Answer Key
11.3 Functions
Pack A — Answers
| # | Question | Answer |
|---|---|---|
| 1 | Let $f(x) = 2x + 3$. Find $f(5)$. | 13 |
| 2 | For $f(x) = 2x + 3$, find $x$ such that $f(x) = 11$. | $x = 4$ |
| 3 | State the largest natural domain over $\mathbb{R}$ of $f(x) = \dfrac{1}{x - 3}$. | $x \in \mathbb{R},\; x \neq 3$ |
| 4 | State the largest natural domain over $\mathbb{R}$ of $g(x) = \sqrt{x - 1}$. | $x \geq 1$ |
| 5 | Let $f(x) = x^2 + -3x + 2$. Find $f(4)$. | 6 |
| 6 | A vertical line drawn through the graph of $y = x^2$ meets the curve in at most one point. Is $y = x^2$ a function of $x$? | Yes — passes the vertical line test. |
| 7 | State the range of $f(x) = 2x + 1$ defined on $\mathbb{R}$. | $\mathbb{R}$ |
| 8 | For $f(x) = 3x - 2$, find the image of $x = -2$. | $f(-2) = -8$ |
| 9 | State the largest natural domain of $h(x) = \dfrac{1}{\sqrt{x - 4}}$. | $x > 4$ |
| 10 | The function $f$ is defined by the table: $f(1) = 4$, $f(2) = 7$, $f(3) = 10$. Find a formula for $f(n)$ if it is linear. | $f(n) = 3n + 1$ |
| 11 | Let $f(x) = 2x + 5$ and $g(x) = x^2 - 3$. Find $f(g( 3))$. | 17 |
| 12 | Let $f(x) = 2x + 5$ and $g(x) = x^2 - 3$. Find a simplified expression for $(f \circ g)(x)$. | $2x^2 - 1$ |
| 13 | Find $f^{-1}(x)$ for $f(x) = 2x + 5$. | $f^{-1}(x) = \dfrac{x - 5}{2}$ |
| 14 | State the domain and range of $f(x) = \dfrac{1}{x - 2}$. | Domain $x \neq 2$; range $y \neq 0$ |
| 15 | For $f(x) = 2x + 5$ and $g(x) = -x + 8$, solve $f(x) = g(x)$. | $x = 1$ |
| 16 | State the domain and range of $f(x) = |x - 3|$. | Domain $\mathbb{R}$; range $y \geq 0$ |
| 17 | For what values of $x$ is $f(x) = \sqrt{2x - 6}$ defined? | $x \geq 3$ |
| 18 | For $f(x) = x^2$ and $g(x) = x + 6$, find all $x$ with $f(x) = g(x)$. | $x = -2$ or $x = 3$ |
| 19 | If $f(3) = 7$ and $f$ is one-to-one, state $f^{-1}(7)$. | 3 |
| 20 | Let $f(x) = \begin{cases} 2x + 1 & x < 0 \\ x^2 & x \geq 0 \end{cases}$. Find $f(-3)$ and $f(2)$. | $f(-3) = -5$; $f(2) = 4$ |
| 21 | Let $f(x) = 2x + 5$. Find $f^{-1}(x)$ and state its domain. | $f^{-1}(x) = \dfrac{x - 5}{2}$; domain $\mathbb{R}$ |
| 22 | Let $f(x) = \sqrt{x}$ and $g(x) = x - 4$. State $(f \circ g)(x)$ and its largest natural domain. | $(f \circ g)(x) = \sqrt{x - 4}$; domain $x \geq 4$ |
| 23 | State the range of $f(x) = |x - 3| + 2$. | $y \geq 2$ |
| 24 | For $f(x) = \lfloor x \rfloor$ (the floor function), find $f(2.7)$ and $f(-1.4)$. | $f(2.7) = 2$; $f(-1.4) = -2$ |
| 25 | $f(x) = x^2$ is not invertible on $\mathbb{R}$. State a domain restriction that makes it invertible, and give the inverse. | Restrict to $x \geq 0$; $f^{-1}(x) = \sqrt{x}$ |
| 26 | A graph passes through $(0, 4)$ and decreases monotonically, approaching $y = 0$ but never reaching it. Is this consistent with $f(x) = 4 \cdot (0.5)^x$? Justify. | Yes — exponential decay with horizontal asymptote $y = 0$ and $f(0) = 4$. |
| 27 | Let $f(x) = 2x + 1$ and $g(x) = \dfrac{1}{x}$ for $x \neq 0$. Find $(g \circ f)(x)$ and state its domain. | $(g \circ f)(x) = \dfrac{1}{2x + 1}$; domain $x \neq -\dfrac{1}{2}$ |
| 28 | For $f(x) = |2x - 4|$, solve $f(x) = 6$. | $x = -1$ or $x = 5$ |
| 29 | The cost (CHF) of producing $n$ bottles is $C(n) = 0.5n + 200$. State (a) the meaning of the gradient, (b) the meaning of $C(0)$. | (a) CHF 0.50 cost per extra bottle; (b) CHF 200 fixed cost. |
| 30 | The graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ in which line? | $y = x$ |
| 31 | Let $f(x) = 2x + 1$ and $g(x) = \dfrac{x}{x - 1}$ for $x \neq 1$. Find $(g \circ f)(x)$ and state its domain. | $(g \circ f)(x) = \dfrac{2x + 1}{2x}$; domain $x \neq 0$ |
| 32 | For $f(x) = \dfrac{1}{x - 1}$, find $(f \circ f)(x)$ and identify the values that are fixed by $f \circ f$. | $(f \circ f)(x) = \dfrac{x - 1}{2 - x}$; defined for $x \neq 1, 2$ |
| 33 | Find the inverse of $f(x) = \dfrac{2x + 3}{x - 1}$ for $x \neq 1$, and state its domain. | $f^{-1}(x) = \dfrac{x + 3}{x - 2}$; domain $x \neq 2$ |
| 34 | State the largest natural domain of $f(x) = \dfrac{1}{\sqrt{x^2 - 4}}$. | $x < -2$ or $x > 2$ |
| 35 | For $f(x) = \begin{cases} 2x + a & x < 1 \\ x^2 & x \geq 1 \end{cases}$, find $a$ such that $f$ is continuous at $x = 1$. | $a = -1$ |
| 36 | Find the range of $f(x) = x^2 - 4x + 7$ on $\mathbb{R}$. | $y \geq 3$ |
| 37 | The graph of $y = f(x)$ passes through $(1, 4)$ and $(3, 10)$. If $f$ is linear, find $f^{-1}(x)$. | $f^{-1}(x) = \dfrac{x - 1}{3}$ |
| 38 | Let $f(x) = 2x - 1$. Solve $f(f(x)) = 11$. | $x = 3.5$ |
| 39 | Let $f(x) = (x - 2)^2$ for $x \geq 2$. Find $f^{-1}(x)$ and its domain. | $f^{-1}(x) = 2 + \sqrt{x}$; domain $x \geq 0$ |
| 40 | A circle of radius 5 centred at the origin has equation $x^2 + y^2 = 25$. Explain why this is **not** a function of $x$, and write the two functions $y = f_1(x)$ and $y = f_2(x)$ that together describe the circle. | Not a function: a vertical line e.g. $x = 0$ meets the circle at $(0, 5)$ and $(0, -5)$. The two function pieces are $f_1(x) = \sqrt{25 - x^2}$ (top) and $f_2(x) = -\sqrt{25 - x^2}$ (bottom). |
Pack B — Answers
| # | Question | Answer |
|---|---|---|
| 1 | Let $f(x) = 3x + -1$. Find $f(4)$. | 11 |
| 2 | For $f(x) = 3x + -2$, find $x$ such that $f(x) = 10$. | $x = 4$ |
| 3 | State the largest natural domain over $\mathbb{R}$ of $f(x) = \dfrac{1}{x - -2}$. | $x \in \mathbb{R},\; x \neq -2$ |
| 4 | State the largest natural domain over $\mathbb{R}$ of $g(x) = \sqrt{x - 5}$. | $x \geq 5$ |
| 5 | Let $f(x) = x^2 + 2x + -5$. Find $f(3)$. | 10 |
| 6 | A vertical line drawn through the graph of $y = x^2$ meets the curve in at most one point. Is $y = x^2$ a function of $x$? | No — a vertical line, e.g. $x = 4$, meets $x = y^2$ at $(4, 2)$ and $(4, -2)$. |
| 7 | State the range of $f(x) = -3x + 4$ defined on $\mathbb{R}$. | $\mathbb{R}$ |
| 8 | For $f(x) = 3x - 2$, find the image of $x = -2$. | $f(3) = -1$ |
| 9 | State the largest natural domain of $h(x) = \dfrac{1}{\sqrt{x - 4}}$. | $x > -1$ |
| 10 | The function $f$ is defined by the table: $f(1) = 4$, $f(2) = 7$, $f(3) = 10$. Find a formula for $f(n)$ if it is linear. | $f(n) = 3n - 1$ |
| 11 | Let $f(x) = 2x + 5$ and $g(x) = x^2 - 3$. Find $f(g( 4))$. | 31 |
| 12 | Let $f(x) = 2x + 5$ and $g(x) = x^2 - 3$. Find a simplified expression for $(f \circ g)(x)$. | $3x^2 + 5$ |
| 13 | Find $f^{-1}(x)$ for $f(x) = 3x + -1$. | $f^{-1}(x) = \dfrac{x + 1}{3}$ |
| 14 | State the domain and range of $f(x) = \dfrac{1}{x - 2}$. | Domain $x \neq -3$; range $y \neq 0$ |
| 15 | For $f(x) = 2x + 5$ and $g(x) = -x + 8$, solve $f(x) = g(x)$. | $x = 4$ |
| 16 | State the domain and range of $f(x) = |x - 3|$. | Domain $\mathbb{R}$; range $y \geq 0$ |
| 17 | For what values of $x$ is $f(x) = \sqrt{2x - 6}$ defined? | $x \leq 5$ |
| 18 | For $f(x) = x^2$ and $g(x) = x + 6$, find all $x$ with $f(x) = g(x)$. | $x = -3$ or $x = 2$ |
| 19 | If $f(3) = 7$ and $f$ is one-to-one, state $f^{-1}(7)$. | $-2$ |
| 20 | Let $f(x) = \begin{cases} 2x + 1 & x < 0 \\ x^2 & x \geq 0 \end{cases}$. Find $f(-3)$ and $f(2)$. | $f(-1) = -1$; $f(3) = 9$ |
| 21 | Let $f(x) = 4x + -3$. Find $f^{-1}(x)$ and state its domain. | $f^{-1}(x) = \dfrac{x + 3}{4}$; domain $\mathbb{R}$ |
| 22 | Let $f(x) = \sqrt{x}$ and $g(x) = x - 4$. State $(f \circ g)(x)$ and its largest natural domain. | $(f \circ g)(x) = \sqrt{2x + 1}$; domain $x \geq -\frac{1}{2}$ |
| 23 | State the range of $f(x) = |x - 3| + 2$. | $y \leq 5$ |
| 24 | For $f(x) = \lfloor x \rfloor$ (the floor function), find $f(2.7)$ and $f(-1.4)$. | $f(3.9) = 3$; $f(-2.1) = -3$ |
| 25 | $f(x) = x^2$ is not invertible on $\mathbb{R}$. State a domain restriction that makes it invertible, and give the inverse. | Restrict to $x \leq 0$; $f^{-1}(x) = -\sqrt{x}$ |
| 26 | A graph passes through $(0, 4)$ and decreases monotonically, approaching $y = 0$ but never reaching it. Is this consistent with $f(x) = 4 \cdot (0.5)^x$? Justify. | Yes — $f(0) = 1 + 3 = 4$, decreasing to $y = 1$. |
| 27 | Let $f(x) = 2x + 1$ and $g(x) = \dfrac{1}{x}$ for $x \neq 0$. Find $(g \circ f)(x)$ and state its domain. | $(f \circ g)(x) = \dfrac{2}{x} + 1$; domain $x \neq 0$ |
| 28 | For $f(x) = |2x - 4|$, solve $f(x) = 6$. | $x = -5$ or $x = 1$ |
| 29 | The cost (CHF) of producing $n$ bottles is $C(n) = 0.5n + 200$. State (a) the meaning of the gradient, (b) the meaning of $C(0)$. | (a) CHF 0.80 per bottle; (b) CHF 150 fixed cost. |
| 30 | The graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ in which line? | Domain of $f^{-1}$ = range of $f$ (and range of $f^{-1}$ = domain of $f$). |
| 31 | Let $f(x) = 2x + 1$ and $g(x) = \dfrac{x}{x - 1}$ for $x \neq 1$. Find $(g \circ f)(x)$ and state its domain. | $(f \circ g)(x) = \dfrac{3x - 1}{x - 1}$; domain $x \neq 1$ |
| 32 | For $f(x) = \dfrac{1}{x - 1}$, find $(f \circ f)(x)$ and identify the values that are fixed by $f \circ f$. | $(f \circ f)(x) = \dfrac{2 - x}{3 - 2x}$ |
| 33 | Find the inverse of $f(x) = \dfrac{2x + 3}{x - 1}$ for $x \neq 1$, and state its domain. | $f^{-1}(x) = \dfrac{2x + 1}{3 - x}$; domain $x \neq 3$ |
| 34 | State the largest natural domain of $f(x) = \dfrac{1}{\sqrt{x^2 - 4}}$. | $x \geq 1$ with $x \neq 3$ |
| 35 | For $f(x) = \begin{cases} 2x + a & x < 1 \\ x^2 & x \geq 1 \end{cases}$, find $a$ such that $f$ is continuous at $x = 1$. | $a = -1$ |
| 36 | Find the range of $f(x) = x^2 - 4x + 7$ on $\mathbb{R}$. | $y \leq 4$ |
| 37 | The graph of $y = f(x)$ passes through $(1, 4)$ and $(3, 10)$. If $f$ is linear, find $f^{-1}(x)$. | $f^{-1}(x) = \dfrac{x + 1}{3}$ |
| 38 | Let $f(x) = 2x - 1$. Solve $f(f(x)) = 11$. | $x = \dfrac{6}{9}$ (i.e. $\frac{2}{3}$) |
| 39 | Let $f(x) = (x - 2)^2$ for $x \geq 2$. Find $f^{-1}(x)$ and its domain. | $f^{-1}(x) = -1 - \sqrt{x}$; domain $x \geq 0$ |
| 40 | A circle of radius 5 centred at the origin has equation $x^2 + y^2 = 25$. Explain why this is **not** a function of $x$, and write the two functions $y = f_1(x)$ and $y = f_2(x)$ that together describe the circle. | Not a function. Top: $f_1(x) = 2\sqrt{1 - x^2/9}$; bottom: $f_2(x) = -2\sqrt{1 - x^2/9}$. |
Problems — Worked Solutions
**Evaluate & solve.** Let $f(x) = 2x + 3$. (a) Find $f(5)$. (b) Find $f(-1)$. (c) Find the value of $x$ for which $f(x) = 11$.
(a) 13. (b) 1. (c) $x = 4$.
**Natural domains.** State the largest natural domain of each function over $\mathbb{R}$. (a) $f(x) = \dfrac{1}{x - 3}$ (b) $g(x) = \sqrt{x - 1}$ (c) $h(x) = \dfrac{1}{\sqrt{4 - x}}$
(a) $x \in \mathbb{R}, x \neq 3$. (b) $x \geq 1$. (c) $x < 4$.
**Vertical line test.** State whether each relation defines $y$ as a function of $x$. Justify. (a) $y = x^2$ (b) $x = y^2$ (c) $x^2 + y^2 = 9$ (d) $y = |x|$
(a) Yes. (b) No. (c) No. (d) Yes.
**Composition & inverse [EXT].** Let $f(x) = 2x + 5$ and $g(x) = x^2 - 3$. (a) Find $f(g(3))$. (b) Find a simplified expression for $(f \circ g)(x)$. (c) Find $f^{-1}(x)$ and state its domain.
(a) 17. (b) $2x^2 - 1$. (c) $f^{-1}(x) = \frac{x - 5}{2}$; domain $\mathbb{R}$.
**Domain of a composition [EXT].** Let $f(x) = \sqrt{x}$ and $g(x) = 5 - x^2$. (a) Find $(f \circ g)(x)$. (b) State its largest natural domain. (c) State its range.
(a) $\sqrt{5 - x^2}$. (b) $-\sqrt{5} \leq x \leq \sqrt{5}$. (c) $0 \leq y \leq \sqrt{5}$.
**Piecewise function.** Let $f(x) = \begin{cases} 2x + 1 & x < 0 \\ x^2 & 0 \leq x \leq 3 \\ 9 & x > 3 \end{cases}$. (a) Find $f(-2)$, $f(0)$, $f(2)$, $f(5)$. (b) Sketch $f$ for $-3 \leq x \leq 5$. (c) State the range of $f$.
(a) $-3, 0, 4, 9$. (b) Linear piece, then upward parabola from $(0,0)$ to $(3,9)$, then horizontal at $y = 9$. (c) Range: $f \geq -5$, but with the linear piece dipping arbitrarily low as $x \to -\infty$; on the stated domain the range is $[-5, 9]$.
**Absolute value [EXT].** Solve and sketch. (a) Solve $|2x - 3| = 7$. (b) Solve $|x - 1| < 4$ and write the answer in interval notation. (c) Sketch $y = |x - 2| - 1$, marking $x$- and $y$-intercepts.
(a) $x = -2$ or $x = 5$. (b) $-3 < x < 5$, i.e. $(-3, 5)$. (c) V-shape with vertex at $(2, -1)$; $x$-intercepts $(1, 0)$ and $(3, 0)$; $y$-intercept $(0, 1)$.
**Modelling with a linear function.** A car rental charges a fixed daily fee $F$ (CHF) plus a charge $c$ (CHF) per km. A driver pays CHF 95 for 200 km in a day, and CHF 125 for 350 km in a day. (a) Write a function $C(d) = cd + F$ for the cost in terms of the distance $d$ (km). (b) Find $c$ and $F$. (c) Predict the cost of a 500 km day.
(a) $C(d) = cd + F$. (b) $c = 0.2$, $F = 55$. (c) CHF 155.
**Finding f from data — quadratic.** A quadratic function $f$ passes through $(0, 3)$, $(1, 6)$, and $(2, 13)$. (a) Assume $f(x) = ax^2 + bx + c$. Set up three equations. (b) Solve to find $a$, $b$, $c$. (c) State $f(-1)$.
(a) $c = 3$; $a + b + c = 6$; $4a + 2b + c = 13$. (b) $a = 2$, $b = 1$, $c = 3$. (c) $f(-1) = 4$.
**Range of a quadratic.** Let $f(x) = x^2 - 4x + 7$. (a) Express $f(x)$ in vertex form. (b) State the minimum value and where it occurs. (c) State the range of $f$.
(a) $(x - 2)^2 + 3$. (b) Min value 3 at $x = 2$. (c) Range $y \geq 3$.
**Inverse from a graph [EXT].** The graph of $f$ is a straight line through $(-2, 1)$ and $(4, 4)$. (a) Find $f(x)$. (b) Find $f^{-1}(x)$. (c) On a single set of axes, sketch $y = f(x)$, $y = f^{-1}(x)$ and $y = x$. State the symmetry.
(a) $f(x) = \frac{1}{2}x + 2$. (b) $f^{-1}(x) = 2x - 4$. (c) $f$ and $f^{-1}$ are reflections of each other in $y = x$.
**Mini-investigation — natural domain.** Three students propose different formulas for the same function: - Anya: $f(x) = \dfrac{x^2 - 1}{x - 1}$ - Bao: $f(x) = x + 1$ - Cara: $f(x) = x + 1$ for $x \neq 1$ (a) Compute $f(2)$ using each formula. (b) State the natural domain of Anya's formula. (c) Are Anya's and Bao's functions equal? Justify. (d) Whose formula is most precise? Justify.
(a) Anya: 3, Bao: 3, Cara: 3. (b) $x \neq 1$. (c) Not strictly equal — they agree everywhere except at $x = 1$. (d) Cara — she states the domain restriction explicitly.