1. No category

2 Sample Exam Questions
Problems marked with an asterisk (*) are particularly challenging and should be given careful consideration.
1. Consider the following graph of f .
4
3
2
1
4
1
2
3
5
t
(a) What is lim+ f (t)? lim− f (t)? lim− f (t)? lim f (t)?
t→0
t→0
t→2
t→−∞
(b) For what values of x does lim f (t) exist?
t→x
(c) Does f have any vertical asymptotes? If so, where?
(d) Does f have any horizontal asymptotes? If so, where?
(e) For what values of x is f discontinuous?
2. Find values for a and b that will make f continuous everywhere, if
⎧
⎪
⎨ 3x + 1
f (x) =
ax + b
⎪
⎩ 2
x
if x < 2
if 2 ≤ x < 5
if 5 ≤ x
−1
3. Find the vertical and horizontal asymptotes for f (x) = a−1 + x−1
, where a is a positive number.
4. Consider the function f (x) =
x+4
.
x2 + 3x − 4
(a) What is the domain of f ?
(b) Compute lim f (x), if this limit exists.
x→−4
(c) Is f continuous at x = −4? Explain your answer by either proving that f is continuous at x = −4 or
telling how to modify f to make it continuous.
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CHAPTER 2 LIMITS AND DERIVATIVES
5. Let f be a continuous function such that f (−1) = −1 and f (1) = 1. Classify the following statements
as
A. Always true
B. Never true
C. Sometimes true
Justify your answers.
(a) f (0) = 0
(b) For some x with −1 ≤ x ≤ 1, f (x) = 0
(c) For all x with −1 ≤ x ≤ 1, −1 ≤ f (x) ≤ 1
(d) Given any y in [−1, 1], then y = f (x) for some x in [−1, 1].
(e) If x < −1 or x > 1, then f (x) < −1 or f (x) > 1.
(f) f (x) = −1 for x < 0 and f (x) = 1 for x > 0.
6. Let f be the function whose graph is given below.
y
2
f
1
0
1
2
(a) Sketch a plausible graph of f .
3
x
(b) Sketch a plausible graph of a function F
such that F = f and F (0) = 1.
2
3
1
2
1
_1
_2
7. Suppose that the line tangent to the graph of y = f (x) at x = 3 passes through the points (−2, 3) and
(4, −1).
(a) Find f (3).
(b) Find f (3).
(c) Find an equation of the line tangent to f at x = 3.
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CHAPTER 2 SAMPLE EXAM QUESTIONS
8. Give examples of functions f (x) and g (x) with lim f (x) = ∞, lim g (x) = ∞ and
x→∞
x→∞
f (x)
(a) lim
=∞
x→∞ g (x)
f (x)
=6
x→∞ g (x)
(b) lim
(c) lim
x→∞
f (x)
=0
g (x)
f (x)
= −1? Either give an example or explain why it is not possible.
x→∞ g (x)
(d) Is it possible to have lim
9. Each of the following limits represent the derivative of a function f at some point a. State a formula for
f and the value of the point a.
(3 + h)2 − 9
h→0
h
2x − 2
x→1 x − 1
(b) lim
(a) lim
(x + 1)3/2 − 8
x→3
x−3
sin (π (2 + h)) − 0
h→0
h
(c) lim
(d) lim
10. Let
⎧ √
⎪
⎨ 3−x
f (x) =
x2
⎪
⎩
27/x
if x ≤ 1
if 1 < x < 3
if x ≥ 3
(a) Evaluate each limit, if it exists.
(i) lim− f (x)
(ii) lim+ f (x)
(iii) lim f (x)
(iv) lim− f (x)
(v) lim+ f (x)
(vi) lim f (x)
(vii) lim f (x)
(viii) lim f (x)
x→1
x→3
x→1
x→1
x→3
x→9
x→3
x→−6
(b) Where is f discontinuous?
11. The graph of f (x) is given below.
For which value(s) of x is f (x) not differentiable? Justify your
answer(s).
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CHAPTER 2 LIMITS AND DERIVATIVES
12. A bicycle starts from rest and its distance travelled is recorded in the following table at one-second
intervals.
t (s)
d (ft)
0
0
1
10
2
24
3
42
4
63
5
84.5
6
107
(a) Compute the average speed of the bicycle from t = 1 to t = 2 and from t = 2 to t = 3.
(b) Estimate the speed after 2 seconds.
(c) Estimate the speed after 5 seconds.
(d) Estimate the speed after 6 seconds.
(e) Can we determine if the cyclist’s speed is constantly increasing? Explain.
13. Referring to the graphs given below, find each limit, if it exists. If a limit does not exist, explain why not.
f (x)
x→0 g (x)
(a) lim
g (x)
x→−1 f (x)
g (x)
(g) lim
x→1 f (x)
(b) lim [g (x) · f (x)]
x→1
(e) lim [g (x) + f (x)] (f) lim− [x + f (x)]
x→−1
(d) lim [x · g (x)]
(c) lim
x→1
x→2
14. The following is a graph of f , the derivative of some function f .
y
2
1
_2
_1
f»
1
2
x
_1
_2
(a) Where is f increasing?
(b) Where does f have a local minimum? Where does f have a local maximum?
(c) Where is f concave up?
(d) Assuming that f (0) = −1, sketch a possible graph of f .
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CHAPTER 2 SAMPLE EXAM SOLUTIONS
2 Sample Exam Solutions
1. (a) lim f (t) = ∞, lim f (t) = 1, lim f (t) = 3, lim f (t) = 1
t→0+
t→0−
t→−∞
t→2−
(b) lim f (t) exists for all x except x = 0 and x = 2.
t→x
(c) There is a vertical asymptote at x = 0.
(d) There is a horizontal asymptote at y = 1.
(e) f is discontinuous at x = 0, 2, and 4.
2. Solve 3 (2) + 1 = 2a + b and 52 = 5a + b to get a = 6, b = −5.
3. Taking lim f (x) gives a horizontal asymptote at y = a. Algebraic simplification gives a vertical
x→∞
asymptote at x = −a. The function is undefined at x = 0, but there is no asymptote there because
lim f (x) = 0.
x→0
4. f (x) =
x+4
(x + 4) (x − 1)
(a) The domain is all values of x except x = 1 and x = −4.
(b) Algebraic simplification gives a limit of − 15 .
(c) f is not continuous at x = −4, for it is not defined there. It can be modified by defining f (−4) to be
− 15 .
5. (a) C. True for f (x) = x, untrue for f (x) = x2 + x − 1
(b) A. True by the Intermediate Value Theorem
(c) C. True for f (x) = x, untrue for f (x) = x2 + x − 1
(d) A. True by the Intermediate Value Theorem
(e) C. True for f (x) = x, untrue for f (x) = x2 + x − 1
(f) B. lim f (x) does not exist, contradicting the continuity of f .
x→0
6. (a) Answers will vary. Look for:
(i) zeros at 1 and 2
(iii) f negative for x ∈ (1, 2)
(ii) f positive for x ∈ (0, 1) and (2, 4)
(iv) f flattens out for x > 2.5
(b) Answers will vary. Look for
(i) F (0) = 1
(iv) F is closest to being flat at x = 2
(ii) F is always increasing
(v) F is concave up for x ∈ (0, 1) and x ∈ (2, 4)
(iii) F is never perfectly flat
(vi) F is concave down for x ∈ (1, 2)
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CHAPTER 2 LIMITS AND DERIVATIVES
7. (a)
3 − (−1)
2
=−
−2 − 4
3
(b) The equation of the tangent line is y − 3 = − 23 (x + 2), so f (3) = − 23 (3 + 2) + 3 = − 13 .
(c) The equation of the tangent line is y − 3 = − 23 (x + 2).
8. Answers will vary; the following are samples only.
(a) f (x) = x2 , g (x) = x
(b) f (x) = 6x, g (x) = x
(c) f (x) = x, g (x) = x2
f (x)
= −1, either f or g would have to be negative for large x. This
g (x)
contradicts the assumption that lim f (x) = lim g (x) = ∞.
(d) This is not possible. For lim
x→∞
x→∞
x→∞
9. Answers will vary.
(a) f (x) = x2 , a = 3
(b) f (x) = 2x , a = 1
(c) f (x) = (x + 1)3/2 , a = 3
√
10. (a) (i) 2
(ii) 1
(iii) Does not exist
(d) f (x) = sin (πx), a = 2
(iv) 9
(v) 9
(vi) 9
(vii) 3
(viii) 3
(b) f is discontinuous at x = 1.
11. f isn’t differentiable at x = 1, because it is not continuous there; at x = −2, because it has a vertical
tangent there; and at x = 4, because it has a cusp there.
12. (a) Answers will vary.
One good answer would be to compute the average speed between 1 and 2
(14 ft/s) and the average speed between 2 and 3 (18 ft/s) and average them to get 16 ft/s. This is also
the answer obtained by computing the average speed between 1 and 3.
(b) Answers will vary. Using reasoning similar to the previous part, we get an estimate of 22 ft/s, but it
could be argued that a number closer to 22.5 would be more accurate.
(c) Answers will vary. The average speed between t = 5 and t = 6 is 22.5 ft/s.
(d) Since we are given information only about the cyclist’s position at one-second intervals, we cannot
determine if the speed is constantly increasing.
13. (a)
1
2
(b) 0
(d) −4
(c) Does not exist, because lim f (x) = 0 while lim g (x) = 1.
x→−1
(e) 1
(f) 2
14. (a) f is increasing on (−1, 1).
(c) f is concave up where
(g) 0
(d)
(b) Local minimum at x = −1; local
maximum at x = 1
f (x)
x→−1
y
_2
_1
0
f
1
_1 Inflection point
is
increasing, that is, on (−2, 0).
_2
152
2 x