Exam practice

A continuous curve has no break in the interval considered: it can be drawn from one end to the other without lifting the pen. Examples: the graph of $y = x^2$, a circle, the graph of the temperature through a day.
A discontinuous curve has a break somewhere: a gap, a jump, or a single missing point, so the pen must be lifted. Examples: the graph of $y = \frac{x^2 - 4}{x - 2}$, which has a hole at x = 2; a graph that jumps from one height to another, like the fare of a ride that rises in steps; the integers marked on a number line.

It means the curve has no break for any real x from −5 to 5: it is defined at every one of those x, including −5 and 5, and it can be drawn from the point at x = −5 to the point at x = 5 without lifting the pen. In symbols, it is continuous on −5 ≤ x ≤ 5.

a. It runs from the filled point at x = −4 to the filled point at x = 5, with one hole at x = 3.
Discontinuous at x = 3; continuous on −4 ≤ x < 3 and 3 < x ≤ 5.

b. A piece from x = −1 to x = 1, ending at the filled dot (1, 1); the next piece starts at the open circle (1, 2) and goes on to x = 5. At x = 1 the graph jumps from height 1 to height 2.
Discontinuous at x = 1; continuous on −1 ≤ x ≤ 1 and 1 < x ≤ 5.

c. The graph starts at an open circle at x = −1 (so −1 is not included) and climbs without end as x nears 3; it is not defined at x = 3. After 3 a second piece comes down from far above, to its end at x = 6.
Discontinuous at x = 3; continuous on −1 < x < 3 and 3 < x ≤ 6.

a. Continuous: its graph is one straight, unbroken line from end to end, drawn without lifting the pen.

b. Continuous at x = 3: the two straight parts meet at the same point, (3, 2). Coming from the left the height heads for 2, from the right it heads for 2, and the wire is there at x = 3: no gap.

c. A point of discontinuity: at x = 4 the graph has a hole (a missing point), so it is discontinuous there.

d. Wire B is continuous and wire C is discontinuous. A bend changes the wire’s direction, not whether it is joined: the two parts still meet at one point, so the pen (or the current) passes from one to the other. A break leaves a gap: there is no point of the wire at x = 4, so the current cannot pass. Continuity asks only “is there a break?”, never “is it smooth?”.

$\lim_{x \to c^-} f(x)$, the left-hand limit, is the value f(x) heads for as x comes closer and closer to c from values less than c. $\lim_{x \to c^+} f(x)$, the right-hand limit, is the value f(x) heads for as x comes closer to c from values greater than c. Neither uses f(c) itself.
Example: for f(x) = 2x at c = 3, f(2.9) = 5.8, f(2.99) = 5.98, f(2.999) = 5.998, so $\lim_{x \to 3^-} f(x) = 6$; and f(3.1) = 6.2, f(3.01) = 6.02, f(3.001) = 6.002, so $\lim_{x \to 3^+} f(x) = 6$.
An example where they differ: f(x) = x for x ≤ 1 and x + 2 for x > 1. From the left f(x) → 1, from the right f(x) → 3: $\lim_{x \to 1^-} f(x) = 1$, $\lim_{x \to 1^+} f(x) = 3$, and there is no limit at 1.

1. The limit exists: $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$, so $\lim_{x \to a} f(x)$ exists.
2. The value f(a) is defined.
3. $\lim_{x \to a} f(x) = f(a)$.

• a.g(2.9) = 5.9, g(2.99) = 5.99, g(2.999) = 5.999.
• b.g(3.1) = 6.1, g(3.01) = 6.01, g(3.001) = 6.001.
• c.g(3) = 3 + 3 = 6.
• d.From b, $\lim_{x \to 3^+} g(x) = 6$; from a, $\lim_{x \to 3^-} g(x) = 6$.
• e.Yes: $\lim_{x \to 3^-} g(x) = \lim_{x \to 3^+} g(x) = g(3) = 6$.

• a.h(0.9) = 1.8 − 1 = 0.8, h(0.99) = 0.98, h(0.999) = 0.998.
• b.h(1.1) = 2.2 − 1 = 1.2, h(1.01) = 1.02, h(1.001) = 1.002.
• c.h(1) = 2 − 1 = 1.
• d.$\lim_{x \to 1^+} h(x) = 1$ (from b) and $\lim_{x \to 1^-} h(x) = 1$ (from a).
• e.Yes: $\lim_{x \to 1^-} h(x) = \lim_{x \to 1^+} h(x) = h(1) = 1$.

• a.0.99 is in 0 ≤ x ≤ 1: k(0.99) = 3 − 0.99 = 2.01.
• b.1.01 is more than 1: k(1.01) = 1.01 + 1 = 2.01.
• c.From the left, 3 − x → 3 − 1 = 2: $\lim_{x \to 1^-} k(x) = 2$. From the right, x + 1 → 2: $\lim_{x \to 1^+} k(x) = 2$.
• d.k(1) = 3 − 1 = 2 (from the first piece, which includes 1). LHL = RHL = k(1) = 2, so k(x) is continuous at x = 1.

• a.p(2.99) = 2 × 2.99 + 1 = 6.98.
• b.p(3.01) = 3.01 + 4 = 7.01.
• c.$\lim_{x \to 3^-} p(x) = 2 \times 3 + 1 = 7$ and $\lim_{x \to 3^+} p(x) = 3 + 4 = 7$.
• d.p(3) = 2 × 3 + 1 = 7. LHL = RHL = p(3) = 7, so p(x) is continuous at x = 3.

For b and c, factorize the top first: $x^2 - 4 = (x - 2)(x + 2)$, $x^3 - 27 = (x - 3)(x^2 + 3x + 9)$.

a. Left: f(0.9) = 6.8, f(0.99) = 6.98, f(0.999) = 6.998. Right: f(1.1) = 7.2, f(1.01) = 7.02, f(1.001) = 7.002. So $\lim_{x \to 1} f(x) = 7$, and f(1) = 2 + 5 = 7. Equal: continuous at x = 1.

b. For x ≠ 2, f(x) = x + 2. Left: f(1.9) = 3.9, f(1.99) = 3.99, f(1.999) = 3.999. Right: f(2.1) = 4.1, f(2.01) = 4.01, f(2.001) = 4.001. So $\lim_{x \to 2} f(x) = 4$. But $f(2) = \frac{4 - 4}{2 - 2} = \frac{0}{0}$, not defined: discontinuous at x = 2.

c. For x ≠ 3, $f(x) = x^2 + 3x + 9$. Left: f(2.9) = 26.11, f(2.99) = 26.9101, f(2.999) = 26.991001. Right: f(3.1) = 27.91, f(3.01) = 27.0901, f(3.001) = 27.009001. So $\lim_{x \to 3} f(x) = 27$. But $f(3) = \frac{27 - 27}{3 - 3} = \frac{0}{0}$, not defined: discontinuous at x = 3.

a. $\lim_{x \to 3^-} f(x) = 7 \times 3 - 2 = 19$ (values 18.3, 18.93, 18.993…), and $\lim_{x \to 3^+} f(x) = 6 \times 3 + 1 = 19$ (19.6, 19.06, 19.006…). So the limit is 19. But f(3) = 5 ≠ 19: discontinuous at x = 3 (the point has been moved off the line).

b. $\lim_{x \to 1^-} g(x) = 1 + 2 = 3$ (2.9, 2.99, 2.999…), and $\lim_{x \to 1^+} g(x) = 3$ (g is 3 for every x > 1). g(1) = 1 + 2 = 3. LHL = RHL = g(1) = 3: continuous at x = 1.

Find LHL, RHL and f(2) in terms of k, then make all three equal.

$\lim_{x \to 2^-} f(x) = 2 \times 2 - 3 = 1$.
$\lim_{x \to 2^+} f(x) = 2 + k$.
f(2) = 2k + 3.
For continuity, LHL = RHL: 2 + k = 1, so k = −1.
Check with f(2): 2(−1) + 3 = 1. ✓ All three are 1, so k = −1.

• a.A curve is continuous at x = c when it has no break there: as x comes towards c from either side, the height heads for one number, and the curve has its point at that height at c (LHL = RHL = f(c)). Example: y = x + 3 at x = 3. It is discontinuous at c when there is a break: a hole ($y = \frac{x^2 - 4}{x - 2}$ at 2), a jump (a function in two pieces that don’t meet), or a point moved off the curve.
• b.For every real number x from −4 to 4, both included, the curve has a point and no break: it can be drawn from x = −4 to x = 4 without lifting the pen, so it is continuous at every point of −4 ≤ x ≤ 4.
• c.g(2.99) = 5.99, g(2.999) = 5.999, so $\lim_{x \to 3^-} g(x) = 6$; g(3.01) = 6.01, g(3.001) = 6.001, so $\lim_{x \to 3^+} g(x) = 6$; and g(3) = 6. LHL = RHL = g(3): continuous at x = 3.

• a.f(1.999) = 7.996 + 1 = 8.996; f(2.001) = 8.004 + 1 = 9.004.
• b.$\lim_{x \to 2^-} f(x) = 9$ and $\lim_{x \to 2^+} f(x) = 9$.
• c.f(2) = 8 + 1 = 9.
• d.Yes: LHL = RHL = f(2) = 9.

• a.$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$.
• b.p(2.9999) = 2.9999 + 5 = 7.9999.
• c.p(3.0001) = 4 × 3.0001 − 4 = 12.0004 − 4 = 8.0004.
• d.$\lim_{x \to 3^-} p(x) = 3 + 5 = 8$, $\lim_{x \to 3^+} p(x) = 12 - 4 = 8$, and p(3) = 3 + 5 = 8. All equal: continuous at x = 3.

• a.A function is continuous (on an interval) when it is continuous at every point of it: at each point c, the left-hand limit, the right-hand limit and f(c) exist and are equal. Its graph has no break there.
• b.$\lim_{x \to 1^-} h(x) = k \times 1 + 3 = k + 3$; $\lim_{x \to 1^+} h(x) = 2 \times 1 - 1 = 1$.
• c.Continuity needs LHL = RHL (= h(1) = k + 3): k + 3 = 1, so k = −2. (Try it in the widget Make the pieces meet.)

a. The corners: A(−4, 3) (where y = 3 meets y = −2x − 5), B(2, 3) (y = 3 and y = x + 1), C(−2, −1) (x + 1 = −2x − 5, so x = −2).
At A: 4(−4) − 5(3) = −31. At B: 8 − 15 = −7. At C: −8 + 5 = −3.
Maximum −3 at C(−2, −1); minimum −31 at A(−4, 3).
b. BC is y = x + 1, slope m₁ = 1; AC is y = −2x − 5, slope m₂ = −2.
$\tan\theta = \pm\frac{m_1 - m_2}{1 + m_1m_2} = \pm\frac{1 + 2}{1 - 2} = \pm 3$.
The acute angle is $\tan^{-1} 3 \approx$ 71.57∘°, which is the triangle’s angle at C; the obtuse angle between the lines is $180° - 71.57° =$ 108.43∘°.

• c.f(2.999) = 3.999; f(3.001) = 4.001.
• d.Yes: $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = 4 = f(3)$.

• a.Look for the x with f(x) = 11. In the first piece, 4x + 3 = 11 gives x = 2, which is in x ≤ 2. ✓ In the second, 5x + 1 = 11 gives x = 2, but that piece is only for x > 2, so it gives nothing new. Only one x gives 11: yes, $f^{-1}(11) = 2$.
• b.$\lim_{x \to 2^-} f(x) = 8 + 3 = 11$ (f(1.99) = 10.96, f(1.999) = 10.996); $\lim_{x \to 2^+} f(x) = 10 + 1 = 11$ (f(2.01) = 11.05, f(2.001) = 11.005); f(2) = 11. Continuous at x = 2.

• a.Coefficient of SD $= \frac{\sigma}{\bar{X}}$: $5x - 1 = \frac{y}{10}$, so y = 50x − 10.
• b.y = 50x − 10 gives $x = \frac{y + 10}{50}$, so $f^{-1}(x) = \frac{x + 10}{50}$ and $f^{-1}(2) = \frac{12}{50} = \frac{6}{25} = 0.24$.
• c.The slope of y = 5x − 1 is 5, so the perpendicular slope is $-\frac{1}{5}$. Through (2, −3): $y + 3 = -\frac{1}{5}(x - 2)$, 5y + 15 = −x + 2, so x + 5y + 13 = 0.

• a.g(2.999) = 14.995 + 2 = 16.995; g(3.001) = 15.005 + 2 = 17.005.
• b.Yes: $\lim_{x \to 3^-} g(x) = \lim_{x \to 3^+} g(x) = 17 = g(3)$.
• c.Slope 5, through (2, −3): y + 3 = 5(x − 2), so 5x − y − 13 = 0.
• d.y = 5x + 2, $x = \frac{y - 2}{5}$: $g^{-1}(x) = \frac{x - 2}{5}$.

a. Draw $y = 3x^2 + 5x + 2$ from a table: $x = -2, -1, -\frac{2}{3}, 0, 1$ give y = 4, 0, 0, 2, 10. The vertex is at $x = -\frac{5}{6}$, $y = -\frac{1}{12}$. The curve cuts the X-axis at x = −1 and $x = -\frac{2}{3}$, the solutions.

• b.f(x) = 3x + 2, so $f^{-1}(x) = \frac{x - 2}{3}$. Then $g \circ f^{-1}(x) = g\left(\frac{x - 2}{3}\right) = \frac{x - 2}{3} + 1 = \frac{x + 1}{3}$. Setting $\frac{x + 1}{3} = 2$: x = 5.
• c.The slope of y = 3x + 2 is 3, so the perpendicular slope is $-\frac{1}{3}$: $y + 2 = -\frac{1}{3}(x + 3)$, 3y + 6 = −x − 3, so x + 3y + 9 = 0.

• a.Slope of $AC = \frac{-3 - 3}{6 - 4} = -3$. Through A(4, 3): y − 3 = −3(x − 4), so 3x + y − 15 = 0.
• b.AB: 3y = x + 5, slope $m_1 = \frac{1}{3}$; AC: slope m₂ = −3. $m_1m_2 = \frac{1}{3} \times (-3) = -1$, so the chords are perpendicular: the angle is 90∘°.
• c.∠BAC, the angle between the chords at A, is 90∘°. An angle of 90∘° on the circle stands on a diameter (the angle in a semicircle is a right angle, and only a diameter’s angle is), so BC is a diameter.
• d.AB: $y = f(x) = \frac{x + 5}{3}$; AC: y = g(x) = −3x + 15.

$f(2) = \frac{7}{3}$, and $g \circ f(2) = g\left(\frac{7}{3}\right) = -7 + 15 =$ 8.

• a.y = 0 is the X-axis (slope 0) and x − y = 0 is y = x (slope 1, at 45∘° to the X-axis). The acute angle is 45∘°.
• b.R₁: (x, y) → (x, −y); then R₂: (x, −y) → (−y, x). So R₂ ∘ R₁: (x, y) → (−y, x), a rotation of +90∘° about O.

A′(−6, −4), B′(10, −6), C′(8, 12).
c. AB = (−6 + 4, −10 − 6) = (−2, −16); BC = (12 + 6, −8 + 10) = (18, 2).
AB · BC = (−2)(18) + (−16)(2) = −36 − 32 = −68.

• a.The angle of elevation of an object above the eye is the angle between the horizontal line from the eye and the line of sight up to the object.
• b.Let AB = h. In △ABC, $\tan 45° = \frac{h}{BC}$, so BC = h. In △ABD, $\tan 30° = \frac{h}{BD}$, so $BD = h\sqrt{3}$.

BD − BC = CD: $h\sqrt{3} - h = 10$, so $h = \frac{10}{\sqrt{3} - 1} = 5(\sqrt{3} + 1) \approx$ 13.66 m.
c. Take B as the origin, BD along the X-axis: $D(h\sqrt{3}, 0)$, A(0, h). Then $\overrightarrow{DA} = (-h\sqrt{3}, h)$ and $\overrightarrow{DB} = (-h\sqrt{3}, 0)$.
$\overrightarrow{DA} \cdot \overrightarrow{DB} = 3h^2 = 3 \times 25(\sqrt{3} + 1)^2 = 300 + 150\sqrt{3} \approx$ 559.81 m².
(Or $|\overrightarrow{DA}|\,|\overrightarrow{DB}|\cos 30° = 2h \times h\sqrt{3} \times \frac{\sqrt{3}}{2} = 3h^2$.)
d. DA · DB = |DB| × (the projection of DA on DB). Here A’s projection on the ground line is B, so the projection is DB itself and the product is $DB^2$, the square of the distance from D to the foot of the house.

• a.The angle of depression of an object below the eye is the angle between the horizontal line from the eye and the line of sight down to the object.
• b.In △ABD, $\tan 45° = \frac{AD}{BD}$, so BD = 80 m; in the same way CD = 80 m. BC = 80 + 80 = 160 m.
• c.Let △PQR be right-angled at P, with M the midpoint of the hypotenuse QR.

Given: $\angle QPR = 90°$, MQ = −MR. To prove: MP = MQ = MR.
PQ = MQ − MP and PR = MR − MP = −MQ − MP [triangle law; M the midpoint].
PQ · PR = 0 [$PQ \perp PR$], so (MQ − MP) · (−MQ − MP) = 0, that is $MP^2 - MQ^2 = 0$.
So MP = MQ, and MQ = MR since M is the midpoint: MP = MQ = MR. Hence proved.
In the tower’s figure, $\angle BAC = 45° + 45° = 90°$ and D is the midpoint of BC: DA = DB = DC = 80 m, as the theorem says.

• a.+90∘° about O sends (x, y) → (−y, x): $T = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.
• b.$T^{\mathsf{T}} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$.
• c.$\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2 & 4 & 1 \\ 1 & 2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ -2 & -4 & -1 \end{pmatrix}$: A′(1, −2), B′(2, −4), C′(3, −1) (a rotation of −90∘°).
• d.|T| = 0 × 0 − (−1)(1) = 1 ≠ 0, so yes, $T^{-1}$ exists; in fact $T^{-1} = T^{\mathsf{T}}$, the rotation of −90∘° that undoes it.

• a.PQ is horizontal (both points have y = 1), so m₁ = 0; PR is upright (both have x = 5), so m₂ is not defined. So m₁m₂ = −1, the usual rule for perpendicular lines, can’t be used here: a horizontal and an upright line are perpendicular, the one case it leaves out.
• b.PQ = (−9, 0), PR = (0, 5): PQ · PR = 0 + 0 = 0 (so $\angle P = 90°$, as given).
• c.Slope of $QR = \frac{6 - 1}{5 + 4} = \frac{5}{9}$: $y - 1 = \frac{5}{9}(x + 4)$, 9y − 9 = 5x + 20, so 5x − 9y + 29 = 0.
• d.By the theorem of Question 9c (the midpoint of the hypotenuse is equidistant from the vertices), with SQ = −SR and PQ · PR = 0: SP = SQ = SR. Check with the coordinates: $S\left(\frac{1}{2}, \frac{7}{2}\right)$, $\overrightarrow{SP} = \left(\frac{9}{2}, -\frac{5}{2}\right)$, $\overrightarrow{SQ} = \left(-\frac{9}{2}, -\frac{5}{2}\right)$, $\overrightarrow{SR} = \left(\frac{9}{2}, \frac{5}{2}\right)$: each has length $\sqrt{\frac{81 + 25}{4}} = \sqrt{26.5} \approx 5.15$. ✓

• a.3x + 2y + 4 = 0, 3y = 6 (that is y = 2) and x = 0.
• b.Test (−1, 0): −3 + 0 + 4 = 1 ≥ 0 ✓, 0 ≤ 6 ✓, −1 ≤ 0 ✓, so the region is on the side of each line that holds (−1, 0): the triangle with vertices A(0, 2), B(0, −2) and $C\left(-\frac{8}{3}, 2\right)$.

• c.At A, the sides lie along y = 2 (horizontal) and x = 0 (upright), so $\angle A = 90°$.
• d.$\begin{pmatrix} 1 & 4 \\ 2 & 5 \end{pmatrix}\begin{pmatrix} 0 & 0 & -\frac{8}{3} \\ 2 & -2 & 2 \end{pmatrix} = \begin{pmatrix} 8 & -8 & \frac{16}{3} \\ 10 & -10 & \frac{14}{3} \end{pmatrix}$: A′(8, 10), B′(−8, −10), $C'\left(\frac{16}{3}, \frac{14}{3}\right)$.

• a.$\frac{\vec{a} + \vec{b}}{2}$.
• b.$\overrightarrow{OD} = \frac{1}{2}\left[(5\vec{i} + 7\vec{j}) + (7\vec{i} + 3\vec{j})\right] =$ 6i + 5j.
• c.AD = (5, 2), BC = (2, −4). AD · BC = 10 − 8 = 2; $|\overrightarrow{AD}| = \sqrt{29}$, $|\overrightarrow{BC}| = \sqrt{20}$.

$\cos\theta = \frac{2}{\sqrt{29}\sqrt{20}} = \frac{2}{\sqrt{580}} = \frac{1}{\sqrt{145}}$, so $\theta \approx 85.24°$.
d. Slope of $AD = \frac{5 - 3}{6 - 1} = \frac{2}{5}$: $y - 3 = \frac{2}{5}(x - 1)$, so 2x − 5y + 13 = 0.

• a.The midpoint of BC is D(7, 5). Slope of $AD = \frac{5 - 3}{7 - 2} = \frac{2}{5}$: $y - 3 = \frac{2}{5}(x - 2)$, so 2x − 5y + 11 = 0.
• b.AD = (5, 2), BC = (2, −4): AD · BC = 2, $\cos\theta = \frac{2}{\sqrt{29}\sqrt{20}} = \frac{1}{\sqrt{145}}$, so $\theta \approx 85.24°$.

a. A parabola opening upward with its vertex at (−2, 5) and axis x = −2; f(0) = 9, f(−4) = 9, f(1) = 14, f(−5) = 14.

• b.Continuous: it is a polynomial, so at every x its value is defined and the values near x head for it. The graph has no break anywhere.
• c.Yes, all are 30. $f(2.99) = 4.99^2 + 5 = 29.9001$, f(2.999) = 29.990001: LHL = 30. $f(3.01) = 5.01^2 + 5 = 30.1001$, f(3.001) = 30.010001: RHL = 30. f(3) = 25 + 5 = 30.