PROBLEM SETS 1 & 2. DUE THURSDAY 7 SEPTEMBER
Problem Set 1. Problems from Lecture 1.
1. Given a quadratic equation of the from ax2 + bx + c = 0, we can solve for x
using the formula
√
−b ± b2 − 4ac
.
x=
2a
Using the above formula, solve the equation 4x2 − 5x − 6 = 0 for x.
In the above equation, a = 4, b = −5, and c = −6. Therefore, the solutions are given
by
p
5 ± (−5)2 − 4 ∗ 4 ∗ (−6)
x =
2∗4
p
5 ± (25 − (−96))
=
√ 8
5 ± 121
=
8
5 ± 11
=
8
3
= {− , 2}
4
These answers can be checked by resubstitution into the original quadratic.
2. Find f (x) if f (x + 1) = x2 − 5x + 3.
We can find f (x) by substituting x − 1 for x in the formula for f (x + 1):
f (x) = f ((x − 1) + 1) = (x − 1)2 − 5(x − 1) + 3 = x2 − 7x + 9
3. Graph the following functions, and give their domain and range.
(a) y = 3x − 2.
Domain: (−∞, ∞). Range: (−∞, ∞)
(b) y = 4x2 + 3.
Domain: (−∞, ∞). Range: [3, ∞)
30
450
400
20
350
10
300
0
250
200
−10
150
−20
100
−30
50
−40
−10
−8
−6
−4
−2
0
2
4
6
8
10
0
−10
−8
−6
y = 3x − 2
−4
−2
0
2
2
4
y = 4x + 3
1
6
8
10
2
PROBLEM SETS 1 & 2
4. Graph the following functions, and give their domain and range.
(a) y = 2x .
Domain: (−∞, ∞). Range: (0, ∞)
(b) y = 2x+3.
Domain: (−∞, ∞). Range: (0, ∞)
16
140
14
120
12
100
10
80
8
60
6
40
4
20
2
0
−4
−3
−2
−1
0
y=2
1
2
3
0
−4
4
−3
−2
−1
x
0
1
2
3
4
x
2 +3
5. Graph the following functions, and give their domain and range.
(a) y = log2 (x).
Domain: (0, ∞). Range: (−∞, ∞).
(b) y = log2 (x) + 5.
Domain: (0, ∞). Range: (−∞, ∞).
2
7
1
6
0
5
−1
4
−2
3
−3
2
−4
1
−5
0
0.5
1
1.5
2
2.5
3
0
0
y = log2 (x)
0.5
1
1.5
2
2.5
3
log2 (x) + 5
6. Graph the following functions, and give their domain and range.
(a) y = sin(x).
Domain: (−∞, ∞). Range: [−1, 1].
(b) y = sin(2x).
Domain: (−∞, ∞). Range: [−1, 1].
PROBLEM SETS 1 & 2
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
−0.2
−0.2
−0.4
−0.4
−0.6
−0.6
−0.8
−0.8
−1
−8
−6
−4
−2
0
2
4
6
8
−1
−8
−6
−4
y = sin(x)
−2
0
3
2
4
6
8
y = sin(2x)
7. Graph the following functions, and give their domain and range.
(a) y = tan(x).
Domain: (− π2 , π2 ) + nπ for all integers n. Range: (−∞, ∞).
(b) y = 4 tan(x).
Domain: (− π2 , π2 ) + nπ for all integers n. Range: (−∞, ∞).
8
30
6
20
4
10
2
0
0
−2
−10
−4
−20
−6
−30
−8
−10
−1.5
−1
−0.5
0
0.5
1
1.5
−40
−1.5
−1
y = tan(x)
−0.5
0
0.5
1
1.5
y = 4 tan(x)
8. Simplify the following expressions.
(a) log10 ( x+y
).
z
) = log10 (x + y) − log10 z
log10 ( x+y
z
x
(b) 25log25 (x+y)+log5 ( y ) .
x
x
25log25 (x+y)+log5 ( y ) = 25log25 (x+y) 25log5 ( y )
x
= (x + y)(52)log5 ( y )
x
= (x + y)(5log5 ( y ) )2
x
= (x + y)( )2
y
9. Make the following computations using right triangles.
(a) For θ = π4 = 45◦ , compute sin(θ), cos(θ), and tan(θ).
One triangle with angle θ = π4 has “adjacent”, “opposite” and “hypotenuse” sides
√
of lengths 1, 1 and 2, respectively. Therefore:
4
PROBLEM SETS 1 & 2
√
1
opposite
2
=√ =
sin(θ) =
hypotenuse
2
2
√
2
adjacent
2
=√ =
cos(θ) =
hypotenuse
2
2
1
opposite
= =1
tan(θ) =
adjacent
1
π
◦
(b) For θ = 6 = 30 , compute sec(θ), csc(θ), and cot(θ).
One triangle with angle θ = π6 has “adjacent”, “opposite” and “hypotenuse” sides
√
of lengths 2, 1 and 5, respectively. Therefore:
√
hypotenuse
1
5
=
=
sec(θ) =
cos(θ)
adjacent
2
1
hypotenuse √
csc(θ) =
=
= 5
sin(θ)
opposite
adjacent
1
=
=2
cot(θ) =
tan(θ)
opposite
10. Simplify the following expressions (i.e. write them in terms of elementary
trig functions sin(φ), sin(θ), etc.).
(a) sin(θ + φ).
sin(θ + φ) = sin(θ) cos(φ) + cos(θ) sin(φ). This is a basic trigonometric identity.
(b) cos(3θ). Using a formula similar to that used in (a), above, we write:
cos(3θ) = cos(θ + 2θ)
= cos(θ) cos(2θ) − sin(θ) sin(2θ)
Using the standard formulas for cos(2θ) and sin(2θ), this can be further simplified
to:
cos(3θ) = cos(θ)(1 − 2 sin2 (θ)) − sin(θ) ∗ 2 sin(θ) cos(θ)
= cos(θ) − 2 sin2 (θ) cos(θ) − 2 sin2 (θ) cos(θ)
= cos(θ)(1 − 4 sin2 (θ))