Chapter Three
Chapter Three
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CHAPTER THREE Hughes Hallett et al ConcepTests and Answers and Comments for Section 3.1
1. The graph of a function f is given in Figure 3.1. If f is a polynomial of degree 3, then the value of f ′′′ (0) is
(a) Positive
(b) Negative
(c) Zero
1
f
−1
1
x
−1
Figure 3.1
ANSWER:
(b). Because the graph of this polynomial of degree 3 is negative for large values of x, the coefficient of x3 will be
negative. (Recall the third derivative of a polynomial of degree 3 is a constant.)
COMMENT:
You could ask students why f (x) could not become positive for x > 1.
2. The graph of a function f is given in Figure 3.2. If f is a polynomial of degree 3, then f ′′′ (0) is
(a) Positive
(b) Negative
(c) Zero
1
f
−1
1
x
−1
Figure 3.2
ANSWER:
(a). Because the graph of this polynomial of degree 3 is positive for large values of x, the coefficient of x3 will be
positive. (Recall the third derivative of a polynomial of degree 3 is a constant.)
COMMENT:
You could ask students if there could be other inflection points for f .
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CHAPTER THREE Hughes Hallett et al 103
3. The graph of a function f is given in Figure 3.3. If f is a polynomial of degree 3, then the values of f ′ (0), f ′′ (0), and
f ′′′ (0) are (respectively)
(a) 0, 0, +
(d) 0, −, −
(b) 0, 0, −
(e) +, −, +
(c) 0, +, −
(f) 0, +, +
2
1
f
−1
1
x
−1
Figure 3.3
ANSWER:
(f). There is a horizontal tangent at the origin, so f ′ (0) = 0. The graph shows that f has horizontal intercepts at −1
and 0, with a double root at 0. Thus f has the form f (x) = k(x + 1)x2 . Because f (x) > 0 for x > 0, then k > 0. So
f ′ (x) = k(3x2 + 2x), f ′′ (x) = k(6x + 2), and f ′′′ (x) = 6k.
COMMENT:
You could ask students why a double root at zero means that f has a factor of x2 .
4. The graph of a function f is given in Figure 3.4. If f is a polynomial of degree 3, then the values of f ′ (0), f ′′ (0), and
f ′′′ (0) are (respectively)
(a) +, 0, +
(d) +, −, −
(b) −, 0, −
(e) +, −, +
(c) +, 0, −
(f) +, +, +
0.5
f
−1
1
x
−0.5
Figure 3.4
ANSWER:
(c). The graph shows that f has horizontal intercepts at x = −1, 0, and 1. Thus f has the form f (x) = kx(x +
1)(x − 1). Because f (x) > 0 for 0 < x < 1, we have k < 0. Then f ′ (x) = k(3x2 − 1), f ′′ (x) = k(6x), and
f ′′′ (x) = 6k.
COMMENT:
You could ask why there could not be another horizontal intercept outside this viewing window.
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CHAPTER THREE Hughes Hallett et al 5. The graph of a function f is given in Figure 3.5. If f is a polynomial of degree 3, then the values of f ′ (0), f ′′ (0), and
f ′′′ (0) are (respectively)
(a) −, +, +
(d) −, +, +
(b) −, −, −
(e) +, −, +
(c) −, +, −
(f) +, +, +
y
f
8
7
6
5
4
3
2
1
−1
1
−1
−2
2
x
Figure 3.5
ANSWER:
(c). At x = 0, the graph is decreasing and concave up, so f ′ (0) < 0 and f ′′ (0) ≥ 0. Because the graph becomes
more negative as x increases beyond 1, the sign of the coefficient of x3 (and thus the sign of f ′′′ (0)) is negative. (Recall
the third derivative of a polynomial of degree 3 is a constant.)
COMMENT:
You could ask why the graph of this function could not become positive for larger values of x.
6. The graph of a function f is given in Figure 3.6. If f is a polynomial of degree 3, then the values of f ′ (0), f ′′ (0), and
f ′′′ (0) are (respectively)
(a) −, −, +
(d) −, +, +
(b) −, 0, −
(e) +, −, +
(c) −, +, −
(f) +, +, +
2
f
1
−2
−1
1
2
x
−1
−2
Figure 3.6
ANSWER:
(d). At x = 0, the graph is decreasing and concave up, so f ′ (0) < 0 and f ′′ (0) ≥ 0. Because the graph is positive
as x increases beyond 0.5, the sign of the coefficient of x3 (and thus the sign of f ′′′ (0)) is positive. (Recall the third
derivative of a polynomial of degree 3 is a constant.)
COMMENT:
You could ask why the graph of this function could not become negative for larger values of x.
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CHAPTER THREE Hughes Hallett et al 105
ConcepTests and Answers and Comments for Section 3.4
1. Given the graphs of the functions f (x) and g(x) in Figures 3.7 and 3.8, which of (a)–(d) is a graph of f (g(x))?
y
y
3
g(x)
2
f (x)
2
1
1
1
2
x
3
1
Figure 3.7
3
x
Figure 3.8
y
(a)
2
y
(b)
6
2
4
1
2
1
2
3
x
y
(c)
2
3
1
2
3
x
y
(d)
3
1
10
8
2
6
4
1
2
1
2
3
x
x
ANSWER:
(c). Because (f (g(x)))′ = f ′ (g(x))g ′ (x), we see f (g(x)) has a horizontal tangent whenever g ′ (x) = 0 or f ′ (g(x)) =
0. Now, f ′ (g(x)) = 0 for 1 < g(x) < 2 and this approximately corresponds to 1.7 < x < 2.5.
COMMENT:
You could have students give specific places in the other choices where there was a conflict with information given
in the graphs of f and g.
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CHAPTER THREE Hughes Hallett et al 2. Given the graphs of the functions f (x) and g(x) in Figures 3.9 and 3.10, which of (a)–(d) is a graph of f (g(x))?
y
y
4
16
g(x)
3
12
2
8
1
4
1
2
3
f (x)
x
4
1
2
Figure 3.9
4
x
Figure 3.10
y
(a)
3
y
(b)
16
6
5
12
4
8
3
2
4
1
1
2
3
4
x
y
(c)
10
10
8
8
6
6
4
4
2
2
2
3
4
x
2
3
4
1
2
3
4
x
y
(d)
1
1
x
ANSWER:
(a). Because (f (g(x)))′ = f ′ (g(x))g ′ (x), we see f (g(x)) has a horizontal tangent whenever g ′ (x) = 0 or f ′ (g(x)) =
0. Now f ′ (x) = 0 only when x = 2, so the composite function only has horizontal tangents when g ′ (x) = 0 or when
g(x) = 2.
COMMENT:
Students may want to check points, which is tedious. This is an opportunity to show the power of reasoning based
on the chain rule.
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CHAPTER THREE Hughes Hallett et al 107
3. Given the graphs of the function g(x) and f (x) in Figures 3.11 and 3.12 , which of (a)–(d) represents f (g(x))?
y
y
3
27
2
18
1
f (x)
9
g(x)
1
2
3
4
5
x
−3
−1
−2
−1
1
−18
−3
−27
Figure 3.11
x
Figure 3.12
y
y
(b)
5
27
1
18
2
3
x
−5
−10
9
−15
1
2
3
4
5
−20
x
−25
−30
−9
y
(d)
y
(c)
3
−9
−2
(a)
2
1
27
24
21
18
15
12
9
6
3
−3
1
2
3
4
5
x
1
2
3
4
5
x
ANSWER:
(c). Because (f (g(x)))′ = f ′ (g(x))g ′ (x), we see f (g(x)) has a horizontal tangent whenever g ′ (x) = 0 or g(x) =
0. This happens when x = 0, 2, and 4. f (g(x)) is also negative for x > 4. Alternatively, f (g(4)) = f (0) = 0 identifies
answer (c).
COMMENT:
You could have students give specific places in the other choices where there was a conflict with information given
in the graphs of f and g.
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CHAPTER THREE Hughes Hallett et al ConcepTests and Answers and Comments for Section 3.5
1. List in order (from smallest to largest) the following functions as regards to the maximum possible value of their slope.
(a) sin x
(b) sin(2x)
(c) sin(3x)
(d) sin(x/2)
ANSWER:
(d), (a), (b), (c). The derivative of sin(αx) is α cos(αx). The largest value of cos(αx) is 1, so the ranking is based
only on the value of α.
COMMENT:
You could mimic this question replacing the sine function by cosine.
2. At which of the following values of x does sin(4πx2 ) attain the largest slope?
(a) x = 0
(b) x = 1/2
(c) x = 1
(d) x = 2
ANSWER:
(d). The derivative of sin(4πx2 ) is 8πx cos(4πx2 ). At these specific values of x, we have the values 0, −4π, 8π,
and 16π.
COMMENT:
You could mimic this question replacing the sine function by cosine.
3. Which of the graphs are those of sin(2x) and sin(3x)?
(a) (I) = sin(2x), (II) = sin(3x)
(c) (II) = sin(2x), (III) = sin(3x)
y
(I)
(b) (I) = sin(2x), (III) = sin(3x)
(d) (III) = sin(2x), (IV) = sin(3x)
y
(II)
1
1
−6 −4 −2
2
−1
(III)
4
6
x
−6 −4 −2
−1
y
(IV)
1
−6 −4 −2
−1
2
4
6
2
4
6
x
y
1
2
4
6
x
−6 −4 −2
x
−1
ANSWER:
d
sin(αx) = α cos(αx), the graph of sin(3x) will have steeper slopes than sin(2x). The first positive
(b). Because
dx
zeros are at x = π/2 for sin(2x) and x = π/3 for sin(3x).
COMMENT:
You may want to emphasize that there are many ways of solving this problem, for example, finding the x-intercepts,
remembering the properties of trigonometric functions, and using calculus to look at the slope of the function at various
points. You could ask for equations for the graphs in choices (II) and (IV).
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CHAPTER THREE Hughes Hallett et al 109
4. Which of the graphs is that of sin(x2 )?
y
(a)
1
1
1
−1
(c)
2
3
4
5
6
x
1
−1
y
(d)
1
x
y
1
1
−1
y
(b)
2
3
4
5
6
x
π/2
π
3π/2
2π
x
−1
ANSWER:
d
(d). Since
sin(x2 ) = 2x cos(x2 ), the maximum value of the slope of the graph increases as x increases. Also,
dx
√
the zeros of sin(x2 ) are x = nπ for n = 0, 1, 2, 3, . . ., which are closer together as x increases.
COMMENT:
You could have students give specific points on the graphs in the other choices which have properties not consistent
with those of y = sin(x2 ).
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CHAPTER THREE Hughes Hallett et al ConcepTests and Answers and Comments for Section 3.8
For Problems 1–2, consider the graphs of f (x) = (1/2)ex and g(x) = (1/2)e−x in Figure 3.13.
y
3
2
g(x)
f (x)
1
−2
−1
1
2
x
−1
Figure 3.13
1. Which of the following functions represents f (x) + g(x)?
y
(a)
−2
−1
4
3
3
2
2
1
1
1
−1
(c)
−2
−1
y
(b)
4
2
x
y
−2
−1
−1
(d)
4
4
3
3
2
2
1
1
1
−1
ANSWER:
(b)
COMMENT:
This is one way to introduce cosh x.
2
x
−2
−1
−1
1
2
1
2
x
y
x
c 2009, John Wiley & Sons
CHAPTER THREE Hughes Hallett et al 2. Which of the following functions represents f (x) − g(x)?
y
(a)
−2
4
4
3
3
2
2
1
1
−1
1
−1
(c)
−2
y
(b)
2
x
−2
−1
−1
y
(d)
4
4
3
3
2
2
1
1
−1
1
−1
ANSWER:
(d)
COMMENT:
This is one way to introduce sinh x.
2
x
−2
−1
−1
1
2
1
2
x
y
x
111
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CHAPTER THREE Hughes Hallett et al ConcepTests and Answers and Comments for Section 3.9
1. In which of the following graphs will using local linearity to approximate the value of the function near x = c give the
least error as ∆x becomes larger?
(a)
(b)
c
x
(c)
c
x
(d)
c
x
c
x
ANSWER:
(d). Local linearity will give exact answers in graph (d) because the tangent line at x = c is identical with the graph
on the interval shown.
COMMENT:
You could lay a pencil on the overhead to illustrate the tangent line in each case.
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CHAPTER THREE Hughes Hallett et al 113
2. To estimate the numerical value of the square root of a number, we use a tangent line approximation about x = a. From
√
Figure 3.14, the graph of x, decide for which number the error in using this approximation has the smallest magnitude.
√
(a) √4.2 about x = 4
(b) √4.5 about x = 4
(c) √9.2 about x = 9
(d) √9.5 about x = 9
(e) √16.2 about x = 16
(f) 16.5 about x = 16
y
5
4
3
2
1
5
10
15
20
x
Figure 3.14
ANSWER:
(e). The least error occurs at the point closest to the perfect square whose tangent line has the flattest slope. Thus the
answer is (e).
COMMENT:
You could ask the students why the approximation is taken about perfect squares. You could also have students graph
√
y = x on the window 0 < x < 17, 0 < y < 4.5 and have their graphing calculator draw tangent lines at the appropriate
points.
ConcepTests and Answers and Comments for Section 3.10
1. A function intersects the x-axis at points a and b, where a < b. The slope of the function at a is positive and the slope at
b is negative. Which of the following is true for any such function? There exists some point on the interval (a, b) where
(a)
(b)
(c)
(d)
The slope is zero and the function has a local maximum.
The slope is zero but there is not local maximum.
There is a local maximum but there does not have to be a point at which the slope is zero.
None of the above have to be true.
ANSWER:
(d). The function f (x) = 1/x2 − 1, where a = −1 and b = 1 is an example where none of the choices, (a), (b), or
(c) are valid.
COMMENT:
Rather than look for an equation, you might ask for an example in the form of a graph. You might also ask if requiring
the function to be bounded would change their answer.
2. A function intersects the x-axis at points a and b, where a < b. The slope of the function at a is positive and the slope at b
is negative. Which of the following is true for any such function for which the limit of the function exists at every point?
There exists some point on the interval (a, b) where
(a) The slope is zero and the function has a local maximum.
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CHAPTER THREE Hughes Hallett et al (b) The slope is zero but there is not local maximum.
(c) There is a local maximum but there does not have to be a point at which the slope is zero.
(d) None of the above have to be true.
ANSWER:
(d). The graph in Figure 3.15 is an example of a function for which none of the choices (a), (b), or (c) are valid.
1
x
a
b
−1
Figure 3.15
COMMENT:
If you graph the function given in the Answer, you might ask students for its equation.
3. A continuous function intersects the x-axis at points a and b, where a < b. The slope of the function at a is positive and
the slope at b is negative. Which of the following is true for any such function? There exists some point on the interval
(a, b) where
(a)
(b)
(c)
(d)
The slope is zero and the function has a local maximum.
The slope is zero but there is not local maximum.
There is a local maximum but there does not have to be a point at which the slope is zero.
None of the above have to be true.
ANSWER:
(c). The Extreme Value Theorem guarantees there will be a maximum. The function
f (x) = (b − a)/2 − |x − (a + b)/2| shows that the slope of such a function need not be zero.
COMMENT:
This answer is more believable if the function is graphed.
4. A continuous and differentiable function intersects the x-axis at points a and b, (a < b). The slope of the function at a is
positive and the slope at b is negative. Which of the following is true for any such function? There exists some point on
the interval (a, b) where
(a)
(b)
(c)
(d)
The slope is zero and the function has a local maximum.
The slope is zero but there is not local maximum.
There is a local maximum but there does not have to be a point at which the slope is zero.
None of the above have to be true.
ANSWER:
(a). The Extreme Value Theorem guarantees there will be a maximum, while the Mean Value Theorem requires a
horizontal tangent on the interval.
COMMENT:
You might review the hypothesis of the Mean Value Theorem.