Free worked solutions · 2026 Edition

Chapter 8: Functions worked solutions

Worked solutions to the practice questions in Chapter 8 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[6 marks]

The line L has equation 4x+3y-7=0.

(a) Find the gradient of L.

[1]

(b) Find the equation of the line through (3,\,2) that is parallel to L. Give your answer in the form ax+by+d=0, where a,\ b,\ d\in\mathbb{Z}.

[2]

(c) Find the equation of the line through (-4,\,1) that is perpendicular to L. Give your answer in the form ax+by+d=0, where a,\ b,\ d\in\mathbb{Z}.

[3]

Show solutionHide solution

(a) 3y=-4x+7, gradient -\dfrac{4}{3}

(b) y-2=-\dfrac{4}{3}(x-3)

4x+3y-18=0

(c) perpendicular gradient \dfrac{3}{4}

y-1=\dfrac{3}{4}(x+4)

3x-4y+16=0

Note: accept any integer multiple. A parallel line keeps the left-hand side 4x+3y, so 4x+3y+d=0 with d=-18 is also valid working.

Final answer

(a) -\frac43
(b) 4x+3y-18=0
(c) 3x-4y+16=0

Question 2

[6 marks]

The function f is defined by f(x)=3+\sqrt{x+5}.

(a) Write down the largest possible domain of f.

[1]

(b) Find the range of f.

[2]

(c) Solve f(x)=7.

[2]

(d) Explain why the equation f(x)=1 has no solution.

[1]

Show solutionHide solution

(a) x\ge -5

(b) \sqrt{x+5}\ge 0, with \sqrt{x+5}=0 at x=-5 (M1)

f(x)\ge 3

(c) \sqrt{x+5}=4 \Rightarrow x+5=16

x=11

(d) 1 is not in the range of f (every output is at least 3)

Final answer

(a) x\ge-5
(b) f(x)\ge3
(c) x=11
(d) every value of f is at least 3, so 1 is not in the range

Question 3

[7 marks]

Let f(x)=2x+1 and g(x)=x^2-3, for x\in\mathbb{R}.

(a) Find f^{-1}(x).

[1]

(b) Find (f^{-1}\circ g)(3).

[2]

(c) Write down an expression for (g\circ f^{-1})(x).

[1]

(d) Solve the equation (g\circ f^{-1})(x)=1.

[3]

Show solutionHide solution

(a) f^{-1}(x)=\dfrac{x-1}{2}

(b) g(3)=6 (A1)

f^{-1}(6)=\dfrac{5}{2}

(c) (g\circ f^{-1})(x)=\left(\dfrac{x-1}{2}\right)^2-3

(d) \left(\dfrac{x-1}{2}\right)^2=4

\dfrac{x-1}{2}=\pm 2 (A1)

x=5 or x=-3

Final answer

(a) \frac{x-1}{2}
(b) \frac52
(c) \left(\frac{x-1}{2}\right)^2-3
(d) x=5 or x=-3

Question 4

[6 marks]

(a) Find the set of values of k for which 2x^2-6x+k=0 has two distinct real roots.

[3]

(b) Find the set of values of k for which x^2+kx+4=0 has no real roots.

[3]

Show solutionHide solution

(a) \Delta=(-6)^2-4(2)k=36-8k (A1)

36-8k>0

k<\dfrac{9}{2}

(b) k^2-16<0

critical values k=\pm 4

-4<k<4

Note: in (b) award A0 for k<\pm 4 or k<4.

Final answer

(a) k<\frac92
(b) -4<k<4

Question 5

[6 marks]

The graph of y=f(x) crosses the x-axis only at x=-6 and x=9.

(a) The graph of y=f(qx), where q>0, has an x-intercept at x=3. Find the value of q and the other x-intercept.

[3]

(b) The graph of y=f(x-a) passes through the origin. Find the two possible values of a.

[2]

(c) The graph of y=f(qx) from part (a) is then reflected in the y-axis. Write down the x-intercepts of the final graph.

[1]

Show solutionHide solution

(a) the x-intercepts of y=f(qx) are x=-\dfrac{6}{q} and x=\dfrac{9}{q}

q>0 so -\dfrac{6}{q}<0; hence \dfrac{9}{q}=3, q=3

other x-intercept x=-2

(b) the zeros move to -6+a and 9+a; one of them must be 0: a=6

or a=-9

(c) the intercepts -2 and 3 change sign: x=2 and x=-3

Final answer

(a) q=3, other intercept x=-2
(b) a=6 or a=-9
(c) x=2 and x=-3

Question 6

[6 marks]

Let f(x)=\dfrac{3x+2}{x-4}, x\ne 4.

(a) Solve the equation f(x)=5.

[2]

(b) Solve the equation f(x)=-1.

[2]

(c) Show that the equation f(x)=3 has no solution, and state what this tells you about the graph of f.

[2]

Show solutionHide solution

(a) 3x+2=5(x-4)=5x-20

x=11

(b) 3x+2=-(x-4)=-x+4

x=\dfrac{1}{2}

(c) 3x+2=3x-12 \Rightarrow 2=-12, impossible

the graph never meets the line y=3, which is its horizontal asymptote

Final answer

(a) x=11
(b) x=\frac12
(c) 3x+2=3x-12 has no solution: y=3 is the horizontal asymptote

Question 7

[10 marks]

Let f(x)=\dfrac{x^2+5x+2}{2x-1}, x \ne \dfrac{1}{2}.

(a) Write down the equation of the vertical asymptote of the graph of f.

[1]

(b) Find the equation of the oblique asymptote of the graph of f.

[3]

(c) Use technology to find the x-intercepts and the coordinates of the two stationary points of the graph of f.

[3]

(d) Sketch the graph of f for -8 \le x \le 8, showing the asymptotes, the intercepts and the stationary points.

[3]

Show solutionHide solution

(a) x=\dfrac{1}{2}

(b) x^2+5x+2=(2x-1)\left(\dfrac{1}{2}x+\dfrac{11}{4}\right)+\dfrac{19}{4} M1A1

y=\dfrac{1}{2}x+\dfrac{11}{4}

(c) x=-4.56 and x=-0.438

local maximum (-1.68,\ 0.821)

local minimum (2.68,\ 5.18)

Note: exact values: x=\frac{-5 \pm \sqrt{17}}{2}; maximum \left(\frac{1-\sqrt{19}}{2},\ 3-\frac{\sqrt{19}}{2}\right), minimum \left(\frac{1+\sqrt{19}}{2},\ 3+\frac{\sqrt{19}}{2}\right). Accept the oblique asymptote as y=\frac{2x+11}{4}.

(d) A1A1A1

(sketch: see the online solutions)

Note: award A1 for the left branch (maximum below the right branch's minimum) and the right branch on the correct sides of the asymptotes, A1 for both asymptotes labelled, A1 for intercepts and stationary points.

Final answer

(a) x=\frac12
(b) y=\frac x2+\frac{11}{4}
(c) x=\frac{-5\pm\sqrt{17}}{2} (-4.56 and -0.438); maximum (-1.68,\,0.821), minimum (2.68,\,5.18)
(d) sketch: see the online solutions

Question 8

[9 marks]

Each function below is one-to-one on the domain given. For each, find an expression for the inverse function and state its domain.

(a) f(x)=(x-5)^2+2, for x\le 5

[3]

(b) g(x)=\dfrac{1}{(x-2)^2}, for x>2

[3]

(c) h(x)=(2x-1)^4, for x\le\dfrac{1}{2}

[3]

Show solutionHide solution

(a) x=(y-5)^2+2\Rightarrow y-5=-\sqrt{x-2} (negative root, as y\le 5)

f^{-1}(x)=5-\sqrt{x-2}

domain x\ge 2

(b) x=\dfrac{1}{(y-2)^2}\Rightarrow y-2=\dfrac{1}{\sqrt{x}} (positive root, as y>2)

g^{-1}(x)=2+\dfrac{1}{\sqrt{x}}

domain x>0

(c) x=(2y-1)^4\Rightarrow 2y-1=-\sqrt[4]{x} (as 2y-1\le 0)

h^{-1}(x)=\dfrac{1-\sqrt[4]{x}}{2}

domain x\ge 0

Note: in each part the M1 requires the correct sign choice; an answer with \pm earns M0A0.

Final answer

(a) 5-\sqrt{x-2}, x\ge2
(b) 2+\frac{1}{\sqrt x}, x>0
(c) \frac{1-x^{1/4}}{2}, x\ge0

Question 9

[8 marks]

Solve each inequality.

(a) \dfrac{x + 7}{x - 1} \le 3

[4]

(b) \dfrac{12}{x} < x + 1

[4]

Show solutionHide solution

(a) \dfrac{x + 7}{x - 1} - 3 \le 0 \Rightarrow \dfrac{x + 7 - 3(x - 1)}{x - 1} \le 0

\dfrac{10 - 2x}{x - 1} \le 0

(sign table: see the online solutions)

x < 1 or x \ge 5 A1A1

(b) \dfrac{12}{x} - x - 1 < 0 \Rightarrow \dfrac{12 - x - x^2}{x} < 0

-\dfrac{(x + 4)(x - 3)}{x} < 0 \Leftrightarrow \dfrac{(x + 4)(x - 3)}{x} > 0

(sign table: see the online solutions)

-4 < x < 0 or x > 3 A1A1

Note: multiplying by x without considering its sign gives x^2 + x - 12 > 0 and x < -4 or x > 3: award M0A0A0A1.

Final answer

(a) x<1 or x\ge5
(b) -4<x<0 or x>3

Question 10

[7 marks]

(a) Solve x^2 - 7\lvert x \rvert + 10 = 0.

[3]

(b) Solve 2x^2 + \lvert x \rvert - 6 = 0.

[3]

(c) Explain why, if x = a is a solution of an equation of the form x^2 + p\lvert x \rvert + q = 0, then x = -a is also a solution.

[1]

Show solutionHide solution

(a) \lvert x \rvert^2 - 7\lvert x \rvert + 10 = 0 (since x^2 = \lvert x \rvert^2) \Rightarrow (\lvert x \rvert - 2)(\lvert x \rvert - 5) = 0

\lvert x \rvert = 2 or \lvert x \rvert = 5

x = \pm 2, x = \pm 5

(b) 2\lvert x \rvert^2 + \lvert x \rvert - 6 = 0 \Rightarrow (2\lvert x \rvert - 3)(\lvert x \rvert + 2) = 0

\lvert x \rvert = -2 is rejected since \lvert x \rvert \ge 0

\lvert x \rvert = \frac{3}{2} \Rightarrow x = \pm \frac{3}{2}

(c) (-a)^2 = a^2 and \lvert -a \rvert = \lvert a \rvert, so the left-hand side takes the same value at -a as at a (it is an even function)

Final answer

(a) x=\pm2, x=\pm5
(b) x=\pm\frac32
(c) (-a)^2=a^2 and |-a|=|a|, so the left side takes the same value at -a

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 11

[5 marks]

The functions f and g are defined by f(x) = \sqrt{x - 1} for x \ge 1, and g(x) = x^2 - 5 for x \in \mathbb{R}.

(a) Find (g \circ f)(x) in its simplest form, and state its domain.

[3]

(b) Find the largest possible domain of f \circ g.

[2]

Show solutionHide solution

(a) (g \circ f)(x) = g\bigl(\sqrt{x-1}\bigr) = \bigl(\sqrt{x-1}\bigr)^2 - 5.

[1]

= (x - 1) - 5 = x - 6.

[1]

The input must first be accepted by f, so the domain of g \circ f is the domain of f: x \ge 1. (The formula x - 6 on its own would suggest all real numbers, which is wrong.)

[1]

(b) (f \circ g)(x) = \sqrt{g(x) - 1} needs g(x) \ge 1: x^2 - 5 \ge 1, so x^2 \ge 6.

[1]

Domain: x \le -\sqrt{6} or x \ge \sqrt{6}.

[1]

Final answer

(a) x-6, x\ge1
(b) x\le-\sqrt6 or x\ge\sqrt6

Question 12

[6 marks]

The functions f and g are defined by f(x) = 3 - (x - 1)^2 for x \ge 1, and g(x) = 2x - 5 for x \in \mathbb{R}.

(a) Find f^{-1}(x) and state its domain.

[3]

(b) Find (f^{-1} \circ g)(x) and write down its largest possible domain.

[3]

Show solutionHide solution

; (a) Swap: x = 3 - (y - 1)^2, so (y-1)^2 = 3 - x.

[1]

Since y \ge 1 take the positive root: f^{-1}(x) = 1 + \sqrt{3 - x}.

[1]

The domain of f^{-1} is the range of f. The greatest value of f is 3 (at x = 1), so x \le 3.

[1]

(b) (f^{-1} \circ g)(x) = 1 + \sqrt{3 - (2x - 5)} = 1 + \sqrt{8 - 2x}.

[1]

The input to f^{-1} must satisfy g(x) \le 3: 2x - 5 \le 3, so x \le 4.

[1]

Domain: x \le 4.

[1]

Final answer

(a) 1+\sqrt{3-x}, x\le3
(b) 1+\sqrt{8-2x}, x\le4

Question 13

[5 marks]

(a) Describe fully the sequence of transformations that maps the graph of y = f(x) onto the graph of y = f(4x + 8).

[3]

The point (12, 5) lies on the graph of y = f(x).

(b) Find the coordinates of the corresponding point on the graph of y = f(4x + 8).

[2]

Show solutionHide solution

; (a) Factor the input: 4x + 8 = 4(x + 2), so y = f(4(x + 2)).

[1]

A horizontal stretch with scale factor \tfrac14 [1], followed by a translation of 2 units to the left [1].

(Equivalent order: translate 8 units left first, then stretch by \tfrac14; the raw constant 8 goes with translate-first.)

(b) We need 4x + 8 = 12, because that input makes f return the known point.

[1]

x = 1, so the corresponding point is (1, 5).

[1]

(Check with the sequence: x = 12 \to \tfrac14 \cdot 12 = 3 \to 3 - 2 = 1.)

Final answer

(a) horizontal stretch, scale factor \frac14, then translation by \binom{-2}{0}
(b) (1,5)

Question 14

[5 marks]

Consider the curve y = x^2 - 2x + 3 and the line y = kx.

(a) Find the exact values of k for which the line is a tangent to the curve.

[3]

(b) Hence find the set of values of k for which the line does not meet the curve.

[2]

Show solutionHide solution

(a) x^2 - 2x + 3 = kx gives x^2 - (2 + k)x + 3 = 0.

[1]

Tangent: \Delta = (2 + k)^2 - 12 = 0, so 2 + k = \pm 2\sqrt{3}.

[1]

k = -2 + 2\sqrt3 or k = -2 - 2\sqrt3.

[1]

(b) The line misses the curve when the equation has no real roots: (2 + k)^2 < 12.

[1]

-2 - 2\sqrt3 < k < -2 + 2\sqrt3.

[1]

Final answer

(a) k=-2\pm2\sqrt3
(b) -2-2\sqrt3<k<-2+2\sqrt3

Question 15

[7 marks]

Let f(x) = \dfrac{5x + 2}{x - 4}, x \ne 4.

(a) Write down the equations of the asymptotes of the graph of y = f(x).

[2]

(b) Use your GDC to find the coordinates of the points where the graph of y = f(x) meets the line y = x + 3.

[3]

(c) Find the range of f for 5 \le x \le 10.

[2]

Show solutionHide solution

(a) Vertical asymptote x = 4 [1]; horizontal asymptote y = 5 [1].

(b) Graph y = \dfrac{5x + 2}{x - 4} and y = x + 3 and use the intersection tool; algebraically 5x + 2 = (x + 3)(x - 4), i.e. x^2 - 6x - 14 = 0.

[1]

x = -1.796\ldots and x = 7.796\ldots (x = 3 \mp \sqrt{23}).

[1]

The points are (-1.80, 1.20) and (7.80, 10.8).

[1]

(c) f(x) = 5 + \dfrac{22}{x - 4} is decreasing on x > 4 (GDC graph, or trace): f(5) = 27 and f(10) = \dfrac{52}{6} = 8.67.

[1]

Range: 8.67 \le f(x) \le 27.

[1]

Final answer

(a) x=4, y=5
(b) (-1.80,\,1.20) and (7.80,\,10.8)
(c) 8.67\le f(x)\le27

Question 16

[7 marks]

Let f(x) = \dfrac{x^2 - 3x + 5}{x + 1}, x \ne -1.

(a) Find the equations of the asymptotes of the graph of y = f(x).

[3]

(b) Use your GDC to find the coordinates of the local minimum point and the local maximum point of the graph.

[2]

(c) Hence write down the range of f.

[2]

Show solutionHide solution

; (a) Vertical asymptote x = -1 (numerator = 9 \ne 0 there).

[1]

Divide: x^2 - 3x + 5 = (x + 1)(x - 4) + 9, so f(x) = x - 4 + \dfrac{9}{x + 1}.

[1]

Oblique asymptote y = x - 4.

[1]

(b) GDC: graph y = \dfrac{x^2 - 3x + 5}{x + 1} and use the minimum and maximum tools on the two branches: local minimum point (2, 1) [1]; local maximum point (-4, -11) [1].

(c) The right-hand branch has least value 1 and the left-hand branch greatest value -11, and each branch is unbounded in the other direction.

[1]

Range: f(x) \le -11 or f(x) \ge 1.

[1]

Final answer

(a) x=-1, y=x-4
(b) minimum (2,\,1), maximum (-4,\,-11)
(c) f(x)\le-11 or f(x)\ge1

Question 17

[5 marks]

The function f is defined by f(x) = \dfrac{kx + 6}{x - 4}, x \ne 4, where k is a constant. The function f is self-inverse.

(a) By using f(f(0)) = 0, find the value of k.

[3]

(b) Hence write down the range of f, giving a reason for your answer.

[2]

Show solutionHide solution

; (a) f(0) = \dfrac{6}{-4} = -\dfrac32.

[1]

f\!\left(-\dfrac32\right) = \dfrac{-\frac{3k}{2} + 6}{-\frac32 - 4} = \dfrac{6 - \frac{3k}{2}}{-\frac{11}{2}}, and this must equal 0.

[1]

6 - \dfrac{3k}{2} = 0, so k = 4.

[1]

(Check: f(x) = \dfrac{4x + 6}{x - 4} has the \dfrac{ax+b}{cx-a} shape with a = 4, which is always self-inverse.)

(b) For a self-inverse function f^{-1} = f, so \operatorname{ran} f = \operatorname{dom} f^{-1} = \operatorname{dom} f.

[1]

The range is f(x) \ne 4. (Equivalently, f(x) = 4 + \dfrac{22}{x-4} never equals 4.)

[1]

Final answer

(a) k=4
(b) f(x)\ne4, the range of f is the domain of f^{-1}=f

Question 18

[5 marks]

Determine whether each function is odd, even or neither. Justify each answer algebraically.

(a) f(x) = 3x^5 - 2x, x \in \mathbb{R}

[2]

(b) g(x) = \dfrac{x^2 + 1}{x^4 + 3}, x \in \mathbb{R}

[2]

(c) h(x) = x^2 + 2x, x \in \mathbb{R}

[1]

Show solutionHide solution

; (a) f(-x) = 3(-x)^5 - 2(-x) = -3x^5 + 2x.

[1]

= -(3x^5 - 2x) = -f(x) for all x, so f is odd.

[1]

(b) g(-x) = \dfrac{(-x)^2 + 1}{(-x)^4 + 3} = \dfrac{x^2 + 1}{x^4 + 3} = g(x).

[1]

So g is even.

[1]

(c) h(1) = 3 and h(-1) = -1. These are neither equal (h(-1) \ne h(1), so not even) nor negatives of each other (h(-1) \ne -h(1), so not odd). So h is neither.

[1]

Final answer

(a) odd
(b) even
(c) neither: h(1)=3 and h(-1)=-1

Question 19

[6 marks]

(a) Solve |2x - 1| < x + 4.

[4]

(b) Hence solve |2|x| - 1| < |x| + 4.

[2]

Show solutionHide solution

; (a) Find where |2x - 1| = x + 4: 2x - 1 = x + 4 gives x = 5; -(2x - 1) = x + 4 gives x = -1. (Both satisfy x + 4 \ge 0.)

[2]

The graph of y = |2x - 1| is a V with vertex \left(\tfrac12, 0\right) and the line y = x + 4 is above it between the two crossing points, so the inequality holds between them.

[1]

-1 < x < 5.

[1]

(b) Replace x by |x|: the inequality holds when -1 < |x| < 5.

[1]

Since |x| \ge 0 always satisfies |x| > -1, the condition is |x| < 5: -5 < x < 5.

[1]

Final answer

(a) -1<x<5
(b) -5<x<5

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 20

[9 marks]

Let f(x) = \dfrac{x}{x^2 + 4}, x \in \mathbb{R}.

(a) Write down the equation of the horizontal asymptote and the coordinates of the axis intercepts.

[2]

(b) Let y = f(x). Show that yx^2 - x + 4y = 0, and use the discriminant of this equation to find the range of f.

[4]

(c) Hence find the coordinates of the local maximum and the local minimum points of the graph of f, and sketch the graph.

[3]

Show solutionHide solution

; (a) The denominator has a higher degree, so f \to 0: horizontal asymptote y = 0.

[1]

f(0) = 0, and f(x) = 0 only when x = 0: the only intercept is the origin (0, 0).

[1]

(b) y = \dfrac{x}{x^2 + 4} gives y(x^2 + 4) = x, so yx^2 - x + 4y = 0.

[1]

For a given y to be in the range this equation must have a real solution for x: \Delta \ge 0.

[1]

\Delta = (-1)^2 - 4(y)(4y) = 1 - 16y^2 \ge 0, so y^2 \le \tfrac1{16}.

[1]

Range: -\tfrac14 \le f(x) \le \tfrac14.

[1]

(c) The extreme values are reached when \Delta = 0. For y = \tfrac14: \tfrac14x^2 - x + 1 = 0, so (x - 2)^2 = 0 and x = 2. For y = -\tfrac14: -\tfrac14x^2 - x - 1 = 0, so x = -2.

[1]

Local maximum \left(2, \tfrac14\right) and local minimum \left(-2, -\tfrac14\right).

[1]

(sketch: see the online solutions)

Odd symmetry about the origin, tending to 0 at both ends and passing through the turning points [1].

Final answer

(a) y=0, (0,0)
(b) -\frac14\le f(x)\le\frac14
(c) maximum (2,\frac14), minimum (-2,-\frac14); sketch: see the online solutions

Question 21

[8 marks]

Let f(x) = x^2 - x - 6, x \in \mathbb{R}.

(a) Find the coordinates of the vertex of the graph of y = f(x).

[2]

(b) Sketch the graph of y = f(|x|), showing the coordinates of all axis intercepts and turning points.

[3]

(c) Find the set of values of k for which the equation f(|x|) = k has exactly four solutions.

[3]

Show solutionHide solution

; (a) f(x) = \left(x - \tfrac12\right)^2 - \tfrac{25}{4}.

[1]

Vertex \left(\tfrac12, -\tfrac{25}{4}\right).

[1]

(b) For x \ge 0 keep the graph of f and for x < 0 add its mirror image in the y-axis. The y-intercept is (0, -6); solving f(x) = 0 gives (x-3)(x+2) = 0, and the root x = 3 lies in x \ge 0, so the x-intercepts are (\pm 3, 0); the minimum points are \left(\pm\tfrac12, -\tfrac{25}{4}\right), and (0, -6) is a local maximum point (a peak) where the two halves meet.

(sketch: see the online solutions)

Symmetric about the y-axis [1]; intercepts (\pm3,0) and (0,-6) [1]; turning points \left(\pm\tfrac12,-\tfrac{25}{4}\right) and the peak (0,-6) [1].

(c) The line y = k meets the graph four times when it is above the two minima but below the peak.

[1]

-\tfrac{25}{4} < k < -6.

[2]

(At k = -\tfrac{25}{4} there are two solutions, at k = -6 there are three, and for k > -6 only two.)

Final answer

(a) \left(\frac12,-\frac{25}{4}\right)
(b) x-intercepts (\pm3,0), y-intercept (0,-6), minima \left(\pm\frac12,-\frac{25}{4}\right)
(c) -\frac{25}{4}<k<-6

Question 22

[8 marks]

Let f(x) = \dfrac{x + 2}{x - 1}, x \ne 1.

(a) Write down the equations of the asymptotes of the graph of y = f(x), and the coordinates of its axis intercepts.

[3]

(b) Sketch the graph of y = |f(x)|.

[2]

(c) Find the values of k for which the equation |f(x)| = k has exactly one solution.

[3]

Show solutionHide solution

; (a) Vertical asymptote x = 1; horizontal asymptote y = 1.

[2]

Intercepts: f(0) = -2, so (0, -2); f(x) = 0 gives x = -2, so (-2, 0).

[1]

(b) f(x) = 1 + \dfrac{3}{x - 1}. For x > 1 the curve is above y = 1; for x < 1 it is below y = 1, crosses the x-axis at x = -2 and falls to -\infty at x = 1^{-}. Reflect the part below the x-axis (for -2 < x < 1) upwards.

[1]

(sketch: see the online solutions)

Correct shape with asymptotes and the corner at (-2, 0) and intercept (0, 2) [1].

(c) Draw the horizontal line y = k on the sketch of y = |f(x)| and count intersections.

[1]

k = 0: the line touches the graph only at the corner (-2, 0): one solution.

[1]

k = 1: the right-hand branch only approaches y = 1 and the far left tail also only approaches it, so the only intersection is on the central part (|f| = 1 gives f = -1, x = -\tfrac12): one solution.

[1]

For 0 < k < 1 and for k > 1 there are two solutions, and for k < 0 none. So k = 0 or k = 1.

Final answer

(a) x=1, y=1; (-2,0), (0,-2)
(b) sketch: see the online solutions
(c) k=0 or k=1

Question 23

[9 marks]

Let f(x) = \dfrac{ax + b}{cx + d}, where c \ne 0 and ad - bc \ne 0.

(a) Show that (f \circ f)(x) = \dfrac{(a^2 + bc)x + b(a + d)}{c(a + d)x + bc + d^2}.

[3]

(b) Prove that f is self-inverse if and only if a + d = 0.

[4]

(c) Find the value of k for which g(x) = \dfrac{kx + 4}{2x - 3} is self-inverse, and write down the range of g.

[2]

Show solutionHide solution

; (a) (f \circ f)(x) = \dfrac{a\cdot\frac{ax + b}{cx + d} + b}{c\cdot\frac{ax + b}{cx + d} + d}.

[1]

Multiply the numerator and denominator by (cx + d): numerator a(ax + b) + b(cx + d) = (a^2 + bc)x + b(a + d);

[1]

denominator c(ax + b) + d(cx + d) = c(a + d)x + bc + d^2.

[1]

(b) (\Leftarrow) If a + d = 0 then (f \circ f)(x) = \dfrac{(a^2 + bc)x}{bc + d^2} = \dfrac{(a^2 + bc)x}{bc + a^2} = x. The denominator is non-zero because ad - bc = -(a^2 + bc) \ne 0. So f \circ f = \mathrm{id} and f is self-inverse.

[2]

(\Rightarrow) Suppose (f \circ f)(x) = x for every x in the domain. Then (a^2 + bc)x + b(a + d) = x\bigl[c(a + d)x + bc + d^2\bigr], so c(a + d)x^2 + (d^2 - a^2)x - b(a + d) = 0.

[1]

A quadratic that is zero for infinitely many x has all its coefficients zero; in particular c(a + d) = 0, and since c \ne 0, a + d = 0.

[1]

(c) a = k, d = -3, so self-inverse needs k - 3 = 0: k = 3.

[1]

Since g is self-inverse, \operatorname{ran} g = \operatorname{dom} g^{-1} = \operatorname{dom} g: g(x) \ne \tfrac32.

[1]

Final answer

(a) shown
(b) proved
(c) k=3; g(x)\ne\frac32

Question 24

[10 marks]

Let g(x) = |x^2 - 2x - 3| and consider the equation g(x) = x + c, where c \in \mathbb{R}.

(a) For c = 2, use your GDC to find all solutions of the equation.

[3]

(b) Find the exact value of c for which the line y = x + c is a tangent to the part of the graph of y = g(x) that lies between its two x-intercepts.

[3]

(c) Hence find the set of values of c for which the equation has exactly four solutions. Justify your answer with a sketch.

[4]

Show solutionHide solution

; (a) For c = 2 graph y = |x^2 - 2x - 3| and y = x + 2 on the GDC; the intersection tool gives four points.

[1]

(Algebraically: x^2 - 3x - 5 = 0 on the outer branches and x^2 - x - 1 = 0 on the middle arch.)

x = -1.19,\ -0.618,\ 1.62,\ 4.19.

[2]

(b) Between the x-intercepts -1 and 3, g(x) = -(x^2 - 2x - 3) = -x^2 + 2x + 3. Equate with the line: -x^2 + 2x + 3 = x + c, i.e. x^2 - x + (c - 3) = 0.

[1]

Tangency: \Delta = 1 - 4(c - 3) = 0.

[1]

c = \dfrac{13}{4}.

[1]

(c) The line y = x + c has gradient 1 and moves upwards as c increases. Key positions: through (-1, 0) when c = 1, and tangent to the arch when c = \tfrac{13}{4}.

[1]

(sketch: see the online solutions)

For 1 < c < \tfrac{13}{4} the line meets the left outer branch once, the arch twice and the right outer branch once: four solutions.

[1]

At c = 1 the line passes through the corner (-1, 0), so two of the intersections coincide (three solutions in all); for c < 1 it meets the left branch and the arch fewer times. At c = \tfrac{13}{4} it touches the arch (three solutions); above that it meets only the two outer branches (two solutions).

[1]

1 < c < \dfrac{13}{4}.

[1]

Final answer

(a) x=-1.19,\ -0.618,\ 1.62,\ 4.19
(b) c=\frac{13}{4}
(c) 1<c<\frac{13}{4}

Question 25

[9 marks]

The function f has domain \mathbb{R} and has an inverse function f^{-1}.

(a) Given that f is an odd function, prove that f^{-1} is also an odd function.

[4]

(b) Prove that f cannot be an even function.

[3]

The function g is defined by g(x) = x^3 + x, x \in \mathbb{R}. It is odd and has an inverse function, and g(2) = 10.

(c) Write down the value of g^{-1}(10), and find the value of g^{-1}(-10), justifying your answer.

[2]

Show solutionHide solution

; (a) Let y be any value in the domain of f^{-1} (the range of f) and write y = f(x), so that f^{-1}(y) = x.

[1]

Since f is odd, f(-x) = -f(x) = -y, so -y is also in the range of f and f^{-1}(-y) = -x.

[2]

Hence f^{-1}(-y) = -x = -f^{-1}(y) for every y in the domain of f^{-1}, so f^{-1} is odd.

[1]

(b) Suppose that f is even. Then f(-1) = f(1).

[1]

But -1 \ne 1, so two different inputs give the same output and f is not one-to-one.

[1]

This contradicts the existence of f^{-1}, so f cannot be even.

[1]

(c) g(2) = 10 gives g^{-1}(10) = 2.

[1]

Since g is odd, g^{-1} is odd by part (a), so g^{-1}(-10) = -g^{-1}(10) = -2.

[1]

Final answer

(a) proof
(b) proof
(c) g^{-1}(10)=2, g^{-1}(-10)=-2

Question 26

[7 marks]

Let f(x)=2x-5 and g(x)=\dfrac{x+1}{3}, where x\in\mathbb{R}.

(a) Find (f\circ g)^{-1}(x).

[3]

(b) Verify that (f\circ g)^{-1}(x)=(g^{-1}\circ f^{-1})(x).

[2]

The function h is defined for x\in\mathbb{R} and satisfies (f\circ h)(3x)=4x+1.

(c) Find h(x).

[2]

Show solutionHide solution

(a) (f\circ g)(x)=2\cdot\dfrac{x+1}{3}-5=\dfrac{2x-13}{3}.

M1A1

Let y=\dfrac{2x-13}{3} and swap: x=\dfrac{2y-13}{3}, so y=\dfrac{3x+13}{2}.

(b) g^{-1}(x)=3x-1 and f^{-1}(x)=\dfrac{x+5}{2}.

g^{-1}(f^{-1}(x))=3\cdot\dfrac{x+5}{2}-1=\dfrac{3x+13}{2}, the same as (a).

(c) 2h(3x)-5=4x+1, so h(3x)=2x+3.

Replace 3x by x: h(x)=\dfrac{2x}{3}+3=\dfrac{2x+9}{3}.

Final answer

(a) \frac{3x+13}{2}
(b) g^{-1}(x)=3x-1, f^{-1}(x)=\frac{x+5}{2}, and g^{-1}(f^{-1}(x))=\frac{3x+13}{2}
(c) h(x)=\frac{2x+9}{3}

Question 27

[8 marks]

Let f(x)=\dfrac{x-2}{x+1}, x\ne -1.

(a) Sketch y=[f(x)]^2, labelling both asymptotes, the y-intercept and the point where the curve touches the x-axis.

[4]

(b) Find all x for which |f(x)|\ge [f(x)]^2.

[4]

Show solutionHide solution

(a) f(x)=1-\dfrac{3}{x+1}, so f\to1 as x\to\pm\infty and [f]^2\to1: the asymptotes of y=[f(x)]^2 are x=-1 and y=1.

At x=0: f(0)=-2, so [f(0)]^2=4, the point (0,4).

f(2)=0, so the curve touches the x-axis at (2,0) and stays above it.

Shape: the left branch lies above y=1 and rises towards +\infty at x=-1; the right branch falls from +\infty to (2,0), then rises towards y=1 without crossing it.

(b) |f|\ge f^2\iff|f|(1-|f|)\ge0\iff|f|\le1 (including the zero of f).

M1R1

|x-2|\le|x+1|\iff(x-2)^2\le(x+1)^2.

-4x+4\le2x+1, so x\ge\dfrac12 (x=-1 is not in the domain and does not satisfy it).

Final answer

(a) asymptotes x=-1 and y=1; y-intercept (0,4); the curve touches the x-axis at (2,0)
(b) x\ge\frac12

Question 28

[12 marks]

The function f is defined by f(x) = x^2 - 6x + 5, x \in \mathbb{R}.

(a) Write f(x) in the form (x - h)^2 + k.

[2]

(b) Sketch the graph of y = f(x), showing the coordinates of its axis intercepts and its vertex.

[3]

The domain of f is now restricted to x \ge 3.

(c) Find f^{-1}(x), and state its domain and its range.

[4]

(d) Solve the equation f(x) = f^{-1}(x), giving the exact solution.

[3]

Show solutionHide solution

; (a) x^2 - 6x + 5 = (x - 3)^2 - 9 + 5.

[1]

f(x) = (x - 3)^2 - 4.

[1]

(b) Vertex (3, -4) [1]; y-intercept (0, 5) [1]; f(x) = (x-1)(x-5) so x-intercepts (1, 0) and (5, 0) [1].

(sketch: see the online solutions)

(c) For x \ge 3 the function is increasing, so it is one-to-one. Swap: x = (y - 3)^2 - 4, so (y - 3)^2 = x + 4.

[1]

Since y \ge 3, take the positive root: f^{-1}(x) = 3 + \sqrt{x + 4}.

[1]

Domain of f^{-1} = range of f on x \ge 3: x \ge -4. [1] Range of f^{-1} = domain of f: f^{-1}(x) \ge 3.

[1]

(d) Both f (on x \ge 3) and f^{-1} are increasing, so their graphs meet on the line y = x: solve f(x) = x, i.e. x^2 - 7x + 5 = 0.

[1]

x = \dfrac{7 \pm \sqrt{29}}{2}; the smaller root (\approx 0.807) is less than 3, outside the domain, and is rejected.

[1]

x = \dfrac{7 + \sqrt{29}}{2} (\approx 6.19).

[1]

Final answer

(a) (x-3)^2-4
(b) (1,0), (5,0), (0,5), vertex (3,-4)
(c) f^{-1}(x)=3+\sqrt{x+4}, domain x\ge-4, range \ge3
(d) x=\frac{7+\sqrt{29}}{2}

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 Where a function meets its inverse

G1 Worked example

[12 marks]

Let f(x)=1-x^3, x\in\mathbb{R}.

(a) Show that f is decreasing, and find f^{-1}(x).

[2]

(b) Use your GDC to sketch the graph of f together with the graph of its inverse, for -1.5\le x\le1.5. Write down the number of points that the two graphs have in common.

[2]

(c) Show that the point (a,b) lies on both graphs if and only if f(a)=b and f(b)=a.

[1]

(d) Suppose that (a,b) lies on both graphs and a\ne b. Show that a^2+ab+b^2=1 and a^3+b^3+a+b=2.

[2]

(e) Let s=a+b. Show that s^3-2s+1=0.

[2]

(f) Hence find every point that is not on the line y=x where the two graphs meet. Give exact coordinates where you can, and the others correct to three significant figures.

[3]

Show solutionHide solution

(a) If a<b then a^3<b^3, so 1-a^3>1-b^3: f is decreasing

x=1-y^3 gives f^{-1}(x)=\sqrt[3]{1-x}

(b) Sketch: two curves, each the reflection of the other in y=x

They meet at 5 points

(c) (a,b) is on y=f^{-1}(x) exactly when b=f^{-1}(a), that is, when f(b)=a; it is on y=f(x) exactly when f(a)=b

(d) b=1-a^3 and a=1-b^3. Subtracting, b-a=b^3-a^3=(b-a)(b^2+ab+a^2), and b-a\ne0

so a^2+ab+b^2=1; adding the two equations, a+b=2-a^3-b^3, so a^3+b^3+a+b=2

(e) a^2+ab+b^2=s^2-ab=1, so ab=s^2-1

a^3+b^3=s^3-3abs=s^3-3s(s^2-1), so s^3-3s^3+3s+s=2, which is s^3-2s+1=0

A1AG

(f) s^3-2s+1=(s-1)(s^2+s-1), so s=1 or s=\dfrac{-1\pm\sqrt5}{2}

s=1: ab=0, giving (0,1) and (1,0)

a and b are the roots of t^2-st+s^2-1=0, with discriminant 4-3s^2. For s=\dfrac{-1+\sqrt5}{2}: (1.15,-0.536) and (-0.536,1.15); for s=\dfrac{-1-\sqrt5}{2} the discriminant is negative, so there are no points

Check: 4 points off the line and 1 on it make the 5 seen in (b).

Final answer

(a) f^{-1}(x)=\sqrt[3]{1-x}
(b) 5
(f) (0,1), (1,0), (1.15,-0.536), (-0.536,1.15)

G1 Investigation

[12 marks]

For k\in\mathbb{R}, let f(x)=k-x^3, x\in\mathbb{R}. This investigation counts the points that the graph of f shares with the graph of its inverse, called meeting points below.

(a) Use your GDC to find the number of meeting points when k=0.5, k=1 and k=1.2. Suggest how the number depends on k.

[2]

(b) Show that, for every k, exactly one meeting point lies on the line y=x.

[2]

(c) Suppose that (a,b) is a meeting point with a\ne b, and let s=a+b. Show that a^2+ab+b^2=1 and that k=2s-s^3.

[3]

(d) Show that a and b are the roots of t^2-st+s^2-1=0. Hence show that -\dfrac{2}{\sqrt3}<s<\dfrac{2}{\sqrt3}.

[2]

(e) You are given that the greatest value of 2s-s^3 for s>0 is \dfrac{4\sqrt6}{9}. Find the exact set of values of k>0 for which the two graphs meet at exactly five points.

[3]

Show solutionHide solution

(a) k=0.5: 3 points; k=1: 5 points; k=1.2: 1 point

For example: one meeting point when k is large, more when k is near 1, and the number is always odd

(b) A meeting point on y=x has f(x)=x, that is x^3+x=k

x^3+x is increasing and takes every real value, so this has exactly one solution

(c) As for any f, (a,b) is a meeting point exactly when b=k-a^3 and a=k-b^3. Subtracting, b-a=(b-a)(a^2+ab+b^2), and a\ne b, so a^2+ab+b^2=1

M1A1

Adding, a+b=2k-(a^3+b^3)=2k-\left(s^3-3s(s^2-1)\right), using ab=s^2-1; so 2k=4s-2s^3, k=2s-s^3

(d) a+b=s and ab=s^2-1, so a and b are the roots of t^2-st+s^2-1=0

They are real and different, so the discriminant s^2-4(s^2-1)=4-3s^2>0, which gives -\dfrac{2}{\sqrt3}<s<\dfrac{2}{\sqrt3}

(e) Each s in this interval with 2s-s^3=k gives two meeting points (a,b) and (b,a). Five points need two such values of s

On 0<s<\dfrac{2}{\sqrt3}, 2s-s^3 rises from 0 to \dfrac{4\sqrt6}{9} and falls back to \dfrac{4}{\sqrt3}-\dfrac{8}{3\sqrt3}=\dfrac{4\sqrt3}{9} at the end, and for s\le0 it is not positive

Two values of s exactly when \dfrac{4\sqrt3}{9}<k<\dfrac{4\sqrt6}{9} (about 0.770<k<1.09)

Note: the end s=\frac{2}{\sqrt3} is excluded because there a=b; at k=\frac{4\sqrt6}{9} the two values of s coincide.

Final answer

(a) 3, 5, 1
(e) \frac{4\sqrt3}{9}<k<\frac{4\sqrt6}{9}

G2 When do two transformations commute?

G2 Worked example

[8 marks]

The transformation T_1 is a vertical stretch with scale factor 2 followed by a translation by \begin{pmatrix}0\\4\end{pmatrix}, so it maps each point (x,y) to (x,2y+4). The transformation T_2 is a vertical stretch with scale factor 3 followed by a translation by \begin{pmatrix}0\\c\end{pmatrix}.

(a) Write down the equation of the image of y=f(x) under T_1 followed by T_2, and under T_2 followed by T_1.

[2]

(b) Find the value of c for which the two orders give the same graph for every function f.

[2]

(c) For this value of c, show that T_1 and T_2 both leave every point of the line y=-4 where it is.

[2]

(d) Show that T_1 and a translation by \begin{pmatrix}0\\d\end{pmatrix} give the same graph in either order only when d=0.

[2]

Show solutionHide solution

(a) T_1 then T_2: y=3\bigl(2f(x)+4\bigr)+c=6f(x)+12+c

T_2 then T_1: y=2\bigl(3f(x)+c\bigr)+4=6f(x)+2c+4

(b) The same graph for every f needs 12+c=2c+4

c=8

(c) T_1: y=-4\mapsto2(-4)+4=-4

T_2: y=-4\mapsto3(-4)+8=-4; x is unchanged by both

(d) T_1 then the translation: y=2f(x)+4+d; the translation then T_1: y=2\bigl(f(x)+d\bigr)+4=2f(x)+2d+4

4+d=2d+4 only when d=0

Final answer

(a) y=6f(x)+12+c and y=6f(x)+2c+4
(b) c=8

G2 Investigation

[12 marks]

For a\ne0, the horizontal change H(a,b) replaces y=f(x) by y=f(ax+b). For p\ne0, the vertical change V(p,q) replaces y=f(x) by y=pf(x)+q. Two changes commute if applying them in either order gives the same graph for every function f.

(a) Show that H(a,b) moves the point (x_0,y_0) of y=f(x) to \left(\dfrac{x_0-b}{a},\,y_0\right).

[1]

(b) Show that H(2,0) followed by H(1,-6) gives y=f(2x-12), and that the other order gives y=f(2x-6).

[2]

(c) Show that H(a_1,b_1) followed by H(a_2,b_2) gives y=f(a_1a_2x+a_1b_2+b_1). Hence prove that the two changes commute if and only if b_1(a_2-1)=b_2(a_1-1).

[4]

(d) For a\ne1, show that H(a,b) leaves exactly one vertical line in place, and find its equation. Use this to describe, in words, when two horizontal changes with a_1\ne1 and a_2\ne1 commute.

[3]

(e) Explain why every horizontal change commutes with every vertical change.

[2]

Show solutionHide solution

(a) The new graph has a point at x when ax+b=x_0, that is x=\dfrac{x_0-b}{a}, with the same height f(x_0)=y_0

A1AG

(b) H(2,0) gives g(x)=f(2x); then H(1,-6) gives g(x-6)=f(2x-12)

H(1,-6) gives f(x-6); then H(2,0) replaces x by 2x: f(2x-6)

(c) H(a_1,b_1) gives g(x)=f(a_1x+b_1); then g(a_2x+b_2)=f\bigl(a_1(a_2x+b_2)+b_1\bigr)=f(a_1a_2x+a_1b_2+b_1)

M1A1

The other order gives f(a_1a_2x+a_2b_1+b_2). They agree for every f exactly when a_1b_2+b_1=a_2b_1+b_2

that is, b_1(a_2-1)=b_2(a_1-1)

(d) A point stays where it is when x_0=\dfrac{x_0-b}{a}, that is x_0(a-1)=-b

the line x=\dfrac{b}{1-a}

Dividing the condition in (c) by (a_1-1)(a_2-1) gives \dfrac{b_1}{1-a_1}=\dfrac{b_2}{1-a_2}: two such changes commute exactly when they leave the same vertical line in place

(e) In either order the result is y=pf(ax+b)+q

A horizontal change acts only on x and a vertical change only on y, so neither affects what the other does

Final answer

(b) f(2x-12), f(2x-6)
(d) x=\frac{b}{1-a}; they commute when they fix the same vertical line

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 19.

Job	TI-84 Plus CE keystrokes
Intersections	`Graph screen: Y1 and Y2, 2nd TRACE (CALC), 5:intersect, three ENTERs per crossing`
Polynomial roots	`APPS, PlySmlt2 (or graph and 2nd TRACE, 2:zero, for real roots)`
Solve an equation	`MATH, Numeric Solver (MATH, 0:Solver on older OS versions): enter the equation, then SOLVE. Or graph both sides and use 2nd TRACE (CALC), 5:intersect`

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL

Chapter 8 teaches functions from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book