Homework 1 solutions
§1.1 # 35 Find the domain of the function h(x) =
1
√
.
4 2
x −5x
The function h(x) is
defined when we can ensure that both the denominator is not zero and that we are
not attempting to take the fourth root of a negative number. Thus we must have
x2 − 5x > 0, i.e., x(x − 5) > 0.
The expression x(x − 5) is positive exactly when x < 0 or x > 5 since otherwise
x(x − 5) is a product of the positive number x with the negative number x − 5.
One could also reason that the expression x(x − 5) is positive exactly when x < 0
or x > 5 by noting that the graph of the function f (x) = x(x − 5) is an upwards
pointing parabola with roots at 0 and 5.
Thus the domain of h(x) is (−∞, 0) ∪ (5, ∞).
§1.3 # 4 The graph of a function f is given.
a) y = f (x) − 2 will shift the graph down 2 units.
b) y = f (x − 2) will shift the graph right 2 units.
c) y = −2f (x) will stretch the graph of f vertically by a factor of two and then
reflect over the x-axis.
d) y = f (x/3) + 1 will stretch the graph of f horizontally by a factor of 3, then
then shift the resulting graph 1 unit upward.
§1.6 # 18 The graph of a function f is given.
a) f is 1-1 because it passes the Horizontal Line Test.
b) Domain of f = [−3, 3]=range of f −1 .Range of f = [−1, 3]=Domain off −1 .
c) Since f (0) = 2, f −1 (2) = 0.
d) Since f (−1.7) ≈ 0, f −1 (0) ≈ −1.7.
1
§2.1 # 6 If a rock is thrown upward on the planet Mars with a velocity of 10 m/s,
its height in meters after t seconds is given by h(t) = 10t − 1.86t2 .
a) Determine the average velocity over the given time intervals.
The average velocity between t and t + h seconds is defined to be
vavg
10(t + h) − 1.86(t + h)2 − (10t − 1.86t2 )
= 10 − 3.72t − 1.86h.
=
h
Therefore, for the intervals given in your book, all you have to do is use the above
formula along with your determined value of h.
i) t = 1, h = 1 ⇒ vavg = 4.42 m/s.
ii) t = 1, h = 0.5 ⇒ vavg = 5.35 m/s.
iii) t = 1, h = 0.1 ⇒ vavg = 6.1 m/s.
iv)) t = 1, h = 0.01 ⇒ vavg = 6.26 m/s.
v) t = 1, h = 0.001 ⇒ vavg = 6.28 m/s.
b) The instantaneous velocity at t = 1 second is (appears to be) 6.28 m/s.
§2.2 #4 For the function f whose graph is given, state the value of the given
quantity. If it does not exist, explain why.
a)As x approaches 2 from the left, the values of f (x) approach 3,
so limx→2− f (x) = 3.
b)As x approaches 2 from the right, the values of f (x) approach 1,
so limx→2+ f (x) = 1.
c) limx→2 f (x) does not exist since the left-hand limit does not equal the right-hand
limit.
d) When x = 2, y = 3, so f (2) = 3.
e) As x approaches 4,the values of f (x) approach 4, so limx→4 f (x) = 4.
f) There is no value of f (x) when x = 4, so f (4) does not exist.
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§2.2 # 30 Determine the infinite limit
x+2
.
x+3
lim −
x→−3
Since for x very close to but less than -3 we have that the denominator (x + 3) is
very small and negative and the numerator is close to −1 we see that the fraction
x+2
x+3
≈
−1
tiny negative #
is huge and positive. Thus we conclude that
lim −
x→−3
x+2
= ∞.
x+3
3