1. Find the domains of the following functions:
(a) f (x) = ln(x − 2)
solution: We need x − 2 > 0, or x > 2. Thus
Df = (2, ∞)
r
(x + 1)(x − 3)
x−1
solution: We need the thing inside the root to be greater than or equal to 0. So
we set up a sign table.
(b) g(x) =
x (−∞, −1) (−1, 1) (1, 3) (3, ∞)
x+1
−
+
+
+
x−1
−
−
+
+
x−3
−
−
−
+
(x + 1)(x − 3)
−
+
−
+
x−1
Thus the expression inside the root is greater than or equal to 0 on [−1, 1)∪[3, ∞).
Thus
Dg = [−1, 1) ∪ [3, ∞)
1
(c) h(x) =
sin(x) cos(x)
solution: Here we need to exclude all the points where cos or sin are 0. sin is 0
at all multiples of π and cos is 0 at π2 plus all multiples of π. Together this means
we need to exclude all integer multiples of π2 :
n
o
nπ
Dh = x ∈ R | x 6=
,n ∈ Z
2
x−1
(d) k(x) = √
2x2 − x + 1
solution: If we calculate the discriminant of the quadratic under the root:
∆ = b2 − 4ac = (−1)2 − 4(2)(1) = −7
Which means that the polynomial has no roots. Since it opens up, that means it
is always positive. Hence its root is always defined, and its root is never 0, and so
the domain is every real number:
Dk = R
(e) r(x) = sin
π
x
solution: The only problem here is that we can’t divide by 0. Hence only x = 0
is excluded:
Dr = {x ∈ R | x 6= 0}
2. Let f (x) =
√
3
3x + 9 and g(x) = 2x + 6. Find f ◦ g(x), g ◦ f (x), g −1 (x), and f −1 (x).
solution:
p
√
f ◦ g(x) = f (g(x)) = f (2x + 6) = 3 3(2x + 6) + 9 = 3 6x + 27
√
√
g ◦ f (x) = g(f (x)) = g( 3 3x + 9) = 2 3 3x + 9 + 6
To find the inverse functions we switch x and y and then solve for y.
p
x = 3 3y + 9
x3 = 3y + 9
x3 − 9
y =
3
Thus f −1 (x) =
x3 − 9
.
3
x = 2y + 6
2y = x − 6
x−6
y =
2
Thus g −1 (x) =
x−6
.
2
3. Solve the following:
(a) 5 ln(x + 3) − 1 = 0 for x
solution:
5 ln(x + 3) − 1 = 0
5 ln(x + 3) = 1
1
ln(x + 3) =
5
x + 3 = e1/5
x = e1/5 − 3
(b) log2 (y) = log2 (y 2 − 1) − log2 (2y) + 2 for y.
solution:
log2 (y) = log2 (y 2 − 1) − log2 (2y) + 2
log2 (y) − log2 (y 2 − 1) + log2 (2y)
y · 2y
log2
y2 − 1
y · 2y
y2 − 1
2y 2
= 2
= 2
= 22
= 4y 2 − 4
2y 2 = 4
√
y = ± 2
√
√
Thus we have two solutions, y = ± 2. However, only y = + 2 is in the domain
of the original expression, for if you plug a negative number into log2 y you get
√
something that is not defined. Thus our only solution is y = 2.
(c) 3t+3 = 5t for t.
solution:
3t+3 = 5t
ln(3t+3 ) = ln(5t )
(t + 3) ln 3 = t ln 5
t ln 5 − t ln 3 = 3 ln 3
t(ln 5 − ln 3) = 3 ln 3
3 ln 3
t =
ln 5 − ln 3
Of course any log works here.
4. Find all solutions of sin(3x) =
1
2
in the interval [0, π].
solution: An angle in the first quadrant that has sin equal to 12 is π6 . sin is also positive
in the second quadrant, meaning our second basic solution is 5π
. Thus
6
3x =
5π
π
+ 2kπ, k ∈ Z or 3x =
+ 2kπ, k ∈ Z
6
6
π
2kπ
5π 2kπ
+
, k ∈ Z or x =
+
,k ∈ Z
18
3
18
3
We solve the following inequalities to find the values of k we need
x=
π
2kπ
+
≤π
18
3
0≤
1
2k
+
≤1
18
3
2k
1
17
1
≤1−
=
− ≤
18
3
18
18
1
17
− ≤ 2k ≤
6
6
17
1
≈ 1.42
− ≤k≤
12
12
The ks in this range are k = 0 and k = 1. Hence
0≤
x=
π 13π
,
18 18
For the other solutions we have
0≤
5π 2kπ
+
≤π
18
3
5
2k
+
≤1
18
3
5
2k
5
13
− ≤
≤1−
=
18
3
18
18
5
13
− ≤ 2k ≤
6
6
5
13
− ≤k≤
≈ 1.08
12
12
The ks in this range are k = 0 and k = 1. Hence
0≤
x=
5π 17π
,
18 18
5. Sketch the following functions:
(a) f (x) =
2x2 +1
x2 −1
solution: This is a rational function. It has no x-intercepts because the top is
never 0. Substituting in 0 for x shows us that the y-intercept is −1.
The horizonal asymptote is at the ratio of the leading coefficients, i.e. at y = 21 = 2.
Since the roots of the denominator are ±1, there are vertical asymptotes at x = ±1.
Looking at a sign table:
x (−∞, −1) (−1, 1) (1, ∞)
x+1
−
+
+
x−1
−
−
+
2x2 + 1
+
+
+
2x2 +1
x2 −1
+
−
+
Since f is positive to the left of the asymptote at −1, it has to approach ∞ as it
approaches −1 from the left. Since f is negative to the right of the asymptote at
−1, it has to approach −∞ as it approaches −1 from the right. Since f is negative
to the left of the asymptote at 1, it has to approach −∞ as it approaches 1 from
the left. Finally, since f is positive to the right of the asymptote at 1, it has to
approach ∞ as it approaches 1 from the right.
This all gets put together to give
(b) g(x) =
3x+5
x+1
solution: Using long division,
3
x+1
3x + 5
− 3x − 3
2
3x + 5
2
=3+
. Hence the graph of g looks like the graph of
x+1
x+1
shifted up by 3, to the left by 1, and stretched vertically by 2.
Thus g(x) =
1
x
(c) h(x) = 3 cos−1 (x)
solution: This is a usual cos−1 graph stretched vertically by 3.
(d) k(x) = 3 sin 2 x −
π
3
+1
solution: This is a sin graph with amplitude 3 and period π shifted up by 1 and
to the right by π3 .
(e) r(x) = 2 − tan(x)
solution: This is a tan graph flipped vertically and shifted up by 2 units.
6. Verify the following identities:
√
(a) cos(sin−1 (x)) = 1 − x2
solution:
LHS = cos(sin−1 (x))
q
1 − sin2 (sin−1 (x))
=
√
=
1 − x2 = RHS
(b)
tan θ
= sec θ − cos θ
csc θ
solution:
RHS = sec θ − cos θ
1
=
− cos θ
cos θ
1 − cos2 θ
=
cos θ
sin2 θ
=
cos θ
sin θ
sin θ
=
cos θ
sin θ 1
=
cos θ csc θ
tan θ
=
= LHS
csc θ
(c) 1 − tan x tan y =
cos(x + y)
cos x cos y
solution:
RHS =
=
=
=
=
cos(x + y)
cos x cos y
cos x cos y − sin x sin y
cos x cos y
sin x sin y
1−
cos x cos y
sin x
sin y
1−
cos x
cos y
1 − tan x tan y = LHS
7. If r = 3% and P = $10, 000, how much do you have after 10 years if interest is compounded monthly? Continuously? If interest is compounded
continuously, how long does it take you to reach $35,000?
solution: The equation for discrete compound interest is
h
r int
A(t) = P 1 +
n
12t
0.03
A(t) = 10000 1 +
12
12(10)
0.03
A(10) = 10000 1 +
12
≈ $13494
The equation for continuous compound interest is
A(t) = P ert
A(t) = 10000e0.03t
A(10) = 10000e0.3
≈ $13498
If interest is compounded continuously to a target of $35,000, we have
A(t) = 10000e0.03t
35000 = 10000e0.03t
3.5 = e0.03t
0.03t = ln 3.5
ln 3.5
t =
≈ 45.8y
0.03
8. Find the exact values of
5π
(a) tan
12
π π
5π
solution:
= + , so
12
6
4
tan
5π
π π
= tan( + )
12
6
4
tan π6 + tan π4
=
1 − tan π6 tan π4
√
1/ 3 + 1
√
=
1 − 1/ 3
√ √
(1 + 3)/ 3
√
= √
( 3 − 1)/ 3
√
(1 + 3)
= √
( 3 − 1)
5π
π
+ cos
12
12
solution: Using the sum-to-product formula,
(b) cos
cos x + cos y =
cos
π
5π
+ cos
=
12
12
=
=
=
x+y
x−y
2 cos
cos
2
2
5π
5π
π π + 12
− 12
12
12
2 cos
cos
2
2
π π 2 cos
cos
4
6
√ ! √ !
2
3
2
2
2
√
6
2
(c) csc 202.5◦
solution: 202.5◦ lies in the third quadrant and 202.5◦ = 180◦ + 22.5◦ . Thus, since
csc is negative in quadrant III, csc 202.5◦ = − csc 22.5◦
csc 202.5◦ = − csc 22.5◦
1
= −
cos 22.5◦
1
= −
cos(45◦ /2)
1
= −q
1+cos 45◦
2
s
= −
2
√
1 + 2/2
s
= −
(2 +
s
= −
2
√
2)/2
4
√
2+ 2
9. If revenue is given by R(x) = 80x − 0.4x2 , where x is the number of units
sold, what is the maximum revenue and how many units do you have to
sell to get it? (Hint: it is a quadratic)
solution: We put this in standard form
80x − 0.4x2 = −0.4(x2 − 200x)
= −0.4(x2 − 200x + 10000 − 10000)
= −0.4(x2 − 200x + 10000) + 4000
= −0.4(x − 100)2 + 4000
This is in standard form. The maximum it attains is 4000 at the value x = 100. Thus
the maximum revenue is $4000 and it as attained when you sell 100 units.
10. (a) If f (x) =
√
x + 1, use the definition of the derivative to find f 0 (x).
solution: We go straight to the definition
f (x + h) − f (x)
h→0
h
√
√
x+h+1− x+1
lim
h→0
h √
√
√
√
x+h+1− x+1
x+h+1+ x+1
√
· √
lim
h→0
h
x+h+1+ x+1
x + h + 1 − (x + 1)
√
lim √
h→0 h( x + h + 1 +
x + 1)
h
√
lim √
h→0 h( x + h + 1 +
x + 1)
1
√
lim √
h→0
x+h+1+ x+1
1
√
√
x+1+ x+1
1
√
2 x+1
f 0 (x) = lim
=
=
=
=
=
=
=
(b) Find the following limits.
√
x−1
• limx→1 x−1
solution: Both the top and bottom are 0 at the limit point, so there is some
hope. Try to rationalize the numerator:
√
√
√
x−1
x−1
x+1
lim
= lim
· √
x→1 x − 1
x→1 x − 1
x+1
x−1
√
= lim
x→1 (x − 1)( x + 1)
1
= lim √
x→1
x+1
1
=
2
2
−8
• limx→2 xx−2
solution: Here the bottom is 0 at the limit point but the top is not. That
means that x = 2 is a vertical asymptote for this function and hence the limit
does not exist.