7.1 Homework Solutions
1.
y = 2x 3 + 9x 2 − 168x
y′ = 6x 2 +18x −168 = 0
c.v.’s: x = 4,−7
3.
extremes:
x2 + 1
Note: Domain x ≠ ±2, 0
x 3 − 4x
x 3 − 4x ( 2x ) − x 2 +1 3x 2 − 4 −x 4 − 7x 2 + 4
y′ =
=
= 0, DNE
2
2
x 3 − 4x
x 3 − 4x
y=
(
)
(
(
)
)(
)
c.v.’s: x = 0,±2,−.729,.729
5.
y = −400, 931
(
) (27x
− 12
)
extremes:
y = .606,−.606
Note: Domain x ∈⎡ − 3, 4 9 ⎤ ∪ ⎡⎣ 3, ∞ ⎤⎦
⎣
⎦
y = 9x 3 − 4x 2 − 27x + 12
1
y′ = 9x 3 − 4x 2 − 27x +12
2
(
2
)
− 8x − 27 =
c.v.’s: x = ± 3, 4 9 , 1.159, − .863
(
27x 2 − 8x − 27
2 9x − 4x − 27x +12
3
2
extremes:
70
)
1
2
= 0, DNE
y = 0, 5.151
7.
( )
y = x2
9 − x2
3
(
1
y′ = x ⋅ 9 − x 2
3
2
)
−2 3
c.v.’s: x = ±3, 0, ±
9.
(
⋅−2x + 9 − x
27
2
2
)
1
3
⋅2x =
extremes:
(
−2x 4x 2 − 27
(
3 9 − x2
)
2
3
) = 0, DNE
y = 0,8.845
3
x + cos x on x ∈ ⎡⎣ −2π , 2π ⎤⎦
2
3
y′ =
− sin x = 0
2
y=
5π
4π π 2π
,−
, ,
3
3 3 3
y = −4.441, 6.441, − 4.034, − 4.128, 1.407, 1.314
c.v.’s: x = −2π , 2π , −
extremes:
71
11. extremes: y = −6, 1, − 8, 3
3
y
2
1
x
−4
−3
−2
−1
1
2
3
4
5
−1
−2
−3
−4
−5
−6
−7
−8
−9
−10
13.
f ( x ) = sin ( 2π x ) , x ∈ ⎡⎣−1,1⎤⎦
f is continuous and differentiable.
f ( −1) = 0 = f (1)∴Rolle’s Theorem conditions are met.
f ′ ( x ) = 2π cos ( 2π x ) = 0
⎧ π
2π x = ⎨ ± ± 2π n
⎩ 2
⎧ 1
x = ⎨± ± n
⎩ 4
x = ± 14, ± 34
72
6
7
8
15.
y
4
3
2
1
x
1
2
3
4
5
6
7
8
9
10
−1
−2
−3
−4
c ≈ 3.4
17. Given that f (x) is a twice differentiable function with f (2) = 6. The
derivative of f (x), f ’(x), is pictured below on the closed interval −1 ≤ x ≤ 5. The
graph of f ’(x) has horizontal tangent lines at x = 1 and at x = 3. Use this
information to answer the questions below.
4
y
2
x
3
−2
−4
−6
−8
−10
a) Find the x-coordinate of each point of inflection on f. Explain your reasoning.
73
x = 1 and x = 3 because these are places where f ’ has a slope of 0. This
means that f ” = 0 at these points. In addition, f ” changes from positive to
negative or negative to positive at these values. Therefore, f has concavity
changes at these x values.
b) Find where the function f attains its absolute maximum value and its absolute
minimum value on the closed interval −1 ≤ x ≤ 5 . Show thework that leads to this
conclusion.
f has critical values at x = –1, 4, and 5, (endpoints of the interval or 0 of f ’).
Maxima occur at x = –1 and 5, with the absolute maximum occurring at x =
–1. This is because f decreases from –1 (because f ’ is negative) and most of
the area under f ’ is negative, meaning the value of f (5) < f (–1). The
absolute minimum occurs at x = 4 because it is the only minimum on the
function (f switches from decreasing to increasing at this point).
c) Let g be defined as the function g ( x ) = x ⋅ f ( x ) . Find the equation of the graph
of the tangent line to g at x = 2.
y −12 = −2 ( x − 2)
74
7.2 Homework Solutions
1.
f ( x) =
f ′( x) =
x
, ⎡0, 2 ⎤⎦
x +1 ⎣
x 2 +1 (1) − x ( 2x )
2
(
)
(x
2
)
+1
2
=
(1− x )(1+ x )
(x
2
)
+1
c.v.’s: x = −1, 1, 0, 2
f ( 0 ) = 0 ← Absolute minimum
f (1) = 1 2 ← Absolute maximum
f (2) = 2 5
3.
2
f ( z ) = ze− z , ⎡⎣0, 2 ⎤⎦
f ′ ( z ) = z ⋅e−z ( −1) + e−z (1) = 0
c.v.’s: x = 0, 1, 2
f ( 0 ) = 0 ← Relative minimum
f (1) = 1 e ← Absolute maximum
f (2) = 1 2
e
5.
f ( x ) = e− x − e−2x , ⎡⎣0, 1⎤⎦
f ′ ( x ) = −e− x − e−2 x ⋅ ( −2 ) = e− x ( 2e− x −1) = 0
c.v.’s: x = 0, ln2, 1
f ( 0 ) = 0 ← Relative minimum
f ( ln2 ) = .25 ← Absolute maximum
f (1) = .233
75
7.
h ( f ) = f 3 − 12 f + 21
h′ ( f ) = 3 f 2 −12 = 0
c.v.’s: f = ±2
h′′ ( f ) = 6 f → h′′ ( −2 ) = −12 < 0∴ minimum
h′′ ( 2 ) = 12 > 0∴ maximum
9.
Let h be a continuous function with h(2) = 5. The graph of the piecewise
linear function below, h’, is shown below on the domain of −3 ≤ x ≤ 7 .
a) Find the x-coordinates of all points of inflection on the graph of y = h ( x ) for the
interval −3 < x < 7 . Justify your answer.
POIs occur where h” = 0 or does not exist, and switches signs. This occurs at
x = 1 and 4; h” does not exist because h’ is not differentiable for those x.
The sign of h” switches because the slopes of h’ switch from positive to
negative or negative to positive.
b) Find the absolute maximum value of h on the interval −3 ≤ x ≤ 7 . Justify your
answer.
−3
Maxima occur at x = –3, 2, and 7. h ( −3) = 5 + ∫ h ( x ) dx = 3.5 , h ( 2) = 5 , and
2
7
h ( 7 ) = 5 + ∫ h ( x ) dx = 3.25. Therefore, the absolute maximum is at x = 2.
2
c) Find the average rate of change of h on the interval −3 ≤ x ≤ 7 .
Average rate of change of h would be the slope of h on that interval. This
would be
h ( 7 ) − h ( −3 )
1
=−
7 − ( −3 )
40
76
d) Find the average rate of change of h’ on the interval −3 ≤ x ≤ 7 . Does the Mean
Value Theorem applied on this interval guarantee a value of c, for −3 < c < 7 , such
that h”(c) is equal to this average rate of change? Explain why or why not.
0.5 − ( −2 ) 1
= ; the Mean Value Theorem does not guarantee this value for c
7 − ( −3)
4
because h’ is not differentiable throughout this interval.
77
7.3 Homework Solutions
1.
Find two positive numbers whose product is 110 and whose sum is a
minimum.
110
xy = 110 → y =
x
110
Sum = S = x + y → S = x +
x
2
110 x −110
S ′ = 1− 2 =
=0
x
x2
x = 0,± 110 → x = 110 ⇒ y = 110
3.
A farmer with 750 feet of fencing material wants to enclose a rectangular
area and divide it into four smaller rectangular pens with sides parallel to one side
of the rectangle. What is the largest possible total area?
Area = xy
A = x ( 750 − 5x ) = 750x − 5x 2
A′ = 750 − 25x = 0
x = 0, 30
Minimum at x = 1
Largest possible area: 18,000 ft2
78
5.
Find the point on the line y = 4x + 7 that is closest to the origin.
D = x 2 + y2
= x 2 + ( 4x + 7 )
2
D = x 2 + ( 4x + 7 ) = 17x 2 + 56x + 49
D ′ = 34x + 56 = 0
x = −28 17 → y = 7 17
−28 , 7
17 17
2
(
)
7.
Find the area of the largest rectangle that can be inscribed in the ellipse
2
x
y2
+
=1
16 9
79
A = 2xy
⎛
x2
x2 ⎞
A = 2x ⋅b 1− 2 = 2bx ⎜ 1− 2 ⎟
a
a ⎠
⎝
1
2
⎡
⎤
⎡
⎤
⎡
⎤
2⎛
x2 2 ⎞ ⎥
⎢
⎢
⎢
⎥
a
1−
2
2
2
2 ⎥
2
2
⎝
⎠ ⎥
−x
−x + a − x ⎥
−2x + a
a
⎢
⎢
⎢
⎥
A′ = 2b
= 2b
= 2b
1 +
1
1
1
⎢ 2⎛
⎥
⎢
⎥
⎢
2
2
2
2
2
2
2
2
2⎛
2⎛
⎞
⎞
⎞ ⎥
a 2 ⎛⎝ 1− x 2 ⎞⎠ ⎥
⎢ a ⎝ 1− x 2 ⎠
⎢ a ⎝ 1− x 2 ⎠ ⎥
⎢ a ⎝ 1− x 2 ⎠ ⎥
a
a
a
a
⎣
⎦
⎣
⎦
⎣
⎦
x = ±a, ± a
2
2⋅
a
⋅b 1−
2
⎛a ⎞
⎜⎝
2 ⎟⎠
2
a2
Max area =
2ab
a 2 1 2ab 1 2ab
1− ⋅ 2 =
=
= ab
2 a
2
2
2 2
9.
You need to enclose 500 cm3 of fluid in a cylinder. If the material you are
using costs 0.001 dollars per cm3, find the size of the cylinder that minimizes the
cost. If the product that you are containing is a sports drink, do you think that the
size that minimizes cost is the most efficient size? Explain.
(
Cost = .001A = .001 2π r 2 + 2π rh
)
There are too many variables to differentiate, but Volume = 500 = π r 2 h∴
Volume = 500 = π r 2 h → h =
⎛
π r2
⎛ 500 ⎞ ⎞
1 −1
2
r
2 ⎟ ⎟ = .002π r +
π
⎝ πr ⎠⎠
Cost = .001⎜ 2π r 2 + 2π r ⎜
⎝
500
1
C′ = .004π r − r −2 = 0
π
80
.004π r =
1
π r2
1
.004π 2
r = 1.363 → h = 85.670
r3 =
r = 4.301 cm, h = 8.603 cm. This seems to be an inefficient size for the
purpose of holding it in one’s hand: it is 8.603 cm tall (almost 3 and a half
inches) and the same distance across – not a comfortable fit for most
people’s hands.
11.
Given that the area of a triangle can be calculated with the formula
1
A = ab sin θ , what value of θ will maximize the area of a triangle (given that a
2
and b are constants)?
1
A′ = abcosθ = 0
2
π
cosθ = 0 ⇒ θ =
2
π
is the only critical value.
2
⎛π⎞
1
1
1
A′′ = − absinθ ⇒ A′′ ⎜ ⎟ = − absinθ = − ab < 0∴max
2
2
2
⎝ 2⎠
π
θ=
2
Implied domain of θ ∈ (0,π ) , therefore, θ =
81
7.4 Homework Solutions
82
83
84
11.
Increasing from ( −∞,5) ∪ (7,10) , decreasing from (5,7 ) ∪ (10,∞ )
13.
Decreasing from ( −∞,−5) ∪ (5,∞ ) , increasing from ( −5,5) , concave down
from ( −∞,−7 ) ∪ ( −3,3) ∪ (7,∞ ) , concave up from ( −7,−3) ∪ (3,7 )
85
15.
86
87
7-5 Homework Solutions
1.
+ 0 − 0+ 0 + 0 − 0 +
− 3.2 −1.6 0 1.6 3.2
− 0 + 0− 0 + 0 − 0 +
y″ ←⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
→
− 2.4 −1 0
1 2.4
x
y′
x
(a)
(b)
(c)
(d)
(e)
←⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→
CV: Max at x = –3.2, 1.6
Min at x = –1.6, 3.2
x = –2.4, –1, 0, 1, 2.4
Inc: x ∈ ( −1.6,0) ∪ ( 0,1.6 )
Dec: x ∈ ( −3, −1.6) ∪ (1.6,3)
Up: x ∈ ( −2.4, −1) ∪ ( 0,1) ∪ ( 2.4,3) Down: x ∈ ( −3, −2.4) ∪ ( −1,0) ∪ (1, 2.4)
Sketch:
y
3
2
1
x
−3
3
−1
−2
−3
88
3.
y′
x
(a)
(b)
(c)
(d)
+ 0 − 0 − 0 +
− 2.5
0
2.5
←⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→
y″
x
− 0 + 0 − 0 +
−1.5
0
1.5
←⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→
CV: Max at x = –2.5
Min at x = 2.5
x = –1.5, 0, 1.5
Inc: x ∈ ( −3, −2.5) ∪ ( 2.5,3) Dec: x ∈ ( −2.5,0) ∪ ( 0, 2.5)
Up: x ∈ ( −1.5,0) ∪ (1.5,3) Down: x ∈ ( −3, −1.5) ∪ ( 0,1.5)
y
3
2
1
x
−3
3
−1
−2
−3
5.
y′
x
+ 0 − 0 + 0 −
−2
0
2
y″
x
←⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→
a)
b)
c)
c.v.s : x = −2, 0, 2
i.v.s : x = ±1
Increasing: x ∈( −∞, − 2 ) ∪ ( 0, 2 )
Decreasing: x ∈( −2, 0 ) ∪ ( 2, ∞ )
d)
Concave Up:
e)
Concave Down:
Sketch:
x ∈( −1, 1)
x ∈( −∞, −1) ∪ (1, ∞ )
89
− 0
−1.5
+
0 −
1.5
←⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→
7.
dy/dx
3
2
1
x
−3
3
−1
−2
−3
6
y
4
2
x
−4
−2
2
−2
90
4