Expand each of the three values, keeping every input wrapped in parentheses while it is substituted:

$f(a + b) = 3(a + b)^2 - 2(a + b) = 3a^2 + 6ab + 3b^2 - 2a - 2b.$

Now subtract $f(a) = 3a^2 - 2a$ and $f(b) = 3b^2 - 2b$. Every squared term and every linear term cancels, and exactly one term survives:

$f(a + b) - f(a) - f(b) = 6ab.$

The linear part $-2x$ cancels on its own, which is the lesson's result about $f(x) = mx$ showing up inside a larger rule. The squared part is what opens the gap.

For a specific failure, take $a = 2$ and $b = 5$:

$f(7) = 133, \qquad f(2) + f(5) = 8 + 65 = 73,$

and the two differ by $60$, which is $6(2)(5)$ as predicted.

Finally, additivity holds at exactly the pairs where the surviving term is zero. A product of two real numbers is zero exactly when one of the factors is zero, so

$6ab = 0 \iff a = 0 \ \text{ or } \ b = 0.$

So the identity does hold at infinitely many pairs, and every one of them contains a zero. No pair of nonzero numbers satisfies it, which is why finding a few pairs that work would be no evidence at all.

Absolute value is defined by cases: $\lvert u \rvert$ is $u$ when $u \ge 0$, and $-u$ when $u < 0$. Take any constant $k \ge 0$ and check the two signs of the input separately.

If $x \ge 0$, then $kx \ge 0$ as well, since multiplying a nonnegative number by a nonnegative number cannot make it negative, so

$g(kx) = kx = k\lvert x \rvert = k\,g(x).$

If $x < 0$, the constant has to be split as well, because $k = 0$ would send $kx$ to $0$, and the definition hands $0$ to its first branch, not its second. Take $k = 0$ first: then $kx = 0$, so

$g(kx) = g(0) = 0 = 0 \cdot \lvert x \rvert = k\,g(x).$

Now take $k > 0$ with $x < 0$. A positive constant times a negative number is strictly negative, so $kx < 0$ and absolute value returns its opposite:

$g(kx) = -(kx) = k(-x) = k\lvert x \rvert = k\,g(x).$

Every case ends at $k\,g(x)$, and every pairing of a nonnegative constant with a real input falls in one of them, so the identity holds at every input as long as the constant is nonnegative.

A negative constant breaks it. Take $k = -1$ and $x = 6$:

$g(-6) = 6, \qquad (-1)\,g(6) = -6.$

The left side is an absolute value and can never be negative, while the right side is negative, so the two sides disagree at every nonzero input once the constant is negative.

Putting $x = 1$ converts a statement about constants into a statement about inputs:

$f(k \cdot 1) = k\,f(1), \qquad \text{that is} \qquad f(k) = f(1)\,k.$

The left side is the value of $f$ at the input $k$. So each constant the identity is granted delivers the value of $f$ at that one input, and no others. For this substitution alone to reach $f(x) = mx$ at every real input, the identity has to be available at every real constant.

Part B's rule is the case that shows what a shortage costs. It satisfies the identity at every constant $k \ge 0$, so the argument does apply there and reports $g(x) = g(1)\,x = x$ for $x \ge 0$, which is correct: $\lvert x \rvert$ really is $x$ on the nonnegative inputs. The argument reaches only the nonnegative inputs, because only the nonnegative constants are available to substitute; nothing forces $g(x) = x$ at a negative input. Indeed,

$g(-6) = 6, \qquad \text{not} \qquad -6.$

So a rule can agree with a through-origin line on half the number line and part company with it on the other half, while satisfying the scaling identity everywhere it was asked to.

None of this weakens the lesson's characterization. It locates the hypothesis the argument leans on: this substitution has to be granted every constant, negative ones included, before it delivers $f(x) = mx$ at every input.

Evaluate the rule at each time, wrapping each input before squaring:

$V(3) = 300 - 2(3)^2 = 282, \qquad V(7) = 300 - 2(7)^2 = 202.$

The volume changed by $202 - 282 = -80$ liters over the $4$ minutes between the two times, so

$\frac{V(7) - V(3)}{7 - 3} = \frac{-80}{4} = -20.$

The average rate of change is $-20$ liters per minute. The sign is part of the answer: a negative rate says the tank held less at the later time, losing $20$ liters in an average minute of that interval.

Copy the whole expression $t + h$ into the slot and square all of it:

$V(t + h) = 300 - 2(t + h)^2 = 300 - 2t^2 - 4th - 2h^2.$

That first step already carries a condition. The rule was declared only for $0 \le t \le 12$, so $V(t + h)$ names a volume only when the later time lies in that range too, which is a second requirement: $0 \le t + h \le 12$.

Subtract $V(t) = 300 - 2t^2$. The constant and the squared term cancel, and every term that survives carries a factor of $h$, which is the check that the expansion was done correctly:

$V(t + h) - V(t) = -4th - 2h^2.$

Dividing by $h$ is a division, legitimate exactly when $h \ne 0$:

$\frac{-4th - 2h^2}{h} = -4t - 2h, \qquad 0 \le t \le 12, \quad 0 \le t + h \le 12, \quad h \ne 0.$

All three conditions travel with the simplified expression, and they come from two different places. The division supplies $h \ne 0$. The model supplies the other two, and they are not decoration: at $t = 11$ with $h = 2$ the expression $-4t - 2h$ still returns $-48$, while the tank was never modelled at $13$ minutes, so there is no average rate there for it to be reporting.

Substituting $t = 3$ and $h = 4$ gives $-4(3) - 2(4) = -20$. Both times, $3$ and $7$, lie inside $0 \le t \le 12$, so that substitution is one the model allows, and it describes the interval running from $3$ minutes to $7$ minutes, so the general expression agrees with the direct calculation.

Two things to separate, and the second is the one that gets skipped.

What the quotient measures. The numerator $V(t + h) - V(t)$ is a change in volume, in liters, and the denominator is the change in time that produced it, in minutes. The quotient is therefore in liters per minute: the average rate of change of the volume over the interval between the times $t$ and $t + h$, both of which the model has to reach, so $0 \le t \le 12$ and $0 \le t + h \le 12$. It is also the slope of the straight line joining the two points $(t, V(t))$ and $(t + h, V(t + h))$, the secant line through them. A negative value records that the tank held less at the later time, which for a draining tank is what we expect.

Why $h = 0$ stays excluded. The simplified expression and the quotient it came from are not the same expression. At $h = 0$ the quotient reads

$\frac{V(t) - V(t)}{0} = \frac{0}{0},$

which is undefined, while the simplified form evaluates there without complaint. The two agree wherever the quotient is defined, and part company at $h = 0$, so the restriction has to travel with the simplified form. Without it, the simplified expression would be claiming a value the quotient never had.

The model's range restricts it the same way, for a different reason. The rule was declared only for $0 \le t \le 12$, so the quotient compares two volumes only when $t$ and $t + h$ both land in that range. At $t = 11$ with $h = 2$ the simplified form returns $-48$ without hesitating, while $V(13)$ was never defined, so once again the simplified expression would be reporting a rate the quotient never had. The full set of conditions is therefore $h \ne 0$ together with $0 \le t \le 12$ and $0 \le t + h \le 12$.

The situation says the same thing. An average rate is a change in volume divided by the time it took, and at $h = 0$ there are not two different times to compare, so there is nothing to average; outside the modelled range there is no volume to compare either.

Expand the product and collect like terms:

$(x - 2)(x^2 + 2x + 4) = x^3 + 2x^2 + 4x - 2x^2 - 4x - 8 = x^3 - 8.$

So for every $x \ne 2$ the factor $x - 2$ cancels and $p(x) = x^2 + 2x + 4 = q(x)$. The two rules agree at every input where both are defined.

That is not enough. The equality test asks for the same domain, the same codomain, and the same value at every input of that domain, so check the domains. The formula for $q$ accepts every real number, so $2$ is in its domain, with $q(2) = 4 + 4 + 4 = 12$. The formula for $p$ carries $x - 2$ in the denominator, so at that input it reads

$p(2) = \frac{0}{0},$

which is undefined, and $2$ is not in the domain of $p$. One rule accepts an input the other refuses, so the two domains are different sets. Both rules were declared into the real numbers, so the codomains match, and the values match wherever both are defined, but a single failed condition is enough: $p$ and $q$ are different functions.

The mismatch is one input wide: $p$ refuses $2$ and $q$ accepts it. A domain is part of a function, and a declared domain travels with it, so declare $q$ on the set $p$ already has:

$q : \{x : x \ne 2\} \to \mathbb{R}, \qquad q(x) = x^2 + 2x + 4.$

Now run the test on the repaired pair. Same domain: both are every real number except $2$, by construction. Same codomain: both were declared into the real numbers. Same values: at every $x \ne 2$ the factorization gives $p(x) = q(x)$, and there is no longer any input at which only one of the two is defined.

All three conditions hold, so the repaired $q$ is the same function as $p$. Notice what the repair did not touch. No formula changed, because the formulas were never the problem; what was wrong was the set of inputs one of them had been handed.

Start with what the shortcut gets right. If one formula simplifies to the other, then at every input where both are defined the two return the same number, so the values condition of the equality test is satisfied. That part is real, and it is why simplification feels conclusive.

Where it breaks is that simplifying is a statement about formulas, while a function is a rule together with a domain. The damage is done by the cancellation itself: to cancel a factor from a fraction, that factor must not be zero, so the cancelling step quietly divides by something the original expression had already refused to divide by. The formula that survives accepts an input the original never did:

$q(2) = 12, \qquad p(2) \ \text{is undefined}.$

So two rules can pass the values condition and still be different functions, which is exactly what part A found.

A correct version has to say what the test says: two functions are equal exactly when they have the same domain, the same codomain, and the same value at every input of that domain. Simplification delivers only the last of the three, so the domains have to be compared separately, before or after, but never skipped. Part B is the same fact read forward: make the domains agree, and the shortcut's conclusion becomes true.

Test the input against the conditions before computing anything. It satisfies $x \le 1$, and it satisfies $1 \le x \le 4$, so two cases claim it. Evaluate both:

$7 - 2(1) = 5, \qquad (1)^2 + 4 = 5.$

The two cases agree. The input $1$ is therefore handed a single output, $5$, by two routes, rather than two different outputs by two routes, so $w(1) = 5$ and nothing about this input stops $w$ being a function. An overlap becomes a defect only when the overlapping cases disagree.

Walk the number line against the three conditions rather than testing scattered inputs. Every $x \le 1$ is claimed by the first case. Every $x$ with $1 \le x \le 4$ is claimed by the second. Every $x > 5$ is claimed by the third. That leaves

$4 < x \le 5$

claimed by no case at all. The inputs $4.5$ and $5$, for instance, receive no output, so $w$ as written does not define a function on all of the real numbers.

The gap sits between where the second case stops and where the third case starts, so move the third case's threshold down to meet the second:

$3x + 8 \quad \text{if } x > 4.$

Check coverage again: every real number is at most $1$, or between $1$ and $4$ inclusive, or greater than $4$, so every input is claimed. Check for conflicts as well: the first two cases share only $x = 1$, where both give $5$; the second and third share nothing, since $x \le 4$ and $x > 4$ cannot hold together; the first and third share nothing either. No input receives two different outputs, so the repaired list does define a function on all of the real numbers.

The repair is not unique. Extending the second case to $1 \le x \le 5$ instead also changes one condition, also covers every real number, and also creates no conflict, since the third case still starts strictly above $5$.

Overlap is not the test; disagreement on an overlap is. First find where both conditions of $z$ hold: $x \le 1$ and $x \ge 0$ hold together at exactly the inputs with $0 \le x \le 1$, a whole interval rather than a single point.

Now compare the formulas there. At $x = 0$,

$7 - 2(0) = 7, \qquad (0)^2 + 4 = 4.$

The list says that $z(0)$ is $7$, and it also says that $z(0)$ is $4$. A function pairs each input with exactly one output, so this list has not defined anything at $x = 0$; it has given two competing definitions, and the same happens at every input of the shared interval where the formulas disagree.

That is precisely what does not happen in $w$. Its first two cases also overlap, but only at the single input $x = 1$, and both cases return $5$ there, so the two routes deliver the same output and every input still has exactly one. Whether an overlap is fatal depends on the values the overlapping cases produce, never on the bare fact that they overlap.

Repairing $z$ takes the same kind of edit as repairing $w$: one of the two conditions has to give up the disputed interval. Requiring $x > 1$ for the second case, for instance, leaves every input claimed by exactly one case.