Download MA107 Precalculus Algebra Solutions to Exam 1 Review Questions

MA107 Precalculus Algebra
Solutions to Exam 1 Review Questions
26 January, 2008
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1. Which (if any, or all) of the following relationships show a function?
a. {(1, 2), (2, 3), (3, 3), (4, 5)}
(*) This is a function. An easy way to tell is to observe that none of the x coordinates are
repeated. Hence, for each x value there is a unique corresponding y value.
b. {(5, 5), (5, 6), (6, 4), (7, 2)}
(*) This is not a function. The x coordinate 5 corresponds to both y = 5 and y = 6.
c. {(1, 1), (2, 1), (3, 1), (4, 1)}
(*) This is a function, for the same reason as in question a.
d. x2 + y2 = 1
(*) Solve for y:
x2 + y2 = 1
2
y2 = 1 −
√x
y = ± 1 − x2
Note
√ that for any value of x there are two corresponding y values, both
− 1 − x2 . Hence, not a function.
e. y + x2 = 1
(*) Solve for y:
y + x2 = 1
y = 1 − x2
This is a function.
2
√
1 − x2 and
2. a. Find the domain of the function f(x) =
4
x−5 .
(*) The underlying assumption for the remainder of the semester is that a function maps a
real number to a real number. So any number which does not map to another real number
is not in the domain.
What can go wrong in a rational expression, such as that above, is that we can be
given a value of x such that the denominator is 0. We can’t divide by 0, so the output of
the function is not a real number.
Solve the denominator for 0. If x − 5 = 0 then x = 5. So f(5) is not a real number.
Hence, 5 cannot be in the domain.
Any other value for x is okay. Hence, the domain is:
All real numbers except 5,
or
{x | x 6= 5},
or
(−∞, 5) ∪ (5, ∞)
b. Find the domain of the function f(x) =
x
.
x2 −1
(*) Again, I will solve the denominator for 0 to describe what is not in the domain.
√
Let x2 − 1 = 0. Then x2 = 1, or x = ± 1 = ±1.
So the domain is:
all real numbers except 1 or −1,
or
{x | x 6= ±1},
or
(−∞, −1) ∪ (−1, 1) ∪ (1, ∞)
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3. The graph of f(x) is given to the right.
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a. What are the x-intercepts?
(*) The graph of f(x) crosses the x axis at x = −11, x = 1, and x = 3.
b. What are the y-intercepts?
(*) The graph crosses the y axis at y = 3.
c. List the coordinates of the relative minima
(*) There is one, at (2, −3).
d. List the coordinates of the relative maxima
(*) There is one, at (0, 3).
e. On what interval(s) is f(x) increasing?
(*) f(x) is increasing on the interval [−1.5, 0), and (2, 3.2].
f. On what interval(s) is f(x) decreasing?
(*) f(x) is decreasing on the interval (0, 2).
h. For what value(s) of x is f(x) = 2
(*) At x = −1.5, x = .5, and x = 3.1, the value of f(x) is 2.
i. What is the value of f(2)?
(*) When x = 2 then y = −3.
j. Is the point (1, 0) a relative maximum?
(*) No. The function is still decreasing around that point. Specifically, to the left of this
point the function is decreasing, and to the right of this point the function is decreasing.
k. What is the domain?
(*) The possible values of x lie in the interval [−1.5, 3.2].
l. What is the range?
(*) The possible values of y lie in the range [−5, 3].
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4. Let f(x) = x2 + x + 1.
a. Find f(x + h).
(*)
f(x + h)
= (x + h)2 + (x + h) + 1
= (x + h)(x + h) + (x + h) + 1
= x2 + 2hx + h2 + x + h + 1
b. Find the difference quotient for f(x).
Recall that f(x + h) = x2 + 2hx + h2 + x + h + 1
and that f(x) = x2 + x + 1.
Then I will substitute (x2 + 2hx + h2 + x + h + 1) for f(x + h),
and (x2 + x + 1) for f(x), and proceed to simplify.
(*)
f(x+h)−f(x)
x
=
(x2 +2hx+h2 +x+h+1)−(x2 +x+1)
h
=
x2 +2hx+h2 +x+h+1−x2 −x−1
h
=
2hx+h2 +h
h
=
h(2x+h+1)
h
= 2x + h + 1
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5. Let f(x) = x2 − 1
a. Find the average rate of change from 1 to 5.
(*) The average rate of change from A to B is given by
f(B)−F(A)
.
B−A
Applying the formula for A = 1 and B = 5, we find:
f(B)−f(A)
B−A
=
f(5)−f(1)
5−1
=
24−0
4
=6
b. Find the equation of the secant line from 1 to 5.
(*) We are looking for a line. Hence, we start with y = mx + b and determine the values
for m and b.
Recall the slope of the secant line is equal to the average rate of change from A to B. Then
we know how to find m. We use the average rate of change formula. We already did that
above, so m = 6.
Now I know y = 6x + b. To find b I pick a point on the line, plug the corresponding x and y values into the equation for the line, and solve for b.
I will let x = 1. (I must choose either x = 1 or x = 5 because those are the two x
values the secant line connects). When x = 1 then y = f(1) = (1)2 − 1 = 0.
Then 0 = 6(1) + b, or b = −6.
Then y = 6x − 6.
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6. The functions S(p) and D(p) model supply and demand of quantities of a particular brand of soup.
S(p) = 5p − 20
D(p) = 30 − 7.5p
a. Find the quantity demanded when the price is 2.
(*) When p = 2, D(p) = D(2) = 30 − 7.5(2) = 15.
b. Find the price when the supply is 15.
(*) We’re asked to find p when S(p) = 15.
I solve the equation 15 = 5p − 20 for p.
Then 5p = 35, or p = 7.
c. Find the equilibrium point. That is, the price for which supply equals demand.
(*) I’m looking for the value of p such that S(p) = D(p).
S(p)
5p − 20
12.5p
p
= D(p)
= 30 − 7.5p
= 50
=4
d. What is the quantity demanded at this point?
(*) When p = 4 then D(p) = D(4) = 0.
This particular brand of soup appears to be lacking something...
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7. Consider the piecewise-defined function below.

if x < −4;
 0
x + 4 if −4 ≤ x ≤ 4;
f(x) =
 2
x − 8 if x > 4.
a. Graph the above function.
b. What is f(−5)?
(*) f(−5) = 0
c. What is f(−2)?
(*) f(−2) = 2
d. What is f(2)?
(*) f(2) = 6
e. What is f(6)?
(*) f(6) = 24
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8. a. Please draw the graph of f(x) = x2 .
(*) You should be able to check the answers to this question and each of the following using
a graphing calculator. Or try:
http://www.webgraphing.com/
http://www.webgraphing.com/graphing_basic.jsp
b. State the equation and draw the graph of f(x) after compressing vertically by a factor of 12 .
(*) 12 f(x) = 12 x2
c. State the equation and draw the graph of f(x) after expanding vertically by a factor of 3.
(*) 3f(x) = 3x2
d. State the equation and draw the graph of f(x) after compressing horizontally by a factor of 12 .
(*) f(2x) = (2x)2 = 4x2
e. State the equation and draw the graph of f(x) after expanding horizontally by a factor of 4.
(*) f( 14 x) = ( 14 x)2 =
1 2
16 x
f. State the equation and draw the graph of f(x) after shifting left by 2 units.
(*) f(x + 2) = (x + 2)2 = x2 + 4x + 4
g. State the equation and draw the graph of f(x) after shifting down by 2 units.
(*) f(x) − 2 = x2 − 2
h. State the equation and draw the graph of f(x) after reflecting over the x-axis.
(*) −f(x) = −x2
i. Is the point (2, 4) on the graph of f(x) after the previous transformation?
(*) When x = 2 then −f(x) = −f(2) = −(2)2 = −4.
x can’t correspond to both 4 and −4, so no.
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