Workshop #5
Professor D. Olles
1. Evaluate arccos − 12 .
2. Evaluate sin(cos−1
2
3)
.
3. Evaluate arctan(−1).
4. Find the exact value of the expressions:
a. log4 64
b. log5
1
125
5. Solve for x:
a. 3x−7 = 8
b. ln(x2 − 2e − 1) = 2
c. e2x + 2e2 = 15
6. Find the domain of the following functions, then sketch the their graphs
labeling two point and the equations of any asymptotes.
a. f (x) = − ln(x − 1) − 2
b. ln(−x) + 2
7. Find the inverse of y =
8. Let f (x) =
a. f
−1
4x+1
3x−2 .
√
1 − x, x ≤ 1.
Find:
(x)
b. The domain of f (x) in interval notation.
c. The domain of f −1 (x) in interval notation.
d. The range of f (x) in interval notation.
e. The range of f −1 (x) in interval notation.
9. Prove that the inverse of a linear function is also linear and the slopes of
the two are reciprocals of each other.
1
Solutions
1. Evaluate arccos − 12 .
1
θ = arccos −
2
1
2
cos θ < 0 in the second and third quadrants. But, 0 ≤ θ ≤ π for inverse cosine.
cos θ = −
So, we will only examine the second quadrant.
cos θ =
2. Evaluate sin(cos−1
1
π
if θ = .
2
3
2π
.
In the second quadrant, θ =
3
1
2π
arccos −
=
2
3
2
3) .
2
−1
θ = cos
3
cos θ =
√
2
3
3
5
θ
2
a2 + 22 = 32
a2 + 4 = 9
a2 = 5
√
a= 5
√
2
5
sin cos−1
= sin θ =
3
3
2
3. Evaluate arctan(−1).
θ = arctan (−1)
tan θ = −1
π
π
< θ < for inverse tangent.
2
2
So, we will only exam the fourth quadrant.
π
π
tan
= 1 in the first quadrant so, tan −
= −1 in the fourth quadrant.
4
4
π
arctan (−1) = −
4
tan θ < 0 in the second and fourth quardants. But, −
4. Find the exact value of the expressions:
a. log4 64
b. log5
= log4 43 = 3
1
125
= log5
1
53
= log5 5−3 = −3
5. Solve for x:
a. 3x−7 = 8
log3 3x−7 = log3 8
x − 7 = log3 8
x = 7 + log3 8
b. ln(x2 − 2e − 1) = 2
eln (x
2
−2e−1)
= e2
x2 − 2e − 1 = e2
x2 = e2 + 2e + 1
x2 = (e + 1)
2
x = ± (e + 1)
c. e2x + 2e2 = 15
e2x = 15 − 2e2
ln e2x = ln 15 − 2e2
2x = ln 15 − 2e2
1
ln 15 − 2e2
2
p
x = ln (15 − 2e2 )
x=
3
6. Find the domain of the following functions, then sketch the their graphs
labeling two point and the equations of any asymptotes.
a. f (x) = − ln(x − 1) − 2
x−1>0
x>1
(1, ∞)
2
0
−2
−4
2
4
6
8
10
b. ln(−x) + 2
−x > 0
x<0
(−∞, 0)
4
2
0
−2
−4
−10
−8
−6
−4
−2
4
0
7. Find the inverse of y =
√
1 − x, x ≤ 1.
√
y = 1−x
p
x= 1−y
x2 = 1 − y
x2 − 1 = −y
y = 1 − x2
0
−20
−40
−60
−80
−100
−60 −40 −20
8. Let f (x) =
4x+1
3x−2 .
0
20
40
60
Find:
a. f −1 (x)
4x + 1
3x − 2
4y + 1
x=
3y − 2
y=
x(3y − 2) = 4y + 1
3xy − 2x = 4y + 1
3xy − 4y = 2x + 1
y(3x − 4) = 2x + 1
2x + 1
3x − 4
2x + 1
f −1 (x) =
3x − 4
y=
b. The domain of f (x) in interval notation.
2 [ 2
−∞,
,∞
3
3
5
c. The domain of f −1 (x) in interval notation.
4 [ 4
−∞,
,∞
3
3
d. The range of f (x) in interval notation.
4 [ 4
−∞,
,∞
3
3
e. The range of f −1 (x) in interval notation.
2 [ 2
−∞,
,∞
3
3
9. Prove that the inverse of a linear function is also linear and the slopes of
the two are reciprocals of each other.
Let f (x) = mx+b be a linear function with slope m and y-intercept (0, b).
y = mx + b
x = my + b
x − b = my
x−b
=y
m
x
b
y=
−
m m
1
b
f −1 (x) = x −
m
m
The slope of f (x) is m and the slope of f −1 (x) is the reciprocal
6
1
.
m