Solution Manual for Thomas' Calculus, SI
Units, 15th edition Joel R. Hass Christopher
E. Heil Maurice D. Weir
Practice Exam: 200 Multiple-Choice Questions (Sample of
20)
Instructions:
• Select the correct answer (A–D) for each question based on Thomas' Calculus, SI Units,
15th Edition, Chapters 1–3.
• Use SI units (e.g., metres, seconds) and show reasoning where applicable.
• Submit as a PDF, including a signed declaration of academic honesty.
Chapter 1: Functions (Sample Questions 1–7)
1. What is the domain of ( f(x) = \frac{2}{x^2 - 4} )?
A) ( (-\infty, \infty) )
B) ( (-\infty, -2) \cup (-2, 2) \cup (2, \infty) )
C) ( [-2, 2] )
D) ( (-\infty, 2) \cup (2, \infty) )
Answer: B
Explanation: The denominator ( x^2 - 4 = (x - 2)(x + 2) = 0 ) when ( x = \pm 2 ). Thus,
the domain excludes ( x = \pm 2 ), so ( (-\infty, -2) \cup (-2, 2) \cup (2, \infty) ) (Hass et
al., 2025, Ch. 1).
2. What is the range of ( f(x) = \frac{1}{x} )?
A) ( (-\infty, \infty) )
B) ( (-\infty, 0) \cup (0, \infty) )
C) ( [0, \infty) )
D) ( (-\infty, 0] )
Answer: B
Explanation: For ( y = \frac{1}{x} ), ( x = \frac{1}{y} ). Since ( x \neq 0 ), ( y \neq 0 ).
As ( x \to \infty ), ( y \to 0^+ ); as ( x \to -\infty ), ( y \to 0^- ). Thus, the range is ( (-\infty,
0) \cup (0, \infty) ) (Hass et al., 2025, Ch. 1).
3. If ( f(x) = 3x + 2 ), what is ( f^{-1}(x) )?
A) ( \frac{x - 2}{3} )
B) ( \frac{x + 2}{3} )
C) ( 3x - 2 )
D) ( \frac{2 - x}{3} )
, Answer: A
Explanation: Solve ( y = 3x + 2 ) for ( x ): ( y - 2 = 3x \implies x = \frac{y - 2}{3} ).
Thus, ( f^{-1}(x) = \frac{x - 2}{3} ) (Hass et al., 2025, Ch. 1).
4. Which function represents a vertical stretch of ( f(x) = x^2 ) by a factor of 3?
A) ( 3x^2 )
B) ( x^2 + 3 )
C) ( (3x)^2 )
D) ( x^3 )
Answer: A
Explanation: A vertical stretch by 3 multiplies the function by 3: ( f(x) = 3x^2 ) (Hass et
al., 2025, Ch. 1).
5. What is the period of ( f(x) = \sin(2x) )?
A) ( \pi )
B) ( 2\pi )
C) ( \frac{\pi}{2} )
D) ( 4\pi )
Answer: A
Explanation: For ( \sin(kx) ), the period is ( \frac{2\pi}{k} ). Here, ( k = 2 ), so period =
( \frac{2\pi}{2} = \pi ) (Hass et al., 2025, Ch. 1).
6. If ( f(x) = \sqrt{x + 3} ), what is the domain?
A) ( (-\infty, \infty) )
B) ( [-3, \infty) )
C) ( (0, \infty) )
D) ( (3, \infty) )
Answer: B
Explanation: The expression ( \sqrt{x + 3} ) requires ( x + 3 \geq 0 \implies x \geq -3 ).
Thus, the domain is ( [-3, \infty) ) (Hass et al., 2025, Ch. 1).
7. Which of the following is an even function?
A) ( f(x) = x^3 )
B) ( f(x) = x^2 + 1 )
C) ( f(x) = 2x )
D) ( f(x) = \sin(x) )
Answer: B
Explanation: An even function satisfies ( f(-x) = f(x) ). For ( f(x) = x^2 + 1 ), ( f(-x) = (-
x)^2 + 1 = x^2 + 1 = f(x) ), so it’s even (Hass et al., 2025, Ch. 1).
Chapter 2: Limits and Continuity (Sample Questions 8–14)
8. Evaluate ( \lim_{x \to 4} \frac{x^2 - 16}{x - 4} ).
A) 4
B) 8
C) 0
D) Does not exist
Answer: B
Units, 15th edition Joel R. Hass Christopher
E. Heil Maurice D. Weir
Practice Exam: 200 Multiple-Choice Questions (Sample of
20)
Instructions:
• Select the correct answer (A–D) for each question based on Thomas' Calculus, SI Units,
15th Edition, Chapters 1–3.
• Use SI units (e.g., metres, seconds) and show reasoning where applicable.
• Submit as a PDF, including a signed declaration of academic honesty.
Chapter 1: Functions (Sample Questions 1–7)
1. What is the domain of ( f(x) = \frac{2}{x^2 - 4} )?
A) ( (-\infty, \infty) )
B) ( (-\infty, -2) \cup (-2, 2) \cup (2, \infty) )
C) ( [-2, 2] )
D) ( (-\infty, 2) \cup (2, \infty) )
Answer: B
Explanation: The denominator ( x^2 - 4 = (x - 2)(x + 2) = 0 ) when ( x = \pm 2 ). Thus,
the domain excludes ( x = \pm 2 ), so ( (-\infty, -2) \cup (-2, 2) \cup (2, \infty) ) (Hass et
al., 2025, Ch. 1).
2. What is the range of ( f(x) = \frac{1}{x} )?
A) ( (-\infty, \infty) )
B) ( (-\infty, 0) \cup (0, \infty) )
C) ( [0, \infty) )
D) ( (-\infty, 0] )
Answer: B
Explanation: For ( y = \frac{1}{x} ), ( x = \frac{1}{y} ). Since ( x \neq 0 ), ( y \neq 0 ).
As ( x \to \infty ), ( y \to 0^+ ); as ( x \to -\infty ), ( y \to 0^- ). Thus, the range is ( (-\infty,
0) \cup (0, \infty) ) (Hass et al., 2025, Ch. 1).
3. If ( f(x) = 3x + 2 ), what is ( f^{-1}(x) )?
A) ( \frac{x - 2}{3} )
B) ( \frac{x + 2}{3} )
C) ( 3x - 2 )
D) ( \frac{2 - x}{3} )
, Answer: A
Explanation: Solve ( y = 3x + 2 ) for ( x ): ( y - 2 = 3x \implies x = \frac{y - 2}{3} ).
Thus, ( f^{-1}(x) = \frac{x - 2}{3} ) (Hass et al., 2025, Ch. 1).
4. Which function represents a vertical stretch of ( f(x) = x^2 ) by a factor of 3?
A) ( 3x^2 )
B) ( x^2 + 3 )
C) ( (3x)^2 )
D) ( x^3 )
Answer: A
Explanation: A vertical stretch by 3 multiplies the function by 3: ( f(x) = 3x^2 ) (Hass et
al., 2025, Ch. 1).
5. What is the period of ( f(x) = \sin(2x) )?
A) ( \pi )
B) ( 2\pi )
C) ( \frac{\pi}{2} )
D) ( 4\pi )
Answer: A
Explanation: For ( \sin(kx) ), the period is ( \frac{2\pi}{k} ). Here, ( k = 2 ), so period =
( \frac{2\pi}{2} = \pi ) (Hass et al., 2025, Ch. 1).
6. If ( f(x) = \sqrt{x + 3} ), what is the domain?
A) ( (-\infty, \infty) )
B) ( [-3, \infty) )
C) ( (0, \infty) )
D) ( (3, \infty) )
Answer: B
Explanation: The expression ( \sqrt{x + 3} ) requires ( x + 3 \geq 0 \implies x \geq -3 ).
Thus, the domain is ( [-3, \infty) ) (Hass et al., 2025, Ch. 1).
7. Which of the following is an even function?
A) ( f(x) = x^3 )
B) ( f(x) = x^2 + 1 )
C) ( f(x) = 2x )
D) ( f(x) = \sin(x) )
Answer: B
Explanation: An even function satisfies ( f(-x) = f(x) ). For ( f(x) = x^2 + 1 ), ( f(-x) = (-
x)^2 + 1 = x^2 + 1 = f(x) ), so it’s even (Hass et al., 2025, Ch. 1).
Chapter 2: Limits and Continuity (Sample Questions 8–14)
8. Evaluate ( \lim_{x \to 4} \frac{x^2 - 16}{x - 4} ).
A) 4
B) 8
C) 0
D) Does not exist
Answer: B
