Functions and Their Graphs: Chapter Test

Nothing in a set of points names the input, so test both readings. With $y$ as the input the rule hands back the single number $y^3 - y$, one output every time. With $x$ as the input, ask which values of $y$ sit above $x = 0$.

$y^3 - y = y(y - 1)(y + 1) = 0 \quad\Longrightarrow\quad y = -1,\ 0,\ 1$

Three outputs for the one input $0$, so with $x$ as the input it is not a function.

The argument is a slot: whatever sits inside is copied into every $x$ of the rule, coefficient included, and then squared.

$f(3t) = 5 - (3t)^2 = 5 - 9t^2$

The $3$ is inside the slot, so it is squared along with $t$.

Two requirements must hold at once, and the domain is their overlap, not either one alone. The square root needs a nonnegative inside, and the denominator factors as $x(x - 4)$.

$x + 6 \ge 0 \quad\text{and}\quad x(x - 4) \ne 0$

So the domain is $x \ge -6$ with $x = 0$ and $x = 4$ removed. The root sits on top, so its inside is allowed to be zero.

No factor has a root inside the interval, so no factor changes sign across it: read each factor's fixed sign there and multiply. On $-5 < x < 1$ the first factor is positive and the other two are negative.

$(+)(-)(-) = +$

Two negatives make a positive, so $f > 0$ throughout the interval.

In $g \circ f$ the function on the right acts first, so $f$ sends $x$ to $x - 7$ and $g$ squares that whole result.

$(g \circ f)(x) = g(x - 7) = (x - 7)^2 = x^2 - 14x + 49$

The cross term $-14x$ is what separates a squared binomial from a term-by-term square.

Match the shape $y = a\,f\big(b(x - h)\big) + k$: here $b = 1$, $h = -2$, $a = 1$ and $k = -5$. A point $(u, v)$ on $f$ lands at $\left(\tfrac{u}{b} + h,\ av + k\right)$.

$(-3,\ 7) \longmapsto (-3 - 2,\ 7 - 5) = (-5,\ 2)$

The inside acts on the number handed to $f$, so finding the new input means solving $x + 2 = -3$, which moves the point left.

The domain is all real numbers, which is symmetric about $0$, so the question can be asked at all. Substitute $-x$ and compare with both targets.

$f(-x) = 2(-x)^6 - (-x)^3 = 2x^6 + x^3$

That is neither $f(x) = 2x^6 - x^3$ nor $-f(x) = -2x^6 + x^3$, so $f$ is neither. Mixed exponent parity is exactly where the exponent shortcut fails.

Write $y = \dfrac{3x}{x + 5}$ and solve for the input: clear the denominator, gather the $x$ terms, factor, then divide.

$y(x + 5) = 3x \ \Longrightarrow\ x(y - 3) = -5y \ \Longrightarrow\ x = \frac{5y}{3 - y}$

Renaming the input gives $f^{-1}(x) = \dfrac{5x}{3 - x}$, and its excluded input $3$ is exactly the one value $f$ never outputs.

Expand, subtract, and check that every surviving term carries a factor of $h$ before dividing.

$\frac{\big(4 - 3(x + h)^2\big) - \big(4 - 3x^2\big)}{h} = \frac{-6xh - 3h^2}{h} = -6x - 3h, \qquad h \ne 0$

Canceling the $h$ is a division, so the identity is claimed only where $h \ne 0$: at $h = 0$ the quotient reads $\tfrac{0}{0}$ and returns nothing for the simplified form to match.

A factor raised to an even power is a square, so it is never negative and holds one sign on both sides of its root; an odd power carries the sign of the single factor and flips.

$(x + 2)^4 > 0 \text{ for every } x \ne -2, \qquad (x - 5) \text{ changes sign at } 5$

So the sign of $f$ is unchanged across $-2$, where the graph meets the axis and turns back, and flips across $5$, where it crosses.

Factor both parts and keep track of which excluded input survives the canceling.

$f(x) = \frac{(x - 1)(x - 5)}{(x - 3)(x - 5)} = \frac{x - 1}{x - 3}, \qquad x \ne 5$

The input $3$ does not cancel, so the graph runs off along a vertical line there rather than losing a single point. The input $5$ does cancel, so evaluating the simplified rule gives the gap's height, $\tfrac{5 - 1}{5 - 3} = 2$.

Factor the inside so the compression and the shift are both visible, then solve for the input that makes the inside equal the old input $8$.

$4x - 12 = 4(x - 3), \qquad 4x - 12 = 8 \ \Longrightarrow\ x = \frac{8}{4} + 3 = 5$

The shift is $3$, not $12$: the factor $4$ divides it, so it has to come out before any shift can be read.

Each test speaks about one reading, and one-to-oneness of a reading is settled by the OTHER test. The horizontal test asks whether $y$ can be the input; the vertical test asks whether $x$ can.

$\text{horizontal passes} \Rightarrow y \text{ is a legal input}, \qquad \text{vertical fails} \Rightarrow x \text{ is not}$

Since the vertical test fails, some value of $x$ carries two points, so two different $y$ inputs share one output and the reading that works repeats a value. One-to-oneness of a reading is a second question, answered by the test the first question did not use.

Two things must be true: $f$ must accept $x$, and $g$ must accept the number $f(x)$ that comes out.

$x + 3 \ge 0 \quad\text{and}\quad \sqrt{x + 3} \ne 6 \ \Longrightarrow\ x + 3 \ne 36$

So the domain is $x \ge -3$ with $x = 33$ removed. The excluded input is found through the values $f$ actually produces, not by intersecting the two domains.

Settle the ends of $p$ first by factoring out the highest power, then apply the outside map to the heights.

$p(x) = x^4\left(3 - \frac{8}{x^3}\right), \qquad \left\lvert \frac{8}{x^3} \right\rvert \le 1 \text{ once } \lvert x \rvert \ge 2$

The bracket stays positive and $x^4$ is never negative, so both ends of $p$ climb without bound. The outside map sends a height $v$ to $-v + 7$, which negates it, so both ends of the transformed graph fall without bound; adding $7$ raises them but cannot stop them.

Put the whole sum into the slot, then subtract the two separate outputs.

$\big(4a^2 + 8ab + 4b^2 - a - b\big) - \big(4a^2 - a\big) - \big(4b^2 - b\big) = 8ab$

The squared term leaves the cross term $8ab$ behind while the linear term cancels exactly, which is why a rule carrying a squared term can never be additive.

Restricting a domain does not repair $f$; it replaces $f$ with a different function, and each of these two is one-to-one, so each earns its own inverse.

$y = (x + 7)^2 \ \Longrightarrow\ x = -7 + \sqrt{y}\ \ (x \ge -7), \qquad x = -7 - \sqrt{y}\ \ (x \le -7)$

Uniqueness says one function has at most one inverse. These are two different functions, so two different answers are exactly what should appear.

Factor first, then test the parity by substituting $-x$.

$f(x) = x^3(x - 3)(x + 3), \qquad f(-x) = -x^5 + 9x^3 = -f(x)$

So $f$ is odd on its symmetric domain. Every factor carries an odd power, $3$, $1$ and $1$, so the sign flips at each of $-3$, $0$ and $3$ and the graph crosses at all three.

Compressing sends an input $u$ to $\tfrac{u}{3}$, and shifting right then sends that to $\tfrac{u}{3} + 6$. Matching the landing rule $\tfrac{u}{b} + h$ reads off $b = 3$ and $h = 6$.

$y = f\big(3(x - 6)\big) = f(3x - 18)$

Composing the same two maps the other way round sends $u$ to $\tfrac{u + 6}{3} = \tfrac{u}{3} + 2$, a different landing at every input. The two maps do not commute, so the order is part of the answer.

The equation says what $f$ does to the number $x + 2$, so name that number and re-express the right side in terms of it. Put $u = x + 2$, so that $x = u - 2$.

$f(u) = 3(u - 2) - 5 = 3u - 11$

Renaming the input gives $f(x) = 3x - 11$, and the check runs forward: $f(x + 2) = 3(x + 2) - 11 = 3x - 5$.