Piecewise Functions: Free Response

Test each input against the three conditions in order, and use only the piece that owns it.

For $x = -10$: this satisfies $x < -4$, so the first piece applies.

$g(-10) = 3(-10) + 2 = -28.$

For $x = -4$: this fails $x < -4$ but satisfies $-4 \le x < 6$ because of the 'or equal to', so the middle piece owns it.

$g(-4) = (-4)^{2} - 1 = 16 - 1 = 15.$

For $x = 2$: this satisfies $-4 \le x < 6$, so the middle piece applies again.

$g(2) = 2^{2} - 1 = 3.$

For $x = 6$: this fails $-4 \le x < 6$ (the strict $<$ excludes $6$) but satisfies $x \ge 6$, so the third piece owns it.

$g(6) = 30.$

List the three conditions and check them against each other directly, rather than testing scattered examples.

Do they overlap anywhere? The first ends with a strict $<-4$ and the second begins with $-4\le$, so $-4$ belongs to the second only. The second ends with a strict $<6$ and the third begins with $6\le$, so $6$ belongs to the third only. No number satisfies two conditions at once.

Do they leave a gap anywhere? Any number below $-4$ satisfies the first. Any number from $-4$ up to just below $6$ satisfies the second. Any number at or above $6$ satisfies the third.

$\{x < -4\} \cup \{-4 \le x < 6\} \cup \{x \ge 6\} = \text{all real numbers.}$

Since the three conditions neither overlap nor leave a gap, they partition the domain, and the domain of $g$ is all real numbers.

Test each length of stay against the three tier conditions in order.

For $t = 2$: this satisfies $0 < t \le 2$ because of the 'or equal to', so the first tier owns it.

$P(2) = 8 \text{ dollars.}$

For $t = 6$: this fails $0 < t \le 2$ but satisfies $2 < t \le 6$, so the second tier owns it.

$P(6) = 14 \text{ dollars.}$

For $t = 6.5$: this fails the first two conditions but satisfies $t > 6$, so the third tier owns it.

$P(6.5) = 22 \text{ dollars.}$

The domain is whatever the three conditions cover: $0 < t \le 2$, $2 < t \le 6$, and $t > 6$ together are exactly $t > 0$, so the domain is $t > 0$.

For the range, take each piece on its own. Every piece here is a constant, and a constant formula returns the same number for every input in its interval, however wide that interval is.

$\text{first piece: always } 8, \quad \text{second piece: always } 14, \quad \text{third piece: always } 22.$

Union the three single values rather than three intervals:

$\text{range} = \{8, 14, 22\}.$

A piece that VARIES with $t$, such as a linear or quadratic piece, would instead sweep through a whole interval of outputs; a constant piece never does, because it ignores $t$ entirely.

Check $t = 2$ against each new condition on its own, rather than assuming it stays with whichever tier it used to belong to.

Against the new first condition, $0 < t < 2$: this is a strict inequality on both sides, and $2$ is not less than $2$, so it fails.

Against the unchanged second condition, $2 < t \le 6$: this is also strict on the left, and $2$ is not greater than $2$, so it fails too.

$0 < 2 < 2 \text{ is false}, \qquad 2 < 2 \le 6 \text{ is false.}$

So $t = 2$ satisfies neither piece's condition, and the rule gives no charge for a customer parked exactly two hours. A function must assign an output to every input in its intended domain, so a pair of conditions that leaves an input with no piece is not a valid description of one; it needs an 'or equal to' on one side of the shared boundary, just as the original pair had.

The two dots are not a separate fact to memorize; they follow directly from which condition contains $b$ itself.

The first condition, $x \le b$, includes $b$, so $b$ belongs to the first piece and

$m(b) = p(b).$

That point is genuinely on the graph, so it gets the closed dot at $\bigl(b, p(b)\bigr)$.

The second condition, $x > b$, excludes $b$, so the second piece never actually supplies a value at $x = b$; it only tells you what the piece is heading toward as $x$ approaches $b$ from the right. That approached-but-not-reached value, $q(b)$, gets the open dot at $\bigl(b, q(b)\bigr)$.

The assignment could not run the other way, because $m(b)$ is a single, definite number (the function has exactly one output at $b$), and that number is whatever the piece owning $b$ says it is, namely $p(b)$.

Start from what the two dots ARE, established in part A: the closed dot is always $\bigl(b, p(b)\bigr)$ and the open dot is always $\bigl(b, q(b)\bigr)$.

$\text{closed dot} = \bigl(b, p(b)\bigr), \qquad \text{open dot} = \bigl(b, q(b)\bigr).$

Connects, forward direction. Suppose $p(b) = q(b)$. Then the closed dot and the open dot share both coordinates, so they are literally the same point. That point is already on the graph (the closed dot says so), so no height is skipped at $x = b$, and the graph connects with no gap.

Connects, converse. Suppose instead that $m$ connects at $b$, meaning the two dots coincide. Two points coincide only when their coordinates match, and both share the first coordinate $b$ already, so their second coordinates must match too: $p(b) = q(b)$.

Jumps, forward direction. Suppose $p(b) \ne q(b)$. The closed dot sits at height $p(b)$ and the open dot sits at the different height $q(b)$. At $x = b$ itself the function equals $p(b)$ only, so the graph shows a gap between the two heights there, a jump.

Jumps, converse. Suppose $m$ jumps at $b$, meaning the two dots sit at different heights. Since those heights are $p(b)$ and $q(b)$ by definition, different heights means $p(b) \ne q(b)$.

$\text{connects} \iff p(b) = q(b), \qquad \text{jumps} \iff p(b) \ne q(b).$

Each direction followed from what the two dots ARE, so the rule holds for every two-piece function built this way, not only the specific ones you have graphed.

Evaluate each formula at the boundary $b = 4$, the same computation the general rule is built on.

$p(4) = \sqrt{4 + 5} = \sqrt{9} = 3, \qquad q(4) = 4 - 1 = 3.$

The two heights agree, $p(4) = q(4) = 3$. Part B proved that this equality is exactly when a two-piece function connects at its boundary, so no picture is needed: $n$ connects at $x = 4$, with the closed dot (from the radical piece, whose condition $x \le 4$ includes the boundary) and the open dot both landing on $(4, 3)$.

Test each input against the three conditions before substituting.

For $x = -3$: this satisfies $x < 1$, so the absolute-value piece applies.

$m(-3) = \lvert -3 - 6 \rvert = \lvert -9 \rvert = 9.$

For $x = 1$: this fails $x < 1$ but satisfies $1 \le x \le 3$, so the quadratic piece owns it.

$m(1) = 2(1)^{2} - 5 = 2 - 5 = -3.$

For $x = 3$: this satisfies $1 \le x \le 3$ (the boundary carries the 'or equal to' here too), so the quadratic piece owns it again.

$m(3) = 2(3)^{2} - 5 = 18 - 5 = 13.$

For $x = 5$: this fails $1 \le x \le 3$ but satisfies $x > 3$, so the constant piece applies.

$m(5) = 13.$

At each boundary, evaluate both pieces, even the one the condition there excludes.

At $x = 1$: the excluded absolute-value piece approaches $\lvert 1 - 6 \rvert = 5$, while the included quadratic piece reaches $2(1)^{2} - 5 = -3$.

$5 \ne -3 \implies m \text{ jumps at } x = 1.$

The quadratic piece includes $x = 1$ (its condition is $1 \le x \le 3$), so it gets the closed dot at $(1, -3)$; the absolute-value piece gets the open dot at $(1, 5)$.

At $x = 3$: the included quadratic piece reaches $2(3)^{2} - 5 = 13$, while the excluded constant piece approaches $13$ as well.

$13 = 13 \implies m \text{ connects at } x = 3.$

The quadratic piece includes $x = 3$, so it again gets the closed dot, this time at $(3, 13)$, and the constant piece's open dot lands on the very same point.

The three conditions $x < 1$, $1 \le x \le 3$, and $x > 3$ cover every real number with no overlap and no gap, so the domain of $m$ is all real numbers.

For the range, restrict each piece to its own interval. The quadratic piece is increasing across $[1, 3]$ (its vertex at $x = 0$ sits to the left of the whole interval), so it sweeps every value from its value at $x = 1$ up to its value at $x = 3$:

$2(1)^{2} - 5 = -3, \qquad 2(3)^{2} - 5 = 13 \implies \text{outputs } [-3, 13].$

For $x < 1$, $\lvert x - 6 \rvert = 6 - x$, which decreases as $x$ increases (larger $x$ means smaller $6 - x$). As $x$ climbs from $-\infty$ toward $1$ (never reaching it), $6 - x$ falls from $+\infty$ toward $5$ (never reaching it either), so this piece contributes every value strictly above $5$: outputs $(5, \infty)$.

The constant piece for $x > 3$ contributes only $\{13\}$, already inside $[-3, 13]$.

Union the three sets:

$[-3, 13] \cup (5, \infty) \cup \{13\} = [-3, \infty),$

since $(5, \infty)$ overlaps $[-3, 13]$ between $5$ and $13$ and extends past it, leaving no gap above $-3$. So the range is $y \ge -3$.

Test each usage against the three tier conditions in order.

For $x = 3$ and $x = 4$, both satisfy $0 \le x \le 4$ (the second by the 'or equal to'), so the flat fee applies to both.

$D(3) = 25, \qquad D(4) = 25.$

For $x = 7$, this satisfies $4 < x \le 10$, the overage tier.

$D(7) = 25 + 6(7 - 4) = 25 + 18 = 43.$

For $x = 10$, this also satisfies $4 < x \le 10$, by the 'or equal to' on the right.

$D(10) = 25 + 6(10 - 4) = 25 + 36 = 61.$

For $x = 12$, this satisfies $x > 10$, the flat cap.

$D(12) = 82.$

Evaluate both pieces at each boundary, even the one the condition there excludes.

At $x = 4$: the included flat fee is $25$, and the excluded overage formula approaches $25 + 6(4 - 4) = 25$ as well.

$25 = 25 \implies D \text{ connects at } x = 4.$

That makes sense: right at the free tier's limit, no extra gigabytes have been used yet, so the overage formula has nothing to add.

At $x = 10$: the included overage formula gives $25 + 6(10 - 4) = 61$, while the excluded flat cap is a fixed $82$.

$61 \ne 82 \implies D \text{ jumps at } x = 10.$

This one is a genuine quirk: the flat cap was not designed to match the formula's value at the crossover, only to stop the bill from climbing further past $10$ gigabytes, so it lands above what the formula alone would have given.

Rule out the constant tiers first: the first tier only ever bills $25$ dollars, and the third only ever bills $82$ dollars, so neither one can produce $49$ dollars. That leaves the second tier, $25 + 6(x - 4)$ for $4 < x \le 10$, whose outputs run from just above $25$ dollars up to $61$ dollars at $x = 10$, an interval that does contain $49$.

Solve the equation for that tier:

$25 + 6(x - 4) = 49 \implies 6(x - 4) = 24 \implies x - 4 = 4 \implies x = 8.$

Check that $8$ actually lies in the tier's own interval, $4 < x \le 10$: it does, so the answer is genuine and not one to discard. The customer used $8$ gigabytes.