CHAPTER 0 PRELIMINARIES
Sample Test Problems
Sample Test Problems, Number 30 – 55
- Find the equation of the line through the indicated point that is parallel to the indicated line, and sketch both lines.
- Write the equation of the line through (–2, 1) that
- Show that (2, –1), (5, 3), and (11, 11) are on the same line.
- Figure 1 can be represented by which equation?
- Figure 2 can be represented by which equation?
- 3y – 4x = 6
- x2 – 2x + y2 = 3
- y=2xx2 + 2
- x = y2 – 3
- Find the points of intersection of the graphs of y = x2 – 2x + 4 and y – x = 4
- Among all lines perpendicular to 4x – y = 2 find the equation of the one that, together with the positive x- and y-axes, forms a triangle of area 8.
- For f(x) = 1/(x + 1) – 1/x, find each value (if possible).
- For g(x) = (x + 1)/x find and simplify each value.
- Describe the natural domain of each function.
- Which of the following functions are odd? Even? Neither even nor odd?
- Sketch the graph of each function.
- Suppose that f is an even function satisfying f(x) = –1 + √x for x ≥ 0. Sketch the graph of f for –4 ≤ x ≤ 4.
- An open box is made by cutting squares of side x inches from the four corners of a sheet of cardboard 24 inches by 32 inches and then turning up the sides. Express the volume V(x) in terms of x. What is the domain for this function?
- Let f(x) = x – (1/x) and g(x) = x2 + 1. Find each value.
- Sketch the graph of each of the following, making use of translations.
- Let f (x) = √(16 – x) and g(x) = x4. What is the domain of each of the following?
- Write F(x) = √(1 + sin2 x) as the complosite of four functions, f ◦ g ◦ h ◦ k.
- Calculate each of the following without using a calculator.
- If sin t = 0.8 and cos t < 0, find each value.
- Write sin 3t in terms of sin t. Hint: 3t = 2t + t.
- A fly sits on the rim of a wheel spinning at the rate of 20 revolutions per minute. If the radius of the whee; is 9 inches, how far does the fly travel in 1 second?
(a)
|
(3, 2): 3x + 2y = 6
|
(b)
|
(1, –1): y
= 2/3 · x + 1
|
(c)
|
(5, 9): y
= 10
|
(d)
|
(–3, 4): x
= –2
|
💬 SOLUTION:
(a)
|
goes through (7, 3);
|
(b)
|
is parallel to 3x – 2y = 5;
|
(c)
|
is perpendicular
to 3x + 4y = 9;
|
(d)
|
is perpendicular
to y = 4;
|
(e)
|
has y-intercept
3.
|
💬 SOLUTION:
💬 SOLUTION:
|
m1
|
=
|
3 + 1
|
=
|
4
|
;
|
||||
|
5 – 2
|
3
|
||||||||
|
m2
|
=
|
11 – 3
|
=
|
8
|
=
|
4
|
;
|
||
|
11 – 5
|
6
|
3
|
|||||||
|
m3
|
=
|
11 + 1
|
=
|
12
|
=
|
4
|
;
|
||
|
11 – 2
|
9
|
3
|
|||||||
m1 = m2
= m3 , so the point lie on the same line.
|
(a)
|
y = x3
|
(b)
|
x = y3
|
(c)
|
y = x2
|
(d)
|
x = y3
|
💬 SOLUTION:
The figure is a
cubic with respect to y.
|
The equation is (b)
x = y3.
|
(a)
|
y = ax2
+ bx + c, with a > 0, b > 0, and c
> 0
|
(b)
|
y = ax2
+ bx + c, with a < 0, b > 0, and c
> 0
|
(c)
|
y = ax2
+ bx + c, with a < 0, b > 0, and c <
0
|
(d)
|
y = ax2
+ bx + c, with a > 0, b > 0, and c <
0
|
💬 SOLUTION:
The figure is a
quadratic, opening downward, with a negative y-intercept.
|
The equation is (c)
y = ax2 + bx
+ c, with a < 0, b > 0 , and c <
0.
|
BACA JUGA:
|
|
In Problems 35-38, sketch the graph of each equation.
💬 SOLUTION:
💬 SOLUTION:
💬 SOLUTION:
💬 SOLUTION:
💬 SOLUTION:
|
y
|
=
|
x2 – 2x + 4
|
and
|
|
y - x
|
=
|
4;
|
|
|
x + 4
|
=
|
x2 – 2x + 4
|
|
|
x2 – 3x
|
=
|
0
|
|
|
x(x
– 3)
|
=
|
0
|
|
|
point of intersection: (0, 4) and (3, 7)
|
💬 SOLUTION:
|
4x – y
|
=
|
2
|
|||||||||||||||
|
y
|
=
|
4x – 2;
|
|||||||||||||||
|
m
|
=
|
–
|
1
|
||||||||||||||
|
4
|
|||||||||||||||||
|
contains (a, 0), (0, b);
|
|||||||||||||||||
|
ab
|
=
|
8
|
|||||||||||||||
|
2
|
|||||||||||||||||
|
ab
|
=
|
16
|
|||||||||||||||
|
b
|
=
|
16
|
|||||||||||||||
|
a
|
|||||||||||||||||
|
b - 0
|
=
|
–
|
b
|
=
|
–
|
1
|
;
|
||||||||||
|
0 – a
|
a
|
4
|
|||||||||||||||
|
a
|
=
|
4b
|
|||||||||||||||
|
a
|
=
|
4 (
|
16
|
)
|
|||||||||||||
|
a
|
|||||||||||||||||
|
a2
|
=
|
64
|
|||||||||||||||
|
a
|
=
|
8
|
|||||||||||||||
|
b
|
=
|
16
|
|||||||||||||||
|
a
|
|||||||||||||||||
|
b
|
=
|
2;
|
|||||||||||||||
|
y
|
=
|
–
|
1
|
x + 2
|
|||||||||||||
|
4
|
(a)
|
f (1)
|
(b)
|
f (–1/2)
|
(c)
|
f (–1)
|
(d)
|
f (t – 1)
|
(e)
|
f (1/t)
|
💬 SOLUTION:
(a)
|
f (1)
|
=
|
1
|
–
|
1
|
=
|
–
|
1
|
|
1 + 1
|
1
|
2
|
(b)
|
f (–1/2)
|
=
|
1
|
–
|
1
|
=
|
4
|
|
–1/2 + 1
|
–1/2
|
(c)
|
f (–1) does not
exist.
|
(d)
|
f (t
– 1)
|
=
|
1
|
–
|
1
|
=
|
1
|
–
|
1
|
|
t – 1 +
1
|
t – 1
|
t
|
t – 1
|
(e)
|
f (1/t)
|
=
|
1
|
–
|
1
|
=
|
t
|
–
|
t
|
|
1/t
+ 1
|
1/t
|
1 + t
|
BACA JUGA:
|
|
(a)
|
g (2)
|
|
(b)
|
g (1/2)
|
|
(c)
|
g(2 + h) –
g(2)
|
|
h
|
💬 SOLUTION:
(a)
|
g (2)
|
=
|
2 + 1
|
=
|
3
|
|||||||||
|
2
|
2
|
||||||||||||
(b)
|
g (1/2)
|
=
|
1/2
+ 1
|
=
|
3
|
|||||||||
|
1/2
|
|||||||||||||
|
|
|
2
+ h + 1
|
–
|
2
+ 1
|
|||||||||
(c)
|
g(2 + h) –
g(2)
|
=
|
2
+ h
|
2
|
||||||||||
|
h
|
h
|
||||||||||||
|
|
|
2h
+ 6 – 3h – 6
|
|||||||||||
|
|
=
|
2
(2 + h)
|
|||||||||||
|
|
h
|
||||||||||||
|
|
|
–
|
h
|
||||||||||
|
|
=
|
2
(2 + h)
|
|||||||||||
|
|
h
|
||||||||||||
|
g(2 + h) –
g(2)
|
=
|
–1
|
|||||||||||
|
h
|
2
(2 + h)
|
(a)
|
g (2)
|
=
|
x
|
|
|
x2 – 1
|
|||
(b)
|
g (x)
|
=
|
√(4 – x2)
|
💬 SOLUTION:
(a)
|
{x ∈ ℝ : x
≠ –1, 1}
|
(b)
|
{x ∈ ℝ : |x|
≤ 2}
|
(a)
|
f (x)
|
=
|
3x
|
||
|
x2 + 1
|
||||
(b)
|
g (x)
|
=
|
|sin x| + cos x
|
||
(c)
|
h (x)
|
=
|
x3 + sin x
|
||
(d)
|
k (x)
|
=
|
x2 + 1
|
||
|
|x| + x4
|
💬 SOLUTION:
(a)
|
f (–x)
|
=
|
3(–x)
|
||||||||||
|
(–x)2 +
1
|
||||||||||||
|
|
=
|
–
|
3(x)
|
; odd
|
||||||||
|
(x)2 +
1
|
||||||||||||
(b)
|
g (–x)
|
=
|
|sin (–x)|
+ cos (–x)
|
|
|||||||||
|
|
=
|
|–sin x|
+ cos x
|
|
|||||||||
|
|
=
|
|sin x| + cos x
|
; even
|
|||||||||
(c)
|
h (–x)
|
=
|
(–x)3
+ sin (–x)
|
|
|||||||||
|
|
=
|
–x3 – sin x
|
; odd
|
|||||||||
(d)
|
k (x)
|
=
|
(–x)2
+ 1
|
||||||||||
|
|–x|
+ (–x)4
|
||||||||||||
|
|
=
|
x2 + 1
|
; even
|
|||||||||
|
|x| + x4
|
(a)
|
f (x)
|
=
|
x2 – 1
|
||||
(b)
|
g (x)
|
=
|
x
|
||||
|
x2 + 1
|
||||||
(c)
|
h (x)
|
= {
|
x2
|
if 0 ≤ x ≤ 2
|
|||
|
6 – x
|
if x > 2
|
💬 SOLUTION:
💬 SOLUTION:
💬 SOLUTION:
|
V(x) = x(32
– 2x)(24 – 2x)
|
|
Domain [0, 12]
|
(a)
|
(f + g)(2)
|
(b)
|
(f ⋅ g)(2)
|
(c)
|
(f ◦ g)(2)
|
(d)
|
(g ◦ f)(2)
|
(e)
|
f 3 (–1)
|
(f)
|
f 2 (2) + g2
(2)
|
💬 SOLUTION:
(a)
|
(f + g)(2)
|
=
|
(2 –
|
1
|
) + (22 + 1)
|
=
|
13
|
|
2
|
2
|
(b)
|
(f ⋅ g)(2)
|
=
|
(
|
3
|
) (5)
|
=
|
15
|
|
2
|
2
|
(c)
|
(f ◦ g)(2)
|
=
|
f(5)
|
=
|
5 –
|
1
|
=
|
24
|
|
5
|
5
|
(d)
|
(g ◦ f)(2)
|
=
|
g(
|
3
|
)
|
=
|
(
|
3
|
)2 + 1 =
|
13
|
|
2
|
2
|
4
|
(e)
|
f 3
(–1)
|
=
|
(–1 +
|
1
|
)3
|
=
|
0
|
|
1
|
(f)
|
f 2 (2) + g2
(2)
|
=
|
(
|
3
|
)2 + (5)2
|
|||||
|
2
|
|||||||||
|
|
=
|
9
|
+ 25
|
||||||
|
4
|
|||||||||
|
|
=
|
109
|
|||||||
|
4
|
BACA JUGA:
|
|
(a)
|
y = 1/4
x2
|
(b)
|
y = 1/4
(x + 2)2
|
(c)
|
y = –1 + 1/4
(x + 2)2
|
💬 SOLUTION:
(a)
|
f
|
(b)
|
f ◦ g
|
(c)
|
g ◦ f
|
💬 SOLUTION:
(a)
|
f = (–∞,
16]
|
(b)
|
f ◦ g
= √(16
– x4); domain [–2, 2]
|
(c)
|
g ◦ f
= [√(16
– x)] 4 = (16 – x)2;
|
|
domain (–∞, 16]
|
|
(note: the simplification [√(16 – x)]
4 = (16 – x)2 is only true given the restricted
domain)
|
💬 SOLUTION:
f(x) = √x, g(x) = 1 + x, h(x) = x2, k(x) = sin x, F(x) = √(1 + sin2 x) = f ◦ g ◦ h ◦ k.
(a)
|
sin 570°
|
(b)
|
cos 9π/2
|
(c)
|
cos
(–13π/6)
|
💬 SOLUTION:
(a)
|
sin (570°) = sin (570°) = –1/2
|
(b)
|
cos (9π/2) = cos (9π/2)
= 0
|
(c)
|
cos
(–13π/6) = cos (–π/6) = √3/2
|
(a)
|
sin (–t)
|
(b)
|
cos (t)
|
(c)
|
sin (2t)
|
(d)
|
tan (t)
|
(e)
|
cos
[(π/6) – t]
|
(f)
|
sin
(π + t)
|
💬 SOLUTION:
(a)
|
sin (–t) = –sin (t) = –0.8
|
(b)
|
sin2 t + cos2 t
= 1
|
|
cos2 t = 1 – (0.8)2
= 0.36
|
|
cos t = –0.6
|
(c)
|
sin (2t) = 2 sin t cos t
= 2(0.8)(–0.6) = –0.96
|
(d)
|
tan (t) = sin t/cos
t = 0.8/–0.6 = –4/3 ≈
–1.333
|
(e)
|
cos
[(π/6) – t] = sin t
= 0.8
|
(f)
|
sin
(π + t) = –sin t
= –0.8
|
💬 SOLUTION:
|
sin 3t = sin (2t + t)
= sin 2t cos t + cos 2t sin t
|
|
= 2 sin t cos2 t +
(1 – 2 sin2 t) sin t
|
|
= 2 sin t (1 – 2 sin2 t)
+ sin t – 2 sin3 t
|
|
= 2 sin t – 2 sin3 t
+ sin t – 2 sin3 t
|
|
= 3 sin t – 4 sin3 t
|
💬 SOLUTION:
|
s = rt
|
|
=
9 (20 rev/min)(2π rad/rev)(1 min/60 sec)(1
sec) = 6π
|
|
≈ 18.85 in.
|
BACA JUGA:
|
|
Demikian soal serta penjelasan untuk Pembahasan Soal Buku Calculus 9th Purcell Chapter 0 - Sample Test Problems - Number 30 – 55. Silahkan untuk berkunjung kembali dikarenakan akan selalu ada update terbaru 😊😄🙏. Silahkan juga untuk memilih dan mendiskusikan di tempat postingan ini di kolom komentar ya supaya semakin bagus diskusi pada setiap postingan. Diperbolehkan request di kolom komentar pada postingan ini tentang rangkuman atau catatan atau soal dan yang lain atau bagian hal yang lainnya, yang sekiranya belum ada di website ini. Terima kasih banyak sebelumnya 👍. Semoga bermanfaat dan berkah untuk kita semua. Amiiinnn 👐👐👐
Jangan lupa untuk 💏 SUBSCRIBE 👪 (Klik lonceng di bawah-kanan layar Anda) dan berikan komentar 💬 atau masukan serta share 👫 postingan ini ke teman-teman untuk berkembangnya https://www.bantalmateri.com/ ini 😀. Terima kasih dan semoga bermanfaat. 😋😆
No comments:
Post a Comment