Download Math 1090-001 Midterm 3

Math 1090-001
Midterm 3
11/20/2014
Instructions: Follow the given instruction for every exercise. You have
60 minutes to complete the exam. Show proof of your work. Notice that
the space left for each question is sufficient, but possibly not necessary,
to answer the question. The number in the square next to the problem
deignates the number of points that problem is worth. The test will be
graded out of 100, but the exercises sum to 105 points, this means 5
points are extra credit.
First and Last Name:
UnID:
1
1. Consider the function
y = 2x2 − 8x + 5.
2
(a) This equation represents a (circle the correct answer):
line
parabola
plane
3
(b) Is it concave up or down? Why?
5
(c) Find the coordinates of the vertex;
5
(d) Find the coordinates of the y and x intercept;
5
(e) You can rewrite the equation as y = 2(x − 2)2 − 3. Looking at the equation in
this form, sketch its graph on the plane below, transforming the graph of y = x2 .
Label at least two points of the graph.
Solution:
1. We are considering a quadratic equation in two variables, it represents a
parabola.
2. The parabola is going to be concave up, since the leading coefficient is positive.
3. we have a formula to find the x coordinate of the vertex: xV =
Plug this in the equation, you get yV = −3.
−b
2a
=
8
4
= 2.
4. if we set x = 0, we find the y intercept (0, 5). Setting y = 0 tells us that
the x-intercept points
are the zeroes of 2x2 − 8x + 5. The quadratic formula
√
yields x1,2 = 4±2 6 .
5. start from y = x2 . Shift by 2 to the right: y = (x − 2)2 . Stretch vertically by
a factor 2, and then shift vertically of 3 units downwards: y = 2(x − 2)2 − 3
(see end of the document).
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2. Solve the following equations:
10
(a) 4, 321e2x−3 = 4, 321;
10
(b) log3 (x) + log3 (x − 2) + 4 log3 (1) = 1
Solution:
1. Divide both members by 4, 321. You’re left with e2x−3 = 1. Take the natural
logarithm of both sides: you get 2x − 3 = 0. The solution is x = 3/2. No
need to check if it is in the domain.
2. This equation makes sense only for x positive and x − 2 > 0, hence we want
x to be bigger than 2. Notice that log3 (1) = 0 by definition of logarithm,
hence the last summand vanishes. Using the logarithm properties we write
log3 (x(x − 2)) = 1. Take the exponential with base 3 on both sides, the
equation then reads x2 − 2x = 3, i.e. x2 − 2x − 3 = 0. This can be solved
factoring, the roots are x = −1 and x = 3. Only the latter is acceptable.
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15
3. A certain investment it is compounded each semester with an interest rate of 2% per
year, how long does it take for the initial amount of money to triple in size?
Solution: We know that such investments follow the rule
r nt
S =P 1+
.
n
(0.1)
In this case, we want S = 3P , and we know r = 0.02 and n = 2. Hence we need
to solve for t the equation
3P = P
0.02 2t
1+
.
2
(0.2)
Cancel out the P and do the sum: you get 3 = (1.01)2t . Take the logarithm (in
base e or 10 is easier) on both sides: ln 3 = ln((1.01)2t ). By the properties of
ln 3
≈ 55, 2.
logarithms 2t ln(1.01) = ln 3, i.e. t = 2 ln(1.01)
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4. Consider the equation given by
y = 3x−2 − 1.
(0.3)
5
(a) This function is (circle the correct answer):
polynomial exponential logarithmic
5
(b) Compute the y-intercept and the x-intercepts;
5
(c) Using the data collected above, sketch the graph of the function
5
(d) Sketch the graph of y = −(3x−2 ) + 1 (Hint: this graph is a transformation of the
graph of the equation given above).
Solution:
1. We have an exponential function;
2. To compute the intersection with the y axis, we set x = 0 in the equation:
we get y = 30−2 − 1 = 3−2 − 1 ∼ −0.889. For the intersections with the x
axis, set y = 0 instead. We have to solve 3x−2 − 1 = 0, i.e. 3x−2 = 1. Take
log3 of both sides, and use some properties of logarithms: log3 3x−2 = x − 2
and log3 1 = 0. Hence we get x − 2 = 0. The x intercept is then the point
(2, 0).
3. put together informations gathered at previous points (see end of the document);
4. the function we have to graph is the opposite (- sign) of the one we started
with: the graph will be reflected across the x-axis (see end of the document).
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5. Consider the functions f (x) = x − 2, and g(x) = x2 + 3x − 4. Compute the following
functions, and state their domain:
5
(a) (f ◦ g)(x);
5
(b) ( fg )(x);
5
(c) f −1 (x)
Solution:
1. (f ◦ g)(x) = g(x) − 2 = x2 + 3x − 4 − 2 = x2 + 3x − 6, this function is defined
for all real numbers;
x−2
x−2
2. ( fg )(x) = x2 +3x−4
= (x+4)(x−1)
. This function is defined on the whole R
except from the points 1, −4.
3. consider y = x − 2. Solve for x: x = y + 2. Swap the role of the variables:
f −1 (x) = x + 2. The domain is the whole set of real numbers.
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6. Use the properties and definition of logarithms to:
10
5
(a) write in condensed form log2 2 + 2 log2 x + 3 log2 y + 12 log2 7 − log2 3;
(b) compute the quantity log(1012345678910111213 ).
Solution:
1. We have log2 2 = 1, 2 log2 x = log2 (x2 ) and 3 log2 y = log2 (y 3 ). Using
properties labeled as 5 and 7 in the book, we then have
"
√ #
2y3 7
1
2x
. (0.4)
log2 2 + log2 (x2 ) + 3 log2 y + log2 7 − log2 3 = log2
2
3
2. Using the definition of logarithm: the exponent to give to 10 in order to get
1012345678910111213 ? It is exactly 12345678910111213.
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Question
Points
1
20
2
20
3
15
4
20
5
15
6
15
Total:
105
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Score
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