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Calculus I: MATH 1351-030
Fall 2011
Exam 1 (B)
Name:
Instructions. (Part 1: Multiple Choice) Solve each of the following problems. Choose the
best solution to each problem and clearly mark your choice on the scantron form.
1. Which of the following statements best describes The Horizontal Line Test?
(a)
4 A function f has an inverse if and only if no horizontal line intersects the graph of
y = f (x) at more than one point.
(b) A curve in the plane is the graph of a function if and only if it intersects any horizontal
line at least once.
(c) The y-intercepts of the graph of f are the points (x, y) that intersect a horizontal
line.
(d) A curve in the plane is the graph of a function if and only if it intersects no horizontal
line more than once.
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(b)
4 f −1 (x) = 21 (x − 3)
(d) The inverse does not exist.
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2. Let f (x) = 2x + 3. Find f −1 if it exists.
(a) f −1 (x) = 2(x + 3)
(c) f −1 (x) = 2x − 3
(e) f −1 (x) = 21 x − 3
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(e) A function f has an inverse if and only if its graph intersects a horizontal line at two
or more points.
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3. What are the exact values of sec−1 (−1)
(a) 3π/2
(c)
4 π
(e) 0, π
(b) π/2, 3π/2
(d) 1/π
4. Find the center and radius of the circle defined by x2 − 8x + y 2 + 2y + 7 = 0.
(a) Center is C = (−4, 1) and radius is R = 10
(b) Center is C = (−4, −1) and radius is R = 10
(c) Center is C = (4, −1) and radius is R = 10
√
(d)
4 Center is C = (4, −1) and radius is R = 10
√
(e) Center is C = (−4, 1) and radius is R = 10
√
5. If f (x) = x + 5 and g(x) = 8x − 9, what is f (g(x))?
√
√
(a) 8√x − 4
(b) 2√2x + 1
(d) 8 x + 5 + 9
(c)
4 2√2x − 1
(e) 8 x + 5 − 9
Calculus I: MATH 1351-030/Exam 1 (B) – Page 2 of 6 – Name:
6. Find the exact value of sec
√
(a) √2/2
(c)
4 √2
(e) 3
π
4
.
(b) 1/2
√
(d) 2 3/3
7. Find the slope m, x-intercept, and y-intercept of the line given by the equation
5x + 3y − 15 = 0.
(a) m = − 53 , x-intercept = (−3, 0), y-intercept = (0, 5)
(b) m = 53 , x-intercept = (−3, 0), y-intercept = (0, −5)
(c)
4 m = − 35 , x-intercept = (3, 0), y-intercept = (0, 5)
(d) m = 3, x-intercept = (3, 0), no y-intercept
e
(e) m = 3, x-intercept = (3, 0), y-intercept = (0, 5)
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8. Given the two points (−1, 2) and (3, −2), what is the distance between them?
(a) 8
(b) 2√
(c) −2
(d) 2 2
√
(e)
4 4 2
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9. Which of the following best describes the curve represented by y − k = m(x − h)?
√
(a) The equation of a circle with center C = (h, k) and radius R = m.
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(b) The equation of a line with slope m and y-intercept (h, k).
(c)
4 The equation of a line with slope m that passes through the point (h, k).
(d) The equation of a line with slope m that passes through the point (−h, −k).
(e) The equation of a line with slope
h
k
and x-intercept (m, 0).
10. Find the domain of the function f defined by
√
f (x) = x2 + 2x
and compute the values f (−1) and f (1) if the x-values are in the domain.
√
S
(a) domain = (−∞, −2) (0, ∞), f (−1) = 0 f (1) = 3.
√
√
(b) domain = (−∞, ∞), f (−1) = − 3, f (1) = 3.
√
S
(c) domain = (−∞, −2) (0, ∞), f (−1) is undefined, f (1) = 3.
S
(d) domain = (−∞, 0] [2, ∞), f (−1) = 1, f (1) is undefined.
√
S
(e)
4 domain = (−∞, −2] [0, ∞), f (−1) is undefined, f (1) = 3.
Calculus I: MATH 1351-030/Exam 1 (B) – Page 3 of 6 – Name:
11. Which of the following are odd functions (i.e., “symmetric with respect to the origin”)?
(I) f (x) =
1
x2
(II) g(x) = x5
(III) h(x) = sin(x)
(IV) s(x) = sin(x) + cos(x)
(b)
4 II and III
(d) I and III
(a) I, II, and III
(c) only II
(e) none are odd
12. Solve the following inequality: −5 ≤ 3 − 2x < 17
(a)
4 (−7, 4]
(b) (−∞, ∞)
S
(c) (−∞, −7) S [4, ∞)
(d) (−7, −4)
(e) (−∞, −7]
[4, ∞)
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13. Solve the following inequality: |6x − 1| > 7
S 4
(a) (−∞, ∞)
(b)
(−∞,
−1]
,
∞
S 3
4
(c) −1, 43
[1, ∞)
(d)
−∞,
−
3
S 4
(e)
4 (−∞, −1)
,∞
3
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c)
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14. Let f be a function with domain D and range R. Which of the following defines “the
inverse of f ”?
(a) The function f −1 with domain D and range R is the inverse of f if f −1 (f (x)) = x
and f (f −1 (y)) = y for all real numbers.
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(b) The function f −1 with domain R and range D is the inverse of f if f −1 (x) = −x.
(c) The function f −1 with domain D and range R is the inverse of f if f −1 (f (x)) = x
for all x in R and f (f −1 (y)) = y for all y in D.
(d)
4 The function f −1 with domain R and range D is the inverse of f if f −1 (f (x)) = x
for all x in D and f (f −1 (y)) = y for all y in R.
(e) The function f −1 with domain R and range D is the inverse of f if f −1 (f (x)) = x
for all x in D and all x in R.
15. Which of the following lines are perpendicular to each other?
(I) y = 21 x + 2
(II) y = 21 x − 4
(III) y = − 12 x − 5
(IV) y = 2x − 4
(a) none are perpendicular
(c) I, II, and III
(e) I and II
(b)
4 III and IV
(d) I and III
Calculus I: MATH 1351-030/Exam 1 (B) – Page 4 of 6 – Name:
16. Which of the following functions are equal?
(I) f (x) =
2x2 +7x−4
2x−1
(II) g(x) = x + 4
(III) h(x) = x + 4, x 6=
1
2
(a) I, II, and III
(c) none are equal
(e) II and III
(b) I and II
(d)
4 I and III
17. Solve the following equation for x:
|1 − 5x| = 2
(a) x = −3/5 and 1/5
(c)
4 x = 3/5 and −1/5
(e) x = 2/5 and 5
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(b) x = 2/5
(d) x = −3/5
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(c) The inverse does not exist.
c)
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18. Let f (x) = tan(x) on the interval − π2 , π2 . Find the inverse f −1 if it exists and state its
domain and range.
(a) f −1 (x) = tan−1 (x) with domain [0, π] and range (−∞, ∞).
(b)
4 f −1 (x) = tan−1 (x) with domain (−∞, ∞) and range − π2 , π2 .
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(d) f −1 (x) = tan−1 (x) with domain (−1, 1) and range − π2 , π2 .
(e) f −1 (x) = tan−1 (x) and domain − π2 , π2 with range (−∞, ∞).
19. Find an equation for the line passing through the point (3, −2) and parallel to the line
2x + 5y − 5 = 0.
(a) 5x + 2y − 16 = 0
(b) 2x − 5y − 4 = 0
(c) 2x + 5y − 4 = 0
(d)
4 2x + 5y + 4 = 0
(e) 5x + 2y + 16 = 0
20. Express the given function h(x) as a composition of two functions f (x) and g(x) such
that h(x) = f (g(x)):
h(x) =
(a) f (x) = x12 , g(x) = x − 7
(c)
4 f (x) = x1 , g(x) = x2 − 7
x2
1
−7
(b) f (x) = 71 , g(x) = x2 − 7
1
(d) f (x) = x−7
, g(x) = x2
Calculus I: MATH 1351-030/Exam 1 (B) – Page 5 of 6 – Name:
Instructions. (Part 2: Written Response) Solve each of the following problems. Show your
work clearly. You must write out all relevant steps. Simply having the correct
answer does not give you credit.
1. Plot the two points (2, −3) and (−4, −1) on a Cartesian plane, calculate the distance
between them, find the coordinates of the midpoint of the line segment between them,
and plot the midpoint on the same graph. Be sure to label all points clearly on your
plot.
√
Solution: Distance is 2 10, midpoint is M = (−1, −2).
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2. Find an equation for the line passing through the point (3, 8) and perpendicular to the
line 9x − 3y − 5 = 0.
Solution: Slope of original line is 3, so slope of perpendicular line is m = −1/3. Use
point-slope or another line form to get: x + 3y − 27 = 0
Calculus I: MATH 1351-030/Exam 1 (B) – Page 6 of 6 – Name:
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3. Sketch a graph of f (x) = cos(x). Clearly indicate a restricted interval on which an
inverse exists. Sketch a graph of cos−1 (x).
Solution: If we restrict cos(x) to the interval [0, π], it is monotonic and the inverse exists.
4. (Bonus) Simplify the expression csc (cos−1 x) Your answer should be completely algebraic; there should be no trigonometric functions in your answer.
Solution:
Let α = cos−1 (x). Then cos(α) = x. Use reference triangle with sides
√
1
x, 1 − x2 , 1 to see that csc (cos−1 x) = csc(α) = √1−x
2