Math 1310 Lab 1.
Name/Unid:
Lab section:
1. (Algebra Review)
(a) Factor the following
x2 + 2x − 8
(b) Evaluate:
(−8)(2/3)
(c) Write the following expression without any negative exponents
1/3 −1/2
4x
y4
(d) Solve for x. Remember when multiplying by a negative, the inequality switches
direction.
x+3
≥1
2x + 2
Solution:
(a) (x − 2)(x + 4)
(b) 4
(c)
y2
2x1/6
(d) If x > −1, then
x + 3 ≥ 2x + 2
1 ≥ x
so x ∈ (−1, 1].
If x < −1, then we get 1 ≤ x, which cannot happen, so we only get the one
interval.
2. (Balancing Levers) Suppose that you have a lever that is on a fulcrum. This lever
is balanced if and only if F1 d1 = F2 d2 , where F1 is the force to the left of the fulcrum
perpendicular to the lever, and d1 is the distance to the left, and F2 , d2 are the symmetric
counterparts.
(a) Suppose that you have a force of 500N 1/2 meter to the left of the fulcrum. How
much force would you have to put 1 meter to the right to balance the fulcrum?
(b) Now suppose that you have a weight of 125N. How far to the right of the fulcrum
would you need to put this weight to balance the lever with the 500N at 1/2 meter
to the left of the fulcrum?
Solution:
(a) 250N
(b)
500/2 = 125x
x=2
so x = 2 meters.
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3. (Logarithms and Exponentials) Solve for x in the following equations.
(a) ln(x) = 4
(b) ex+1 = 2e2
(c) ln(2x2 ) − ln(x) = 1
(d) ln(x − 1) + ln(x + 2) = ln(10)
Solution:
(a) x = e4
(b)
e−2 ex+1 = 2
ex−1 = 2
x = ln(2) + 1
(c)
ln(2x2 ) − ln(x) = 1
ln(2x) = 1
e
x =
2
(d)
ln(x2 − x − 2) = ln(10)
x2 + x − 12 = 0
(x + 4)(x − 3) = 0
x = 3, x = −4 are solutions of the quadratic equation. x = 3 is the only solution
of the problem, since x = −4 is not admissible for the domain of the logarithms.
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4. (Fun With Functions)
(a) Consider the following graph of f (x) = x3 . Without using technology, sketch the
graph of f (x − 3) and f (x) + 2.
(b) Consider the following function
(
1 − x2
f (x) =
2x + 1
if x ≤ 0
if x > 0.
Evaluate f (−1) and f (5).
(c) Let f (x) = x2 − 2, and g(x) = x − 4. Find f ◦ g(x), and evaluate f ◦ g(2).
Solution:
(a) The graph of f (x − 3) will be the original graph of f (x) shifted to the right by
three units.
The graph of f (x) + 2 will be the original graph shifted up by two units.
(b) f (−1) = 0, and f (5) = 11.
(c)
f ◦ g = (x − 2)2 − 2
= x2 − 8x + 14
and we have f ◦ g(2)=2.
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5. (Trigonometry Review)
(a) Convert from degrees to radians
• 200 degrees
• 60 degrees
Convert from radians to degrees
• 1/2
• π/6
(b) Find the numerical value of the following without use of a calculator:
•
•
•
•
sin(π/4)
cos(5π/6)
cos(−π/3)
tan(7π/6)
(c) Find all values of x such that cos(2x) = cos(x) for 0 ≤ x ≤ 2π.
Solution:
(a) Convert from degrees to radians
2π
360
2π
• 60 degrees = 60 ·
360
• 200 degrees = 200 ·
Convert from radians to degrees
180
2π
• π/6 = 30
• 1/2 =
(b)
√
2/2
√
• cos(5π/6) = − 3/2
• sin(π/4) =
• cos(−π/3) = 1/2
√
• tan(7π/6) = 3/3
(c) x = 0, 2π/3, 4π/3, 2π
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