Mathematic 108, Fall 2015: Solutions #1
Problem #1. Determine the (largest) domain and range of the function given by f (x) =
the graph of this function.
√ 1
.
1−x2
Sketch
Solution #1. We need to
work backwards:
√ look at this as the composite function k ◦ g ◦ h(x), and
2 is defined and nonzero. This occurs when 1 − x2 is strictly positive (since
√ 1
is
defined
when
1
−
x
2
1−x
it can’t be 0, and the square root is undefined for negative arguments). Therefore, solving this inequality,
we get −1 < x < 1.
√
1
Graphing, we see that the range of 1 − x2 is (0, 1], so the range of √1−x
is [1, ∞) (graph not included).
2
Problem #2. Determine the (largest) domain of the function given by f (x) =
this is not the same function as g(x) = cos2 (x).
1
1+tan2 (x) .
Explain why
Solution #2. Once again, this is a composite function, so in order to find the domain, it suffices to find
the set where 1 + tan2 (x) is defined and nonzero. Fortunately, tan2 (x) is a perfect square, so we have the
inequality 1 + tan2 (x) ≥ 1 > 0. Therefore, the domain is the set of all points where tan2 (x) is defined. We
know that cos(x) = 0 whenever x is of the form (n + 12 )π, for all integers n, so the largest domain of f (x) is
the set of all real numbers except for these points.
We know that cos2 (x) is defined for all x, so these functions are not defined on the same domain and
therefore are different functions (they are equal on the points where they are defined; however, that’s
immaterial here).
Problem #3. Ann leaves Baltimore at 7:00AM and drives at a constant speed south along I-95. She
passes Washington, DC, which is 40 mi from Baltimore, at 8:30AM
a) Express the distance traveled (in miles) in terms of the time traveled (in hours).
b) Express the distance traveled (in kilometers) in terms of the time of day (in hours).
c) How are these two functions related?
Solution #3.
40
a) Ann travels 40 miles in 1.5 hours, so m(t) = 1.5
t, where m(t) is the number of miles travelled at time
t.
b) At time t = 7 Ann has travelled 0 kilometers. At time t = 8.5 Ann has travelled approximately
64.37 kilometers (or 64.37376 kilometers to be exact). Therefore, using point slope form, we have
k(t) ≈ 64.37
1.5 (t − 7). (An exact answer would be k(t) = 42.91584t − 300.41088).
c) We have the relation k(t) ≈ 1.6 ∗ m(t − 7). Graphically, this represents a horizontal shift followed by
a vertical stretch, where the horizontal shift (along the time axis) reflects the fact that we changed
the way we’re enumerating the hours (i.e. changing the departure time from t = 0 to t = 7) and the
vertical stretch reflects the fact that we’re changing the units of measurement.
Problem #4. Let f (x) =
1
x
and
g(x) =
0
1
x
1≤x≤2
x > 2.
Determine the formulas for the following functions and their domains:
a) f g.
b) f ◦ g.
c) g ◦ f .
Solution #4.
a) f g is defined on points where both f and g are defined, since as long as they are defined, their product
is defined. f is defined for all x 6= 0 and g is defined for all x ≥ 1. The domain of their product is the
intersection of their domains, which is the set [1, ∞). On this domain,
0 1≤x≤2
f g(x) =
1
x>2
x2
b) We know that f ◦ g is defined whenever g(x) is defined and nonzero. This is true for x > 2 (since if
1 ≤ x ≤ 2, then f ◦ g(x) = 10 , and if x < 1, then g(x) is undefined. When x > 2, g(x) = x1 , and
f ◦ g(x) = x.
c) We know that g ◦ f (x) is defined whenever f (x) is defined and f (x) ≥ 1. Both of these hold for
0 < x ≥ 1 (if x > 1, f (x) < 1 so g ◦ f (x) is undefined; similarly, if x < 0, the same thing happens, and
if x = 0, f (x) is undefined). The behavior of the function depends on the value of f (x): if 1 ≤ f (x) ≤ 2,
1
= x. Looking at the behavior of f (x) = x1 , we see that:
g ◦ f (x) = 0, and if f (x) > 2, g ◦ f (x) = f (x)
g ◦ f (x) =
x
0
if 0 < x < 21
if 12 ≤ x ≤ 1
Problem #5. Express the following functions in the form f ◦ g where f is a rational function and g is a
trigonometric function:
a) u(t) =
b) w(t) =
cos(t)
1+cos(t) .
cos(t)
.
sin2 (t)
Solution #5.
t
a) This one is fairly straightforward: we set g(t) = cos(t), then we see that f (t) must be 1+t
b) This one is a little bit trickier: We can use trig identities to express numerator and denominator
in terms of the same trig function, and since trig identities deal with squares of trig functions, we
set g(t) = cos(t) and try to express w(t) as a rational function of cos(t). We see from the identity
cos(t)
t
sin2 (t) + cos2 (t) = 1 that we can rewrite w(t) as 1−cos
2 (t) , so f (t) = 1−t2 .
Problem #6. Show that if f is an even function and g is an odd function, then f ◦ g is an even function.
Solution #6. To verify this, it suffices to show that f ◦ g(−x) = f ◦ g(x) for all x. We do this by applying
the definitions of even and odd functions to f and g.
f ◦ g(−x) = f (g(−x))
= f (−g(x))
= f (g(x))
Problem #7. Explain why a periodic function cannot be one-to-one.
Solution #7. If a function f has period a, then by definition f (x + a) = f (x) but x + a 6= x, so by
definition f is therefore not one-to-one.
Problem #8. Determine the largest value L so that the f (x) = (x − 2)2 + 2 is one-to-one on the interval
(−L, L). Find the formula for f −1 and its domain.
Solution #8. If we graph f we see that the vertex is at the point (2, 2). Since the function is symmetric
over the line x = 2, f is not one-to-one on any interval of the form (−L, L) for L > 2. Additionally, the
function is strictly decreasing (and therefore one-to-one) on any interval to the left of x = 2, so the largest
value L is 2.
To find f −1 (x), we first need to note the domain and range of f −1 . The domain of f (x) is (−2, 2) and
the range is the interval (2, 18). We need to flip these to find the domain and range of f −1 (x): the domain
is (2, 18) and the range is (−2, 2). To find the formula for f −1 , we need to write y = f (x), flip x and y, and
solve for y:
x = (y − 2)2 + 2
x − 2 = (y − 2)2
p
√
x − 2 = (y − 2)2
√
x − 2 = |y − 2|
√
− x−2=y−2
√
2− x−2=y
Note that y − 2 is negative, so in taking the absolute value of |y − 2|, we flipped signs.
Problem #9. Find a formula for f −1 and determine its domain when f (x) = 1 +
√
1 − 2x.
Solution #9. The domain of f −1 is the range of f . We see that f (x) ≥ 1 (since the square root of a
number is always positive). There is no upper bound, so the range of f (x) is [1, ∞).
To find the inverse, we set y = f (x), flip x and y, and solve for y:
p
x = 1 + 1 − 2y
p
x − 1 = 1 − 2y
(x − 1)2 = 1 − 2y
1
y = (1 − (x − 1)2 )
2
Problem #10.
a) Verify that if f and g are one-to-one, then so is f ◦ g and that (f ◦ g)−1 = g −1 ◦ f −1 .
b) Use this to find a formula for h−1 when
h(x) =
1 + ex
.
1 − ex
Solution #10.
a) By definition, f ◦ g is one-to-one if, whenever f ◦ g(x1 ) = f ◦ g(x2 ), it follows that x1 = x2 .
Assume f ◦ g(x1 ) = f ◦ g(x2 ). Then, since f is one-to-one, it follows from the definition that g(x1 ) =
g(x2 ). Since g is one-to-one, we can similarly conclude that x1 = x2 . Therefore, f ◦ g is one-to-one.
To show that (f ◦ g)−1 is g −1 ◦ f −1 , it suffices to show that (f ◦ g) ◦ (g −1 ◦ f −1 )(x) = x and that
((g −1 ◦ f −1 ) ◦ (f ◦ g)(x) = x. The first follows from the fact that, by the definition of the inverse of g,
f ◦ (g ◦ g −1 ) ◦ f −1 (x) is f ◦ f −1 (x), which by the definition of f −1 , is equal to x. The second follows
similarly, since g −1 ◦ (f −1 ◦ f ) ◦ g(x) is g −1 ◦ g(x), which is equal to x. Therefore, (f ◦ g)−1 is g −1 ◦ f −1 .
1+x
and g(x) = ex . In this case, g −1 (x) is ln(x)
b) We can decompose h(x) into f ◦ g(x), where f (x) = 1−x
−1
and f −1 (x) is x−1
= ln x−1
x+1 . Therefore, (f ◦ g)
x+1 .
Book Problems.
a) Section 1.1: # 4, # 14
b) Section 1.2: # 10
c) Section 1.3: # 4, # 32, # 34
d) Section 1.4: # 20
e) Section 1.5: # 10, # 30, # 56.
Solution #1.1.4.
a)
b)
c)
d)
e)
f)
From examination of the graph, f (−4) ≈ −2 and g(3) ≈ 4.
From examination of the graph, f (x) = g(x) at x ≈ −2 and x ≈ 2.
x ≈ −2 and x ≈ 2.8.
f is decreasing on [0, 4].
f has domain [−4, 4] and range [−2, 3] (approximately).
g has domain [−4, 3] and range [0.5, 4] (approximately).
Solution #1.1.14.
Each runner finished the race, since there exists one t for which y = 100 for each
runner(so there exists a time for which each runner crossed the finish line). Runner A started out the slowest,
but won the race. Runner B started out the fastest, but stayed still for a few seconds and came in second.
Runner C ran at a roughly constant pace, and came in last pace.
Solution #1.2.10.
The first graph has a double root at x = 3 (since it’s tangent to the x-axis there).
Therefore, f (x) = c(x − 3)2 . Plugging in the point (4, 2), we see that c = 2, so f (x) = 2(x − 3)2 . For the
second graph, we can’t say anything about the roots, only that the function is of the form g(x) = ax2 +bx+c.
Setting x = 0 and g(x) = 1, we see that c = 1, since a · 02 + b · 0 + c = 1. We then have the system of
equations
a(−2)2 + b(−2) + 1 =
2
a(1)2 + b(1) + 1 = −2.5
The solution to this system is a = −1, b = −2.5, c = 1, which corresponds with the equation
g(x) = −x2 − 2.5x + 1.
Solution #1.3.4.
a)
b)
c)
d)
This
This
This
This
would
would
would
would
be
be
be
be
(Graphs not included)
the
the
the
the
original
original
original
original
graph,
graph,
graph,
graph,
shifted down 3 units.
shifted 1 unit to the left.
”squished” towards the x-axis by a factor of 21 .
reflected over the x-axis.
Solution #1.3.32.
√
√
a) 3 − x + x2 − 1, The domain is the set (−∞, −1] ∪ [1, 3], as that’s the domain where f and g are
defined.
√
√
b) p3 − x − x2 − 1, with the same domain as the previous problem
c) q(3 − x)(x2 − 1), with the same domain.
d)
3−x
x2 −1 ,
with the domain (−∞, −1) ∪ (1, 3], as that’s the domain where f and g are defined and g is
nonzero (since otherwise, you’d be dividing by 0).
Solution #1.3.34.
a)
b)
c)
d)
The domain for all of these is R.
(1 − 4x)3 − 2
1 − 4(x3 − 2)
(x3 − 2)3 − 2
1 − 4(1 − 4x)
Solution #1.4.20.
a) We need 10t − 100 to be greater than or equal to 0, which occurs on the interval [2, ∞).
b) The domain of this function is R, as the domains of et − 1 and sin(t) are R (so no matter what et − 1
exists and is in the domain of sin).
Solution #1.5.10.
This is not one-to-one, since f (1) = f (−1), for instance.
Solution #1.5.30.
included).
The graph of this is a curve, where f (−1) ≈ 4, f (0) ≈ 2, and f (1) ≈ 0 (Graph not
Solution #1.5.56.
a) Since ex is a strictly increasing function, we have that the natural logarithm preserves inequalities, so
0 < 3x − 1 < ln(2). Therefore, 31 < x < 1+ln(2)
.
3
b) We can add and subtract to get −2 < 2 ln(x), so ln(x) > −1. Therefore, x > e−1 .