Math 10250 Final Exam Solutions Fall 2009
1. The slope is m =
q = 20p + 190.
q
50 90
40
=
=
= 20, and the equation is q 90 = 20(p 5), or
p
75
2
2. The revenue function is R = q p = q (0.1q + 40) = 0.1q 2 + 40q , while the cost function is
C = 5q . Therefo
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Date
Math 10250 Activity 1: Functions and their Geometric Properties1 (Sec. 0.20.3)
GOAL: Understand the fundamental concept of function as a relation between variables expressed by
a formula, a graph, or a table and use it to model change.
Q1: What
Math 10250 Exam 1 Solutions Fall 2008
1 At t = 0 wind energy production was w = 16.8 MW and was growing at the rate (slope) 5.2 MW/year. Using
the point-slope formula we obtain the equation w 16.8 = 5.2(t 0), or w = 5.2t + 16.8.
2 At t = 0 wind energy pro
Math 10250 Exam 2 Solutions Fall 2008
1. Applying the product rule we obtain f (x) = 2x ln x + x = x(2 ln x + 1). Setting f (x) = 0 gives
x(2 ln x + 1) = 0. Since x is in the domain of ln x we must have x > 0. Thus, we get the equation
2 ln x + 1 = 0, or
Math 1250 Final Exam Solutions Fall 2008
1.) We need to nd the equation of the line passing from the points (t1 = 0, E1 = 20, 000) and
,
(t2 = 5, E2 = 135, 000). The slope of this line is E = 135,000020,000 = 1155000 = 23, 000. Therefore
t
5
busing the po
Math 10250 Exam 3 Solutions Fall 2006
1. R(q ) =
72q
,
q +2
q 0; R (q ) =
144
(q +2)2
288
(q +3)3
> 0, R (q ) =
72q
2. P (q ) = q+2 4q , q 0; P (q ) =
answer is (a).
144
(q +2)2
< 0; the answer is (c).
4 and P is maximum at q = 4 with P (4) = 32; the
3.
Math 10250 Exam 3 Solutions Fall 2006
1. R(q ) =
72q
,
q +2
q 0; R (q ) =
144
(q +2)2
288
(q +3)3
> 0, R (q ) =
72q
2. P (q ) = q+2 4q , q 0; P (q ) =
answer is (a).
144
(q +2)2
< 0; the answer is (c).
4 and P is maximum at q = 4 with P (4) = 32; the
3.
Math 10250 Exam 1 Solutions Fall 2006
q
5, 000
=
= 100. Therefore using the point-slope formula for the equation of the (demand) line
p
50
having slope 100 and passing through (200, 25000) gives q 25, 000 = 100(p 200), or q = 100p +45, 000.
1. We have
2.
Math 1250 Final Exam Solutions Fall 2008
1.) We need to nd the equation of the line passing from the points (t1 = 0, E1 = 20, 000) and
,
(t2 = 5, E2 = 135, 000). The slope of this line is E = 135,000020,000 = 1155000 = 23, 000. Therefore
t
5
busing the po
Math 10250,
Exam 3 Solutions,
Fall 2008
1. If f (x) = 3x5 10x4 , then f (x) = 15x4 40x3 and f (x) = 60x3 120x2 = 60x2 (x 2). The
second derivative changes sign ONLY at x = 2. So f (x) has ONE inection point.
2. If f (x) = 2x3 + 3x2 72x + 5, then f (x) = 6
Math 10250 Exam 2 Solutions Fall 2008
1. Applying the product rule we obtain f (x) = 2x ln x + x = x(2 ln x + 1). Setting f (x) = 0 gives
x(2 ln x + 1) = 0. Since x is in the domain of ln x we must have x > 0. Thus, we get the equation
2 ln x + 1 = 0, or
Math 10250 Exam 1 Solutions Fall 2008
1 At t = 0 wind energy production was w = 16.8 MW and was growing at the rate (slope) 5.2 MW/year. Using
the point-slope formula we obtain the equation w 16.8 = 5.2(t 0), or w = 5.2t + 16.8.
2 At t = 0 wind energy pro
Name
Date
Math 10250 Activity 2: Linear and Quadratic Functions (sect. 0.4 and 0.5)
GOAL: Understand the concept of slope for lines and linear functions and learn how to visualize
quadratic functions by completing the square.
y
A linear function is a func
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Date
Math 10250 Activity 4: Limits (Sect. 1.1)
GOAL: To obtain an intuitive understanding of the fundamental concept of limit and learn rules for
computing it.
x2 2x 3
Q1: Using your intuition, how would you interpret the statement: The function f (x
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Date
Math 10250 Activity 3: Polynomial, Rational, and Power Functions (sect. 0.6)
GOAL: To learn the basic properties and beheavior of Polynomial, Rational, and Power Functions.
Q1: Write an example of a function which is:
linear:
quadratic:
cubic
Math 10250 Exam 3 Solutions Fall 2013
1.) Since f (x) > 0, f is increasing in the interval [0, 1]. On the other hand, by f (x) < 0, the graph of f
is concave DOWN.
2.) Dierentiating the equation xey + y = 1 with respect to the independent variable x, usin
Math 1250 Final Exam Solutions Fall 2008
1.) We need to nd the equation of the line passing from the points (t1 = 0, E1 = 20, 000) and
,
(t2 = 5, E2 = 135, 000). The slope of this line is E = 135,000020,000 = 1155000 = 23, 000. Therefore
t
5
busing the po
Math 10250,
Exam 3 Solutions,
Fall 2008
1. If f (x) = 3x5 10x4 , then f (x) = 15x4 40x3 and f (x) = 60x3 120x2 = 60x2 (x 2). The
second derivative changes sign ONLY at x = 2. So f (x) has ONE inection point.
2. If f (x) = 2x3 + 3x2 72x + 5, then f (x) = 6
Math 10250 Exam 2 Solutions Fall 2013
1.) y = x ln(2x) = x(ln(2) + ln(x), so from the product and sum rules we get y = 1(ln(2) + ln(x) + x(0 +
1
1
x ) = ln(x) + ln(2) + 1 = ln(2x) + 1. Alternatively, by the product and chain rule y = ln(2x) + x 2x 2 =
ln(
Math 10250 Exam 1 Solutions Fall 2008
1 At t = 0 wind energy production was w = 16.8 MW and was growing at the rate (slope) 5.2 MW/year. Using
the point-slope formula we obtain the equation w 16.8 = 5.2(t 0), or w = 5.2t + 16.8.
2 At t = 0 wind energy pro
Math 10250 Exam 1 Solutions Fall 2013
1.) We are given that V (0) = 12, 000 and V (5) = 2, 000. Therefore the slope is m =
2000. Therefore, V (t) = 2000t + 12, 000.
V
t
= 10, 000/5 =
2.) Since the minimum cost occurs at q = 10, we know that h = 10 (this i
Math 10250 Exam 2 Solutions Fall 2008
1. Applying the product rule we obtain f (x) = 2x ln x + x = x(2 ln x + 1). Setting f (x) = 0 gives
x(2 ln x + 1) = 0. Since x is in the domain of ln x we must have x > 0. Thus, we get the equation
2 ln x + 1 = 0, or
Name
Date
Math 10250 Activity 9: Compound Interest and the Number e (Sec. 2.2)
Last time: Let A(t) be the balance at time t (years) of a bank account earning interest at an annual
rate r (in decimals) compounded n times a year. Then we have:
A(t) = P 1 +
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Date
Math 10250 Activity 8: Exponential Functions (sect. 2.1)
GOAL: Learn exponential functions with dierent bases and use them to model real-world situtations.
Exponential functions are of the form : f (x) = bx , where b > 0 is called base, like f (
Name
Date
Math 10250 Activity 7: Continuity (Sec. 1.3)
GOAL: Understand the concept of continuity and its basic properties, including the intermediate value
theorem.
y
Idea of Continuity: A function is continuous if you
never have to lift your pencil whil
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Date
Math 10250 Activity 6: Limits.(sect. 1.2 cont.), and Continuity (sect. 1.3)
GOAL: Understand behavior of functions at and horizontal asymptotes. For rational functions
the behavior at is determined by the leading tearms.
Limits at innity and hor
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Date
Math 10250 Activity 5: One-sided and Innite Limits (sect. 1.1 cont. & sect. 1.2)
GOAL: To learn about the limit of a function f (x) as x approaches to a number a from one side (left
or right), get an understanding of innite limits and relate the
Test! UBC Calculus Online Course Notes
Composite Functions
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UBC Calculus Online Course Notes Equations of Straight Lines
A Review of Lines and Slopes
This page serves as a quick review of straight lines and their important features. Many of these features are fundamental to a mathematical understanding of Calculus