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Math 141 - Midterm Review Solutions
(1) Chapter 1
(a) 1.1: Functions
(i) Write the function f (x) = 2x with domain {1, 2, 3, 4} using dots and arrows, and
using pairs of numbers.
We get {(1, 2), (2, 4), (3, 6), (4, 8)} and
1
2
2
3
4
4
6
8
(ii) Express the domain of (f + g)(x) and (f /g)(x) in terms of the domains of f and
g.
The domain of f + g is
domain of f ∩ domain of g.
The domain of f /g is
domain of f ∩ domain of g
except for x-values where g(x) = 0.
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and g(x) =
(iii) If f (x) = x−1
domain of (2f + 2g)(x)?
We get
√
x, write a formula for (2f + 3g)(x). What is the
(2f + 3g)(x) = 2
1
x−1
√
+ 3 x.
The domain is
domain of f ∩ domain of g.
(iv) If f (x) = x2 − x, find (and simplify) and expression for the function
f (x + h) − f (x)
.
h
f (x + h) − f (x)
((x + h)2 − (x + h)) − (x2 − x)
=
h
h
2
2
x + 2xh + h − x − h − x2 + x
=
h
2
2xh + h − h
=
h
= 2x + h − 1.
(b) 1.2: Graphs of functions
1
2
(i) Given a function f (x) with domain D (a collection of real numbers), the graph
of f consists of all points (x,
), where x is in
.
all points (x, f (x)), where x is in the domain D.
(ii) Which of the following curves are graphs of functions? Why?
The curve that starts in the bottom left is the graph of a function, because it
passes the vertical line test. Every vertical line intersects the curve at most once.
The curve that starts in the top left is not the graph of a function. There is a
vertical line (specifically, x = 3) that intersects the curve more than once.
(iii) The graph of a function f (x) is pictured below. Is it true that f (2) = 1? Is it
true that f (0) = −3? Are there any numbers x so that f (x) = 5?
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It is not true that f (2) = 1. It looks like f (2) is about −1. On the other hand,
f (0) is indeed −3. To see if there are any numbers x such that f (x) = 5, we can
imagine the horizontal line y = 5 and see where it intersects the curve. It looks
like it intersects at about x = ±4.
(c) 1.3: Properties of Functions
(i) Describe what feature the graph of a function f (x) must have if it has even
symmetry. Write the definition of even symmety (that is, write the equation
that f (x) must satisfy if it is to be even). Repeat for f (x) odd.
To be even, the graph must by symmetric about the y-axis (that is, reflection
across the y-axis must send the graph back to itself), and the formula for f (x)
must satisfy f (−x) = f (x). To be odd, the graph must be symetric about the
origin (that is, a half turn rotation around the origin must send the graph back
to itself), and the function must satisfy f (−x) = −f (x).
(ii) Determine if the following functions are even or odd or neither:
x4
,
x2 + 1
x3 + 1,
ln x,
ln |x|,
(−x)4
x4
=
(−x)2 + 1
x2 + 1
x4
6= − 2
,
x +1
so the first function is even and not odd.
x+
√
3
x.
4
(−x)3 + 1 = −x3 + 1
6= −(x3 + 1)
and also
6= x3 + 1.
So the second function is neither even nor odd.
ln x cannot be even or odd because its domain is not symmetric about the y-axis.
ln | − x| = ln |x|
6= − ln |x|,
so ln |x| is even and not odd.
(−x) +
So x +
√
3
√
3
√
√
−x = −x − 3 x = −(x + 3 x)
√
6= x + 3 x.
x is odd but not even.
(d) 1.4 Library of Functions
(i) For each of the following functions, what is the function’s domain, range, and
graph? Is the function even or odd or neither?
x, xn for n odd, xn for n even, c for c a constant,
1
x,
√
x,
√
3
x, |x|.
See section 1.4.
(ii) Sketch a graph of the following function. What is its domain?
(
−x − 3,
f (x) = 1
x,
x < −2
−1 < x < 3.
The graph looks like the line y = −x − 3 to the left of x = −2, and it looks like
1
x for −1 < x < 3. So it looks like this:
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Note that my picture should probably indicate that the endpoints are not included. Just looking at the inequalities in the multipart rule, you can see that
the domain of f can be no larger than (−∞, −2) ∪ (−1, 3). But there is another
hole in the domain at x = 0, because 1/x is not defined there. So the domain of
f is
(−∞, −2) ∪ (−1, 0) ∪ (0, 3).
(e) 1.5 Graphical Transformations
(i) For each graphical operation, describe how to modify the formula for the function
f (x) to affect the graph of f (x) in the indicated way:
• Shift the graph up by c units.
Replace f (x) by f (x) + c.
• Stretch the graph vertically by a factor of a.
Replace f (x) by af (x).
• Shift the graph to the right by c units.
Replace x by x − c, yielding f (x − c).
• Flip the graph horizontally.
Replace x by −x, yielding f (−x).
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• Shift the graph to the left by c units.
Replace x by x + c, yielding f (x + c).
• Stretch the graph horizontally by a factor of a.
Replace x by a1 , yielding f ( a1 x).
• Compress the graph vertically by a factor of a.
Replace x by ax, yielding f (ax).
(ii) Sketch a graph of the following function by starting with the graph of
using graphical operations:
1
x
and
−2
− 1.
3x + 2
Use the following graphical operations:
1
• Replace x by x + 2, giving x+2
, shifting the graph to the left by 2.
• Replace x by 3x, giving
of 3.
1
3x+2 ,
compressing the graph horizontally by a factor
−2
• Multiply the function by −2, giving 3x+2
, flipping the graph vertically and
stretching vertically by a factor of 2.
−2
• Subtract 1 from the function, giving 3x+2
− 1, shifting the graph down by
1.
You can do the operations in a different order, but remember that if you switch
the order of the two horizontal operations or the two vertical operations, the
numbers change.
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(iii) Sketch a graph of − ln(−x).
We can use the following grapical operations, starting with the graph of ln(x):
• Replace x by −x, giving ln(−x), flipping the graph horizontally.
• Multiply the function by −1, giving − ln(−x), flipping the graph vertically.
This gives the following:
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(2) Chapter 2: Linear and Quadratic Functions
(a) 2.1: Linear Functions
(i) What is the general form of a linear function?
f (x) = mx + b for constants m and b.
(ii) If f (x) = mx + b is a linear function, what is the domain and range of f ? Do
they depend on m or b?
The domain is always all real numbers. If m = 0, the range is just the single
number b. If m 6= 0, the range is all real numbers.
(iii) If f (x) is linear, how many possible solutions are there to an equation f (x) = c,
where c is some constant real number? Does such an equation always have a
solution?
Such an equation does not always have a solution. Let f (x) = mx + b. If m = 0,
then the equation f (x) = c has a solution only if c = b (in which case any x is
a solution). If m = 0 and c 6= b, then there are no solutions. If m 6= 0, then the
equation f (x) = c has exactly one solution, no matter what c is.
(iv) When is a linear function mx + b even? When is it odd?
A linear function is even when m = 0 and odd when b = 0. (I’ll let you check
that this is true.)
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(v) Is the following data represented by a linear function? If so, what is the function?
x f (x)
1 2.5
3
4
5 5.5
7
7
Taking the first two data points, we get
∆y
1.5
=
.
∆x
2
∆y
Checking the other data points gives the same ratio for ∆x
. So the function is
linear with m = 1.5/2 = 3/4. Now we can plug in a data point to solve for b:
4 = f (3) =
3
4
· 3 + b ⇒ b = 74 .
So we have f (x) = 34 x + 74 .
(b) 2.3 Quadratic functions
(i) When is a function of the form ax2 + bx + c not a quadratic function?
When a = 0. Then it is a linear function.
(ii) If c is a real number and f (x) is quadratic, how many solutions could the equation
f (x) = c have? Draw pictures demonstrating each possibility.
It might have 0, 1 or 2 solutions.
(iii) Find roots of f (x) = x2 − x − 3 by completing the square.
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To find roots, solve f (x) = 0:
0 = x2 − x − 3
= (x − 12 )2 −
= (x −
13
4
r
±
= (x −
13
=x−
4
1 2
2)
1 2
2)
x=
±
3
1
2
r
1
2
−
1
4 −
13
4
13
.
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(iv) Find the intersection points of the graphs of f (x) = 12 x2 − 2x − 1 and g(x) =
−x2 + 2.
To find intersection points, just solve f (x) = g(x):
1 2
2x
3 2
2x
− 2x − 1 = −x2 + 2
− 2x − 3 = 0.
From here we can complete the square, or just use the quadratic formula:
√
2 ± 4 + 18
x=
.
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(v) What is the range of f (x) = 2x2 − 4x + 5.
Since this quadratic opens up, the range is every y-value above the y-coordinate
of the vertex. The x-coordinate of the vertex is
x=
−b
= 1.
2a
So the y-coordinate of the vertex is f (1) = 3. Thus, the range of f (x) is y >= 3.
(c) 2.4 Properties of Quadratic Functions
(i) When is a quadratic function ax2 + bx + c even? When is it odd?
It is even when b = 0 and odd when a = c = 0. (I’ll let you check that this is
true.)
(ii) What is the minimum value taken by the function g(x) = 3x2 + 2x − 1?
Since this quadratic opens up, the minimum value is taken at the vertex. This
is at
−b
x=
= − 31 ,
g(− 13 ) = − 43 .
2a
So − 43 is the minimum value.
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(iii) What is the minimum value taken by the function h(x) = −2x2 + 50x?
This quadratic opens down, so it has no minimum value.
(d) 2.5 Quadratic inequalities
(i) Solve −x2 + 3x − 1 ≤ 0.
First solve the equation −x2 + 3x − 1 = 0:
√
−3 ± 9 − 4
x=
=
−2
3
2
√
±
5
2 .
Since the quadratic opens down and the inequality is ≤, solutions to the inequality are outside solutions to the equation. So the set of solutions is
−∞, 32 −
√
5
2
i
∪
h
3
2
√
+
5
2 ,∞
.
(ii) Solve x2 − x + 2 < 21 x2 + 10.
Rewriting the inequality, we get 21 x2 −x−8 < 0. Solving the associated equation,
we get
x=
1±
√
√
1 + 16
= 1 ± 17.
1
Since the quadratic opens up and the inquality is <, solutions to the inequality
are between solutions to the equation. So the set of solutions is
(1 −
√
17, 1 +
√
17).
(iii) Is it possible for an inequality of the form ax2 + bx + c > 0 to have (1, 2) ∪ (3, 4)
as its set of solutions? How about [−10, 10]? How about {5}?
(1, 2) ∪ (3, 4) is impossible, but the other two are possible.
(e) 2.8 Equations and Inequalities involving the Absolute Value
(i) Sketch a graph of f (x) = |x + 1| − 1 and use your picture to solve the inequality
f (x) > 3.
We can sketch the graph using graphical transformations: a shift to the left by
1 and down by 1. We get
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The set of solutions to the inequality is thus
(−∞, −5) ∪ (3, ∞).
(ii) Solve |x2 − 1| = 1.
Removing the absolute value signs, we get two equations:
x2 − 1 = 1,
x2 − 1 = −1.
Solving the first gives
√ x = ±2, and solving the second gives x = 0. So the set of
solutions is x = 0, ± 2.
(iii) Solve |x2 + 6x − 2| = −3.
Since the right-hand side of the equation is negative, there are no solutions. The
absolute value of something can’t be negative.