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Adams Calculus 8th Edition Solution
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Adams Calculus 8th Edition Solution

Adams Calculus 8th Edition Solution   Instant Download - Complete Test Bank With Answers     Sample Questions Are Posted Below   INSTRUCTOR’S SOLUTIONS MANUAL SECTION 5.7 (PAGE 328)24. For intersections: |x| = √2 cos(π x/4). Thus x = ±1.Area A = 2∫ 10(√2 cos π x4 − x)d x=(8√2π sin π x4 − x2)∣∣∣∣10= …

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Adams Calculus 8th Edition Solution

 

Instant Download – Complete Test Bank With Answers

 

 

Sample Questions Are Posted Below

 

INSTRUCTOR’S SOLUTIONS MANUAL SECTION 5.7 (PAGE 328)
24. For intersections: |x| = √2 cos(π x/4). Thus x = ±1.
Area A = 2
∫ 1
0
(√2 cos π x
4 − x
)
d x
=
(
8√2
π sin π x
4 − x2
)∣
∣
∣
∣
1
0
= 8
π − 1 sq. units.
y
x
y=√2 cos π x
4
A
1
y=|x|
Fig. 5.7.24
25. For intersections: x = sin(π x/2). Thus x = ±1.
Area A = 2
∫ 1
0
(
sin π x
2 − x
)
d x
=
(
− 4
π cos π x
2 − x2
)∣
∣
∣
∣
1
0
= 4
π − 1 sq. units.
y
x
y=x
y=sin π x
2
1
A
Fig. 5.7.25
26. For intersections: ex = x + 2. There are two roots, both
of which must be found numerically. We used a TI-85
solve routine to get x1 ≈ −1.841406 and x2 ≈ 1.146193.
Thus
Area A =
∫ x2
x1
(x + 2 − ex ) d x
=
( x2
2 + 2x − ex
)∣
∣
∣
∣
x2
x1
≈ 1.949091 sq. units.
y
x
y=x+2
y=exA
x1 x2
Fig. 5.7.26
27. Area of R = 4
∫ 1
0
√
x2 − x4 d x
= 4
∫ 1
0
x
√
1 − x2 d x Let u = 1 − x2
du = −2x d x
= 2
∫ 1
0
u1/2 du = 4
3 u3/2
∣
∣
∣
∣
1
0
= 4
3 sq. units.
y
x
y2=x2 −x4
RR
Fig. 5.7.27
28. Loop area = 2
∫ 0
−2
x2√2 + x d x Let u2 = 2 + x
2u du = d x
= 2
∫ √2
0
(u2 − 2)2u(2u) du = 4
∫ √2
0
(u6 − 4u4 + 4u2) du
= 4
( 1
7 u7 − 4
5 u5 + 4
3 u3
)∣
∣
∣
∣
√2
0
= 256√2
105 sq. units.
y
x
−2 A
y2=x4 (2+x)
Fig. 5.7.28
29. The tangent line to y = ex at x = 1 is y − e = e(x − 1),
or y = ex. Thus
Area of R =
∫ 1
0
(ex − ex) d x
=
(
ex − ex2
2
)∣
∣
∣
∣
1
0
= e
2 − 1 sq. units.
207Copyright © 2014 Pearson Canada Inc.
SECTION 5.7 (PAGE 328) ADAMS and ESSEX: CALCULUS 8
y
x
(1,e)
y=ex
R
y=ex
Fig. 5.7.29
30. The tangent line to y = x3 at (1, 1) is y −1 = 3(x −1), or
y = 3x − 2. The intersections of y = x3 and this tangent
line occur where x3 − 3x + 2 = 0. Of course x = 1
is a (double) root of this cubic equation, which therefore
factors to (x − 1)2(x + 2) = 0. The other intersection is
at x = −2. Thus
Area of R =
∫ 1
−2
(x3 − 3x + 2) d x
=
( x4
4 − 3x2
2 + 2x
)∣
∣
∣
∣
1
−2
= − 15
4 − 3
2 + 6 + 2 + 4 = 27
4 sq. units.
y
x
(1,1)
y=3x−2R
(−2,−8)
y=x3
Fig. 5.7.30
Review Exercises 5 (page 329)
1. 1
j2 − 1
( j + 1)2 = j2 + 2 j + 1 − j2
j2( j + 1)2 = 2 j + 1
j2( j + 1)2
n∑
j=1
2 j + 1
j2( j + 1)2 =
n∑
j=1
( 1
j2 − 1
( j + 1)2
)
= 1
12 − 1
(n + 1)2 = n2 + 2n
(n + 1)2
2. The number of balls is
40 × 30 + 39 × 29 + · · · + 12 × 2 + 11 × 1
=
30∑
i=1
i (i + 10) = (30)(31)(61)
6 + 10 (30)(31)
2 = 14,105.
3. xi = 1 + (2i /n), (i = 0, 1, 2, . . . , n), 1xi = 2/n.
∫ 3
1
f (x) d x = lim
n→∞
n∑
i=1
(x2
i − 2xi + 3) 2
n
= lim
n→∞
2
n
n∑
i=1
[(
1 + 4i
n + 4i 2
n2
)
−
(
2 + 4i
n
)
+ 3
]
= lim
n→∞
2
n
n∑
i=1
[
2 + 4
n2 i 2
]
= lim
n→∞
( 4
n n + 8
n3
n(n + 1)(2n + 1)
6
)
= 4 + 8
3 = 20
3
4. Rn = ∑n
i=1(1/n)√1 + (i /n) is a Riemann sum for
f (x) = √1 + x on the interval [0, 1]. Thus
lim
n→∞ Rn =
∫ 1
0
√1 + x d x
= 2
3 (1 + x)3/2
∣
∣
∣
∣
1
0
= 4√2 − 2
3 .
5.
∫ π
−π
(2 − sin x) d x = 2(2π ) −
∫ π
−π
sin x d x = 4π − 0 = 4π
6.
∫ √5
0
√
5 − x2 d x = 1/4 of the area of a circle of radius √5
= 1
4 π(√5)2 = 5π
4
7. ∫ 3
1
(
1 − x
2
)
d x = area A1 − area A2 = 0
y
x
y=1− x
2
A1
A2
3
1
Fig. R-5.7
8. ∫ π
0 cos x d x = area A1 − area A2 = 0
y
x
y=cos x
A1
A2
π
Fig. R-5.8
9. ¯f = 1
2π
∫ π
−π
(2 − sin(x3)) d x = 1
2π [2(2π ) − 0] = 2
10. ¯h = 1
3
∫ 3
0
|x − 2| d x = 1
3
5
2 = 5
6 (via #9)
208Copyright © 2014 Pearson Canada Inc.
INSTRUCTOR’S SOLUTIONS MANUAL REVIEW EXERCISES 5 (PAGE 329)
11. f (t) =
∫ t
13
sin(x2) d x, f ′(t) = sin(t2)
12. f (x) =
∫ sin x
−13
√
1 + t2 dt, f ′(x) =
√
1 + sin2 x(cos x)
13. g(s) =
∫ 1
4s
esin u du, g′(s) = −4esin(4s)
14. g(θ ) =
∫ ecos θ
esin θ
ln x d x
g′(θ ) = (ln(ecos θ ))ecos θ (− sin θ ) − (ln(esin θ ))esin θ cos θ
= − sin θ cos θ (ecos θ + esin θ )
15. 2 f (x) + 1 = 3
∫ 1
x
f (t) dt
2 f ′(x) = −3 f (x) H⇒ f (x) = Ce−3x/2
2 f (1) + 1 = 0
− 1
2 = f (1) = Ce−3/2 H⇒ C = − 1
2 e3/2
f (x) = − 1
2 e(3/2)(1−x).
16. I =
∫ π
0
x f (sin x) d x Let x = π − u
d x = −du
= −
∫ 0
π
(π − u) f (sin(π − u)) du (but sin(π − u) = sin u)
= π
∫ π
0
f (sin u) du −
∫ π
0
u f (sin u) du
= π
∫ π
0
f (sin x) d x − I.
Now, solving for I , we get
∫ π
0
x f (sin x) d x = I = π
2
∫ π
0
f (sin x) d x.
17. y = 2 + x − x2 and y = 0 intersect where 2 + x − x2 = 0,
that is, where (2 − x)(1 + x) = 0, namely at x = −1 and
x = 2. Since 2 + x − x2 ≥ 0 on [−1, 2], the required area
is
∫ 2
−1
(2 + x − x2) d x =
(
2x + x2
2 − x3
3
)∣
∣
∣
∣
2
−1
= 9
2 sq. units..
18. The area bounded by y = (x − 1)2, y = 0, and x = 0 is
∫ 1
0
(x − 1)2 d x = (x − 1)3
3
∣
∣
∣
∣
1
0
= 1
3 sq. units..
19. x = y − y4 and x = 0 intersect where y − y4 = 0, that
is, at y = 0 and y = 1. Since y − y4 ≥ 0 on [0, 1], the
required area is
∫ 1
0
(y − y4 − 0) d y =
( y2
2 − y5
5
)∣
∣
∣
∣
1
0
= 3
10 sq. units.
20. y = 4x − x2 and y = 3 meet where x2 − 4x + 3 = 0, that
is, at x = 1 and x = 3. Since 4x − x2 ≥ 3 on [1, 3], the
required area is
∫ 3
1
(4x − x2 − 3) d x =
(
2x2 − x3
3 − 3x
)∣
∣
∣
∣
3
1
= 4
3 sq. units.
21. y = sin x and y = cos(2x) intersect at x = π/6, but
nowhere else in the interval [0, π/6]. The area between
the curves in that interval is
∫ π/6
0
(cos(2x) − sin x) d x = ( 1
2 sin(2x) + cos x)∣
∣
∣
∣
π/6
0
=
√3
4 +
√3
2 − 1 = 3√3
4 − 1 sq. units..
22. y = 5 − x2 and y = 4/x2 meet where 5 − x2 = 4/x2, that
is, where
x4 − 5x2 + 4 = 0
(x2 − 1)(x2 − 4) = 0.
There are four intersections: x = ±1 and x = ±2. By
symmetry (see the figure) the total area bounded by the
curves is
2
∫ 2
1
(
5 − x2 − 4
x2
)
d x = 2
(
5x − x3
3 + 4
x
)∣
∣
∣
∣
2
1
= 4
3 sq. units.
y
x
y = 4
x2
y = 5 − x2
1 2
A A
Fig. R-5.22
23.
∫
x2 cos(2x3 + 1) d x Let u = 2x3 + 1
du = 6x2 d x
= 1
6
∫
cos u du = sin u
6 + C = sin(2x3 + 1)
6 + C
209Copyright © 2014 Pearson Canada Inc.
REVIEW EXERCISES 5 (PAGE 329) ADAMS and ESSEX: CALCULUS 8
24.
∫ e
1
ln x
x d x Let u = ln x
du = d x/x
=
∫ 1
0
u du = u2
2
∣
∣
∣
∣
1
0
= 1
2
25.
∫ 4
0
√
9t2 + t4 dt
=
∫ 4
0
t
√
9 + t2 dt Let u = 9 + t2
du = 2t dt
= 1
2
∫ 25
9
√u du = 1
3 u3/2
∣
∣
∣
∣
25
9
= 98
3
26.
∫
sin3(π x) d x
=
∫
sin(π x)
(
1 − cos2(π x)
)
d x Let u = cos(π x)
du = −π sin(π x) d x
= − 1
π
∫
(1 − u2) du
= 1
π
( u3
3 − u
)
+ C = 1
3π cos3(π x) − 1
π cos(π x) + C
27.
∫ ln 2
0
eu
4 + e2u du Let v = eu
dv = eu du
=
∫ 2
1
dv
4 + v2
= 1
2 tan−1 v
2
∣
∣
∣
∣
2
1
= π
8 − 1
2 tan−1 1
2
28.
∫ 4√e
1
tan2(π ln x)
x d x Let u = π ln x
du = (π/x) d x
= 1
π
∫ π/4
0
tan2 u du = 1
π
∫ π/4
0
(sec2 u − 1) du
= 1
π (tan u − u)
∣
∣
∣
∣
π/4
0
= 1
π − 1
4
29.
∫ sin √2s + 1
√2s + 1 ds Let u = √2s + 1
du = ds/√2s + 1
=
∫
sin u du = − cos u + C = − cos √2s + 1 + C
30.
∫
cos2 t
5 sin2 t
5 dt = 1
4
∫
sin2 2t
5 dt
= 1
8
∫ (
1 − cos 4t
5
)
dt
= 1
8
(
t − 5
4 sin 4t
5
)
+ C
31. F(x) =
∫ x2 −2x
0
1
1 + t2 dt.
Since 1/(1 + t2) > 0 for all t, F(x) will be minimum
when
x2 − 2x = (x − 1)2 − 1
is minimum, that is, when x = 1. The minimum value is
F(1) =
∫ −1
0
dt
1 + t2 = tan−1t
∣
∣
∣
∣
−1
0
= − π
4 .
F has no maximum value; F(x) < π/2 for all x, but
F(x) → π/2 if x2 − 2x → ∞, which happens as
x → ±∞.
32. f (x) = 4x − x2 ≥ 0 if 0 ≤ x ≤ 4, and f (x) < 0
otherwise. If a < b, then ∫ b
a f (x) d x will be maximum if
[a, b] = [0, 4]; extending the interval to the left of 0 or to
the right of 4 will introduce negative contributions to the
integral. The maximum value is
∫ 4
0
(4x − x2) d x =
(
2x2 − x3
3
)∣
∣
∣
∣
4
0
= 32
3 .
33. The average value of v(t) = d x/dt over [t0, t1] is
1
t1 − t0
∫ t1
t0
d x
dt dt = 1
t1 − t0
x(t)
∣
∣
∣
∣
t1
t0
= x(t1) − x(t0)
t1 − t0
= vav.
34. If y(t) is the distance the object falls in t seconds from
its release time, then
y′′(t) = g, y(0) = 0, and y′(0) = 0.
Antidifferentiating twice and using the initial conditions
leads to
y(t) = 1
2 gt2.
The average height during the time interval [0, T ] is
1
T
∫ T
0
1
2 gt2 dt = g
2T
T 3
3 = gT 2
6 = y
( T
√3
)
.
35. Let f (x) = ax3 + bx2 + cx + d so that
∫ 1
0
f (x) d x = a
4 + b
3 + c
2 + d.
We want this integral to be
(
f (x1) + f (x2)
)
/2 for all
choices of a, b, c, and d. Thus we require that
a(x3
1 + x3
2 ) + b(x2
1 + x2
2 ) + c(x1 + x2) + 2d
= 2
∫ 1
0
f (x) d x = a
2 + 2b
3 + c + 2d.
210Copyright © 2014 Pearson Canada Inc.
INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 5 (PAGE 330)
It follows that x1 and x2 must satisfy
x3
1 + x3
2 = 1
2 (1)
x2
1 + x2
2 = 2
3 (2)
x2 + x2 = 1. (3)
At first glance this system may seem overdetermined;
there are three equations in only two unknowns. How-
ever, they do admit a solution as we now show. Squar-
ing equation (3) and subtracting equation (2) we get
2x1 x2 = 1/3. Subtracting this latter equation from
equation (2) then gives (x2 − x1)2 = 1/3, so that
x2 − x1 = 1/√3 (the positive square root since we want
x1 < x2). Adding and subtracting this equation and equa-
tion (3) then produces the values x2 = (√3 + 1)/(2√3)
and x1 = (√3 − 1)/(2√3). These values also satisfy
equation (1) since
x3
1 + x3
2 = (x2 + x2)(x2
1 − x1x2 + x2
2 ) = 1 ×
( 2
3 − 1
6
)
= 1
2 .
Challenging Problems 5 (page 330)
1. xi = 2i/n , 0 ≤ i ≤ n, f (x) = 1/x on [1, 2]. Since f is
decreasing, f is largest at the left endpoint and smallest
at the right endpoint of any interval [2(i−1)/n , 2i/n ] of the
partition. Thus
U ( f, Pn ) =
n∑
i=1
1
2(i−1)/n (2i/n − 2(i−1)/n )
=
n∑
i=1
(21/n − 1) = n(21/n − 1)
L( f, Pn ) =
n∑
i=1
1
2i/n (2i/n − 2(i−1)/n )
=
n∑
i=1
(1 − 2−1/n ) = n(1 − 2−1/n ) = U ( f, Pn )
21/n .
Now, by l’Hˆopital’s rule,
lim
n→∞ n(21/n − 1) = lim
x→∞
21/x − 1
1/x
[ 0
0
]
= lim
x→∞
21/x ln 2(−1/x2)
−1/x2 = ln 2.
Thus limn→∞ U ( f, Pn ) = limn→∞ L( f, Pn ) = ln s.
2. a) cos
(
( j + 1
2 )t
)
− cos
(
( j − 1
2 )t
)
= cos( j t) cos( 1
2 t) − sin( j t) sin( 1
2 t)
− cos( j t) cos( 1
2 t) − sin( j t) sin( 1
2 t)
= −2 sin( j t) sin( 1
2 t).
Therefore, we obtain a telescoping sum:
n∑
j=1
sin( j t)
= − 1
2 sin( 1
2 t)
n∑
j=1
[
cos
(
( j + 1
2 )t
)
− cos
(
( j − 1
2 )t
)]
= − 1
2 sin( 1
2 t)
[
cos
(
(n + 1
2 )t
)
− cos( 1
2 t)
]
= 1
2 sin( 1
2 t)
[
cos( 1
2 t) − cos
(
(n + 1
2 )t
)]
.
b) Let Pn = {0, π
2n , 2π
2n , 3π
2n , . . . nπ
2n } be the partition of
[0, π/2] into n subintervals of equal length
1x = π/2n. Using t = π/2n in the formula ob-
tained in part (a), we get
∫ π/2
0
sin x d x
= lim
n→∞
n∑
j=1
sin
( jπ
2n
) π
2n
= lim
n→∞
π
2n
1
2 sin(π/(4n))
(
cos π
4n − cos (2n + 1)π
4n
)
= lim
n→∞
π/(4n)
sin(π/(4n)) lim
n→∞
(
cos π
4n − cos (2n + 1)π
4n
)
= 1 ×
(
cos 0 − cos π
2
)
= 1.
3. a) sin
(
( j + 1
2 )t
)
− sin
(
( j − 1
2 )t
)
= sin( j t) cos( 1
2 t) + cos( j t) sin( 1
2 t)
− sin( j t) cos( 1
2 t) + cos( j t) sin( 1
2 t)
= 2 cos( j t) sin( 1
2 t).
Therefore, we obtain a telescoping sum:
n∑
j=1
cos( j t)
= 1
2 sin( 1
2 t)
n∑
j=1
[
sin
(
( j + 1
2 )t
)
− sin
(
( j − 1
2 )t
)]
= 1
2 sin( 1
2 t)
[
sin
(
(n + 1
2 )t
)
− sin( 1
2 t)
]
.
211Copyright © 2014 Pearson Canada Inc.
CHALLENGING PROBLEMS 5 (PAGE 330) ADAMS and ESSEX: CALCULUS 8
b) Let Pn = {0, π
3n , 2π
3n , 3π
3n , . . . nπ
3n } be the partition of
[0, π/3] into n subintervals of equal length
1x = π/3n. Using t = π/3n in the formula ob-
tained in part (a), we get
∫ π/2
0
cos x d x
= lim
n→∞
n∑
j=1
cos
( jπ
3n
) π
3n
= lim
n→∞
π
3n
1
2 sin(π/(6n))
(
sin (2n + 1)π
6n − sin π
6n
)
= lim
n→∞
π/(6n)
sin(π/(6n)) lim
n→∞
(
sin (2n + 1)π
6n − sin π
6n
)
= 1 ×
(
sin π
3 − sin 0
)
=
√3
2 .
4. f (x) = 1/x2, 1 = x0 < x1 < x2 < · · · < xn = 2. If
ci = √xi−1 xi , then
x2
i−1 < xi=1 xi = c2
i < x2
i ,
so xi−1 < ci < xi . We have
n∑
i=1
f (ci ) 1xi =
n∑
i=1
1
xi−1 xi
(xi − xi−1)
=
n∑
i=1
( 1
xi−1
− 1
xi
)
(telescoping)
= 1
x0
− 1
xn
= 1 − 1
2 = 1
2 .
Thus
∫ 2
1
d x
x2 = lim
n→∞
n∑
i=1
f (ci ) 1xi = 1
2 .
5. We want to prove that for each positive integer k,
n∑
j=1
jk = nk+1
k + 1 + nk
2 + Pk−1 (n),
where Pk−1 is a polynomial of degree at most k − 1.
First check the case k = 1:
n∑
j=1
j = n(n + 1)
2 = n1+1
1 + 1 + n
2 + P0(n),
where P0(n) = 0 certainly has degree ≤ 0. Now assume
that the formula above holds for k = 1, 2, 3, . . . , m. We
will show that it also holds for k = m + 1. To this end,
sum the the formula
( j+1)m+2− jm+2 = (m+2) jm+1+ (m + 2)(m + 1)
2 jm +· · ·+1
(obtained by the Binomial Theorem) for j = 1, 2, . . . , n.
The left side telescopes, and we get
(n + 1)m+2 − 1m+2 = (m + 2)
n∑
j=1
jm+1
+ (m + 2)(m + 1)
2
n∑
j=1
jm + · · · +
n∑
j=1
1.
Expanding the binomial power on the left and using the
induction hypothesis on the other terms we get
nm+2 + (m + 2)nm+1 + · · · = (m + 2)
n∑
j=1
jm+1
+ (m + 2)(m + 1)
2
nm+1
m + 1 + · · · ,
where the · · · represent terms of degree m or lower in the
variable n. Solving for the remaining sum, we get
n∑
j=1
jm+1
= 1
m + 2
(
nm+2 + (m + 2)nm+1 + · · · − m + 2
2 nm+1 − · · ·
)
= nm+2
m + 2 + nm+1
2 + · · ·
so that the formula is also correct for k = m + 1. Hence
it is true for all positive integers k by induction.
b) Using the technique of Example 2 in Section 6.2 and
the result above,
∫ a
0
xk d x = lim
n→∞
a
n
n∑
j=1
( a
n
) j
= ak+1 lim
n→∞
1
nk+1
n∑
j=1
jk
= ak+1 lim
n→∞
( 1
k + 1 + 1
2n + Pk−1 (n)
nk+1
)
= ak+1
k + 1 .
6. Let f (x) = ax3 + bx2 + cx + d. We used Maple to
calculate the following:
The tangent to y = f (x) at P = ( p, f ( p)) has equation
y = g(x) = ap3 + bp2 + cp + d + (3ap2 + 2bp + c)(x − p).
This line intersects y = f (x) at x = p (double root) and
at x = q, where
q = − 2ap + b
a .
212Copyright © 2014 Pearson Canada Inc.
INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 5 (PAGE 330)
Similarly, the tangent to y = f (x) at x = q has equation
y = h(x) = aq3 + bq2 + cq + d + (3aq2 + 2bq + c)(x − q),
and intersects y = f (x) at x = q (double root) and
x = r , where
r = − 2aq + b
a = 4ap + b
a .
The area between y = f (x) and the tangent line at P is
the absolute value of
∫ q
p
( f (x) − g(x) d x
= − 1
12
( 81a4 p4 + 108a3 bp3 + 54a2 b2 p2 + 12ab3 p + b4
a3
)
.
The area between y = f (x) and the tangent line at
Q = (q, f (q)) is the absolute value of
∫ r
q
( f (x) − h(x) d x
= − 4
3
( 81a4 p4 + 108a3bp3 + 54a2 b2 p2 + 12ab3 p + b4
a3
)
,
which is 16 times the area between y = f (x) and the
tangent at P.
7. We continue with the calculations begun in the previous
problem. P and Q are as they were in that problem, but
R = (r, f (r )) is now the inflection point of y = f (x),
given by f ′′(r ) = 0. Maple gives
r = − b
3a .
Since
p − r = b + 3ap
a and r − q = 2(b + 3ap)
a
have the same sign, R must lie between Q and P on the
curve y = f (x). The line Q R has a rather complicated
equation y = k(x), which we won’t reproduce here, but
the area between this line and the curve y = f (x) is
the absolute value of ∫ q
r ( f (x) − k(x)) d x, which Maple
evaluates to be
− 4
81
( 81a4 p4 + 108a3 bp3 + 54a2b2 p2 + 12ab3 p + b4
a3
)
,
which is 16/27 of the area between the curve and its
tangent at P. This leaves 11/27 of that area to lie be-
tween the curve, Q R, and the tangent, so Q R divides the
area between y = f (x) and its tangent at P in the ratio
16/11.
8. Let f (x) = ax4 + bx3 + cx2 + d x + e. The tangent to
y = f (x) at P = ( p, f ( p)) has equation
y = g(x) = ap4+bp3+cp2+d p+e+(4ap3+3bp2+2cp+d)(x− p),
and intersects y = f (x) at x = p (double root) and at the
two points
x = −2ap − b ± √b2 − 4ac − 4abp − 8a2 p2
2a .
If these latter two points coincide, then the tangent is a
“double tangent.” This happens if
8a2 p2 + 4abp + 4ac − b2 = 0,
which has two solutions, which we take to be p and q:
p = −b + √3b2 − 8ac
4a
q = −b − √3b2 − 8ac
4a = − p − b
2a .
(Both roots exist and are distinct provided 3b2 > 8ac.)
The point T corresponds to x = t = ( p + q)/2 = −b/4a.
The tangent to y = f (x) at x = t has equation
y = h(x) = − 3b4
256a3 + b2c
16a2 − bd
4a +e+
( b3
8a2 − bc
2a + d
) (
x + b
4a
)
and it intersects y = f (x) at the points U and V with
x-coordinates
u = −b − √2√3b2 − 8ac
4a ,
v = −b + √2√3b2 − 8ac
4a .
Q
P
A
R
U
B
S
V
T
Fig. C-5.8
213Copyright © 2014 Pearson Canada Inc.
CHALLENGING PROBLEMS 5 (PAGE 330) ADAMS and ESSEX: CALCULUS 8
a) The areas between the curve y = f (x) and the lines
P Q and U V are, respectively, the absolute values of
A1 =
∫ q
p
( f (x)−g(x)) d x and A2 =
∫ v
u
(h(x)− f (x)) d x.
Maple calculates these two integrals and simplifies
the ratio A1/ A2 to be 1/√2.
b) The two inflection points A and B of f have x–
coordinates shown by Maple to be
α = −3b − √3(3b2 − 8ac)
12a and
β = −3b + √3(3b2 − 8ac)
12a .
It then determines the four points of intersection of
the line y = k(x) through these inflection points and
the curve. The other two points have x-coordinates
r = −3b − √15(3b2 − 8ac)
12a and
s = −3b + √15(3b2 − 8ac)
12a .
The region bounded by RS and the curve y = f (x)
is divided into three parts by A and B. The areas of
these three regions are the absolute values of
A1 =
∫ α
r
(k(x) − f (x)) d x
A2 =
∫ β
α
( f (x) − k(x)) d x
A3 =
∫ s
β
(k(x) − f (x)) d x.
The expressions calculated by Maple for k(x) and
for these three areas are very complicated, but Maple
simplifies the rations A3/ A1 and A2/ A1 to 1 and 2
respectively, as was to be shown.
214Copyright © 2014 Pearson Canada Inc

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