MA 123 - Elementary Calculus and Its Applications
Chapter 1 Homework - Answers
1. Suppose f (x) = 3x2 − 5x + 2. Determine the following:
a.) f (−3)
= 3(−3)2 − 5(−3) + 2
= 3(9) − (−15) + 2
= 27 + 15 + 2
= 44
b.) f (2x)
= 3(2x)2 − 5(2x) + 2
= 3(4x2 ) − 10x + 2
= 12x2 − 10x + 2
c.) f (x + h) = 3(x + h)2 − 5(x + h) + 2
= 3(x2 + 2xh + h2 ) − (5x + 5h) + 2
= 3x2 + 6xh + 3h2 − 5x − 5h + 2
2. a.) Solve y 2 + y = 5x − 2y + 8 for x as a function of y.
y 2 + y = 5x − 2y + 8
⇒
5x = y 2 + 3y − 8
⇒
1
3
8
x = y2 + y −
5
5
5
b.) Find the domain for x(y).
Since x(y) is a polynomial, its domain is (−∞, ∞) .
3. a.) Solve 3x3 y − 6x + 3y = 7 for y as a function of x.
3x3 y − 6x + 3y = 7
⇒
y(3x3 + 3) = 6x + 7
⇒
y=
6x + 7
3x3 + 3
b.) Find the domain for y(x).
Since y(x) is a rational function, it is undefined when 3x3 + 3 = 0, that is, when x = −1. So, its
domain is (−∞, −1) ∪ (−1, ∞) .
4. Find the domain for the following functions:
1
a.) f (x) = − x + 2
3
f (x) is a polynomial, so its domain is (−∞, ∞) .
b.) g(x) = x2 − 4x + 1
g(x) is a polynomial, so its domain is (−∞, ∞) .
1
c.) s(t) =
√
9 − t2
Since s(t) contains a square root, it is undefined when (9 − t2 ) = (3 − t)(3 + t) < 0, that is, when
t < −3 or t > 3. So, its domain is [−3, 3] .
1
−s
Since k(s) is a rational function, it is undefined when (s2 − s) = s(s − 1) = 0, that is, when
x = 0 or x = 1. So, its domain is (−∞, 0) ∪ (0, 1) ∪ (1, ∞) .
d.) k(s) =
s2
5. Sketch graphs for functions (a), (b), and (c) from problem 4.
(b) g(x)
(a) f (x)
(c) s(t)
6. Use the graphs sketched in problem 5 to determine the range of (a), (b), and (c) from problem 4.
1
a.) f (x) = − x + 2 has a range of (−∞, ∞) .
3
b.) g(x) = x2 − 4x + 1 has a range of [−3, ∞) .
c.) s(t) =
√
9 − t2 has a range of [0, 3] .
2
7. Consider the graphs of f and g given below.
a.) Determine the values of f (−3) and g(1).
f (−3) = −1 and g(1) = 1
b.) For what value(s) of x is f (x) = g(x)?
f (x) = g(x) occurs at x = −2 and at x = 2 .
8. Factor the following quadratic equations and determine their roots.
a.) x2 + 2x − 15 = (x + 5)(x − 3) has roots at −5 and 3 .
b.) x2 − 36 = (x + 6)(x − 6) has roots at −6 and 6 .
c.) 3x2 − 20x − 7 = (3x + 1)(x − 7) has roots at − 13 and 7 .

3x − 3 x < 2





1 2
2≤x≤4
9. Graph y = f (x), where f (x) =
2x





4
x>4
(Hint: You may want to find f(-1), f(0), f(1), f(2), etc.)
3
10. Find the equation of the line going through the points (−1, −2) and (1, 1):
a.) in point-slope form.
The slope of the line containing these two points is
m =
3
1 − (−2)
= . Then, the point1 − (−1)
2
3
slope form using (−1, −2) is y + 2 = (x + 1) . If we use (1, 1) instead, the equation becomes
2
3
y − 1 = (x − 1) .
2
b.) in slope-intercept form.
From the first point-slope equation of part (a), we get
y+2 =
3
3
3
(x + 1) = x +
, which
2
2
2
3
1
means y = x − .
2
2
11. Let f (x) = −7(2x + 1) and g(x) = −3x(x − 2). Determine the point(s) of intersection of f (x) and
g(x), and write them in the form (x, y).
f (x) = g(x) means that −7(2x + 1) = −14x − 7 = −3x2 + 6x = −3x(x − 2), that is, 3x2 − 20x − 7 = 0.
We know from 8(c) that 3x2 − 20x − 7 has roots at − 31 and 7. So, evaluating f (− 13 ) and f (7) gives
us that the intersection points are (− 13 , − 73 ) and (7, −105) .
12. A ladder is 13 feet long. One end of the ladder is on the ground, and the other end rests on a vertical
wall. How far is the base of the ladder from the wall when the top of the ladder is 12 feet from the
ground?
The problem describes a right triangle with hypotenuse 13 and one base length of 12. Using the
Pythagorean
Theorem, we must have the remaining base length x such that x2 + 122 = 132 , that is,
√
2
x = 13 − 122 = 5 feet.
13. A man is six feet tall, and he stands a distance of 10 feet from a lightpost. If the top of the light is
12 feet from the ground, how long is the man’s shadow?
The diagram above describes the situation in the problem with the variable x measuring the length of
the man’s shadow. Since the large and small triangles share the angle A and both have right angles,
the two triangles must be similar. So, the 12 and 6 are corresponding side lengths. This means that
the side lengths of the large triangle are twice as large as the side lengths of the smaller triangle. The
base of the smaller triangle has length x, and the base of the larger triangle has length 10 + x. Since
10 + x = 2x because of the scale factor, we must have that the man’s shadow has a length of 10 feet .
4
p
14. Recall that the distance formula is (x2 − x1 )2 + (y2 − y1 )2 . Find the distance between the points
(−2, 0) and (7, 4).
√
p
p
√
√
D = (x2 − x1 )2 + (y2 − y1 )2 = (7 − (−2))2 + (4 − 0)2 = 92 + 42 = 81 + 16 = 97
√
[f (2 + h) − f (2)] · [f (2 + h) + f (2)]
.
h
p
p
p
√
√
First, f (2+h) = (2 + h)2 − 1 = (4 + 4h + h2 ) − 1 = 3 + 4h + h2 , while f (2) = 22 − 1 = 3.
15. Suppose that f (x) =
x2 − 1. Write a simplified expression for
Now,
f (2 + h)2 + f (2)f (2 + h) − f (2)f (2 + h) − f (2)2
[f (2 + h) − f (2)] · [f (2 + h) + f (2)]
=
h
h
f (2 + h)2 − f (2)2
=
h
√
√
( 3 + 4h + h2 )2 − ( 3)2
=
h
3 + 4h + h2 − 3
=
h
4h + h2
=
h
= 4+h
5