Corrected exercises: functions, graphs and transformations (MATH 140)

a) Domain: the graph starts with a FILLED dot at $x = -3$ and ends with an OPEN dot at $x = 4$, and there is no gap in between, since the second segment starts at $x = 1$ exactly where the first one stops. So the domain is $[-3, 4)$. Value at $1$: above $x = 1$ there are two dots, $(1, 3)$ open and $(1, 4)$ filled; only the filled one is on the graph, so $f(1) = 4$. Range: the first segment, $y = x + 2$ for $-3 \le x < 1$, produces every value of $[-1, 3)$; the second, $y = 5 - x$ for $1 \le x < 4$, produces every value of $(1, 4]$. The range is the union, $[-1, 3) \cup (1, 4] = [-1, 4]$. The trap is to copy the open dots into the range: $3$ is missed at the open dot $(1, 3)$ but attained on the second segment at $x = 2$, where $5 - 2 = 3$; and $1$, missed at $(4, 1)$, is attained at $x = -1$ on the first segment. An open dot removes a POINT of the graph, not automatically a value of the range.

b) Two restrictions, taken one at a time. The even root needs $x + 3 \ge 0$, so $x \ge -3$. The denominator needs $x^2 - 4x = x(x - 4) \ne 0$, so $x \ne 0$ and $x \ne 4$. Both must hold at once: the domain is $[-3, \infty)$ with $0$ and $4$ removed, that is $[-3, 0) \cup (0, 4) \cup (4, \infty)$. Check with a test value: $g(-3) = \frac{0}{21} = 0$ exists, so $-3$ belongs to the domain and the bracket is closed there. Forgetting to factor the denominator and removing only $x = 4$ costs the mark.

c) The cube root accepts every real number, including negative ones, so the numerator imposes NOTHING: $\sqrt[3]{-8} = -2$ is a real number. The denominator is an even root, and it is a denominator, so what is under it must be STRICTLY positive: $x^2 - 5x + 4 = (x - 1)(x - 4) > 0$. A product of two factors is positive when both have the same sign: $x < 1$ or $x > 4$. The domain is $(-\infty, 1) \cup (4, \infty)$. Two classic errors: writing $\ge 0$, which would accept $x = 1$ and divide by $0$; and requiring $x - 8 \ge 0$ as if the cube root were a square root, which would give the wrong domain $[8, \infty)$.

d) Complete the square: $x^2 + 4x + 1 = (x^2 + 4x + 4) - 3 = (x + 2)^2 - 3$. On $[-3, 1]$, the quantity $x + 2$ runs over $[-1, 3]$, so its square runs over $[0, 9]$: it reaches $0$ at $x = -2$, which IS in the interval, and its largest value $9$ at $x = 1$. Hence $q$ runs over $[0 - 3, 9 - 3] = [-3, 6]$. The trap is to evaluate the two ends only, $q(-3) = -2$ and $q(1) = 6$, and answer $[-2, 6]$: the vertex $(-2, -3)$ lies inside the interval, lower than both ends. The ends give the range only when the function moves in one direction over the whole interval.

e) No. Two functions are equal when they have the same domain AND the same values. $F$ is not defined at $x = 3$ (denominator $0$), while $G(3) = 6$. Everywhere else, $x^2 - 9 = (x - 3)(x + 3)$ and the factor $x - 3 \ne 0$ can be cancelled: $F(x) = x + 3$ for $x \ne 3$. So the graph of $F$ is the line $y = x + 3$ with ONE point removed, an open dot at $(3, 6)$. The only value the line would take there is $6$, and no other $x$ gives $x + 3 = 6$, so the range of $F$ is $\mathbb{R} \setminus \{6\}$. Simplifying first and reading the domain on the simplified formula is the error the whole chapter warns against: the domain is read on the formula AS GIVEN.

a) $h(-3) = 0$: the line passes through $(-3, 0)$. At $x = -1$ the figure shows an open dot at $(-1, 2)$ and a filled dot at $(-1, 1)$; the filled one is the graph, so $h(-1) = 1$. At $x = 2$ the filled dot is $(2, 4)$ and the open one $(2, 1)$, so $h(2) = 4$; and $h(4) = 1$ on the ray. Formula, piece by piece. The line goes through $(-3, 0)$ and $(-2, 1)$: slope $1$, so $y = x + 3$, and it stops at $x = -1$ with an open dot, so the interval is $x < -1$. The arc passes through $(0, 0)$, $(1, 1)$ and $(2, 4)$: $y = x^2$ on $-1 \le x \le 2$, both ends filled. The ray is $y = 1$ for $x > 2$. The inequalities, strict or not, ARE the open and filled dots, and they are part of the answer: each input must belong to exactly one piece.

b) $2x - 6 \ge 0$ exactly when $x \ge 3$, so $f(x) = 2x - 6$ for $x \ge 3$ and $f(x) = -(2x - 6) = 6 - 2x$ for $x < 3$. For $k$, the sign of $x^2 - 4 = (x - 2)(x + 2)$ decides: it is $\ge 0$ when $x \le -2$ or $x \ge 2$, and negative when $-2 < x < 2$. So $k(x) = x^2 - 4$ if $x \le -2$ or $x \ge 2$, and $k(x) = 4 - x^2$ if $-2 < x < 2$. Graphically, the part of the parabola below the $x$-axis is reflected upward, and nothing else moves. The sign to study is the sign of what is INSIDE the bars, never the sign of $x$.

c) Split on the sign of $x - 1$. Case $x \ge 1$: $x - 1 = 2x + 4$ gives $x = -5$, which is NOT $\ge 1$: rejected. Case $x < 1$: $-(x - 1) = 2x + 4$ gives $-3x = 3$, $x = -1$, which is $< 1$: kept. Check: $|-1 - 1| = 2$ and $2(-1) + 4 = 2$. A faster filter was available: an absolute value is never negative, so the right side needs $2x + 4 \ge 0$, that is $x \ge -2$, which eliminates $-5$ at once, since $|-6| = 6$ while $2(-5) + 4 = -6$. Keeping $x = -5$ is the classic lost mark: the candidate solved its equation but not its case.

d) $x^2 - 6x + 9 = (x - 3)^2$, so $\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = |x - 3|$, NOT $x - 3$. For $x < 3$, $x - 3$ is negative, so $|x - 3| = 3 - x$. Check at $x = 0$: $\sqrt{9} = 3$, and $3 - 0 = 3$, while the student's $x - 3$ gives $-3$, a negative square root, impossible. The rule $\sqrt{a^2} = a$ holds only for $a \ge 0$; in general $\sqrt{a^2} = |a|$. This is the inside-outside reading again: the square root sees $(x - 3)^2$, which has forgotten the sign of $x - 3$.

e) Solve on each piece, then check each candidate against its interval. First piece: $x + 3 = 2$ gives $x = -1$, but the first piece is $x < -1$: rejected (and indeed $h(-1) = 1$, not $2$; the point $(-1, 2)$ is the open dot). Second piece: $x^2 = 2$ gives $x = \sqrt 2$ or $x = -\sqrt 2$; the piece is $-1 \le x \le 2$, and $\sqrt 2 \approx 1.41$ is in it, while $-\sqrt 2 \approx -1.41$ is not: only $\sqrt 2$ is kept. Third piece: $1 = 2$ has no solution. Answer: $x = \sqrt 2$, and only that. On the figure, the horizontal line $y = 2$ meets the graph once, on the arc; it only touches the open dot at $(-1, 2)$, which is not a point of the graph.

a) $f_1(-x) = (-x)^4 - 3(-x)^2 + 1 = x^4 - 3x^2 + 1 = f_1(x)$: even, every power of $x$ is even. $f_2(-x) = (-x)^3 - 4(-x) = -x^3 + 4x = -(x^3 - 4x) = -f_2(x)$: odd. $f_3(-x) = -x^3 + x^2$, which is neither $x^3 + x^2$ nor $-x^3 - x^2$. To be sure it is NEITHER, one counterexample for each symmetry is enough: $f_3(1) = 2$ and $f_3(-1) = 0$, and $0 \ne 2$ (not even), $0 \ne -2$ (not odd). A sum of an even part and an odd part, both nonzero, is neither. The two domains are $\mathbb{R}$, symmetric, as the test requires.

b) $k(-x) = \frac{-x}{(-x)^2 + 1} = -\frac{x}{x^2 + 1} = -k(x)$: odd. For $m$, the look is misleading: it IS a square, but a square of $x + 1$, not of $x$. $m(1) = 4$ and $m(-1) = 0$, so $m$ is neither even nor odd; its graph is symmetric about the line $x = -1$, not about the $y$-axis. For $n$: $n(-x) = (-x)|-x| = -x|x| = -n(x)$, since $|-x| = |x|$: odd. The minus sign inside the absolute value disappears, the one outside stays.

c) Even: each point $(a, b)$ of the graph has its mirror $(-a, b)$. The vertices $(1, 2)$, $(3, -1)$, $(4, 0)$ give $(-1, 2)$, $(-3, -1)$, $(-4, 0)$, joined by segments. Odd: each point $(a, b)$ has its image $(-a, -b)$, a half-turn about the origin: $(-1, -2)$, $(-3, 1)$, $(-4, 0)$. The solution figure shows both completions. For the last question, take $x = 0$ in $f(-x) = -f(x)$: $f(0) = -f(0)$, so $2f(0) = 0$ and $f(0) = 0$. The drawn graph passes through the origin, so the odd completion is possible; if it had started at $(0, 1)$, no odd function could extend it.

d) A single point proves nothing about a statement that must hold for EVERY $x$. Here $p(1) = 0$ and $p(-1) = 0$ by coincidence, because $\pm 1$ are roots. At $x = 2$: $p(2) = 6$ and $p(-2) = -8 + 2 = -6 \ne 6$, so $p$ is NOT even. The algebra decides: $p(-x) = -x^3 + x = -(x^3 - x) = -p(x)$ for every $x$, so $p$ is odd. Since $p$ is odd and $p(1) = 0$, also $p(-1) = -0 = 0$: the coincidence of the student is exactly what oddness predicts at a root.

e) Parity requires a domain symmetric about $0$: if $x$ is in the domain, $-x$ must be too. Here $2$ is in $[-1, 3]$ but $-2$ is not, so $s(-2)$ does not exist and $s$ is neither even nor odd, although its formula is $x^2$. Parity belongs to the function, formula AND domain. For the product, let $E$ be even and $O$ odd with the same symmetric domain: $(EO)(-x) = E(-x)\,O(-x) = E(x)\,(-O(x)) = -E(x)O(x) = -(EO)(x)$, so $EO$ is odd. Check with $E(x) = x^2$, $O(x) = x$: $x^3$ is odd. The signs multiply like $+1$ and $-1$: even times odd is odd, odd times odd is even.

a) Inside, $x - 3 = a$ gives $x = a + 3$: each point moves $3$ units RIGHT. Outside, $+1$: each $y$ goes up $1$. So $(a, b) \mapsto (a + 3, b + 1)$: $A(-2, 0) \mapsto (1, 1)$, $B(0, 4) \mapsto (3, 5)$, $C(2, 2) \mapsto (5, 3)$, $D(4, 2) \mapsto (7, 3)$. The domain moves with the $x$-coordinates: $[-2 + 3, 4 + 3] = [1, 7]$; the range with the $y$-coordinates: $[0 + 1, 4 + 1] = [1, 5]$. Check one point in the formula: at $x = 3$, $f(3 - 3) + 1 = f(0) + 1 = 5$, as found. Moving LEFT for a minus sign is the error this check catches.

b) Both act only on $y$, but in a different order. $y = -f(x) + 2$: first the minus sign, then $+2$, so $b \mapsto -b + 2$; $B$ goes to $(0, -4 + 2) = (0, -2)$, and the range $[0, 4]$ becomes $[-4 + 2, 0 + 2] = [-2, 2]$. $y = -(f(x) + 2) = -f(x) - 2$: first $+2$, then the minus sign, so $b \mapsto -(b + 2)$; $B$ goes to $(0, -6)$ and the range becomes $[-6, -2]$. Outside the parentheses, the order is the ORDER OF OPERATIONS of the formula, exactly as when evaluating a number: the parentheses of the second formula force the addition first. Two graphs $4$ units apart for the same three symbols in a different order.