1. This graph shows the relationship between the temperature of water (in
◦
C) and its density, at normal atmospheric pressure. (The inset shows a
zoomed-in version of a region of the larger graph.)
(a) Is this the graph of a function? Why or why not?
Solution: It is not the graph of a function because it fails the vertical
line test – it has lots of y-coordinates corresponding to x = 0
(b) What are the possibilities for the temperatures of a glass of water if
its density is 990kg/m3 ?
Solution: The temperature of the glass of water could be 0◦ C or about
50◦ C
(c) Draw a graph showing the relationship between the temperature of
water in ◦ F and its density at atmospheric pressure. To find the
temperature of water in ◦ F from its temperature in ◦ C, you can use
the formula
9
TF = TC + 32
5
where TF is the temperature in ◦ F and TC is the temperature in ◦ C.
Solution: This isn’t quite a function, but it’s almost a function, so
one way to think about it is using transformations of function graphs:
we have the graph of D where D(TC ) equals the density of water at
temperature TC , and we want a new function E(TF ), where you first
find the temperature in degrees C and then find D(that value). So in
other words we find that
TC =
1
TF − 32
9/5
and then that means
E(TF ) = D
TF − 32
5/9
and you graph that. The other option is just to re-label the axes on
the existing graph, to replace the numbers on the x-axis by the corresponding Fahrenheit temperatures:
Old x-coord New x-coord
9
−150
5 (−150) + 32 = −238
9
−100
5 (−100) + 32 = −148
9
−50
5 (−50) + 32 = −58
9
0
5 (0) + 32 = 32
9
50
5 (50) + 32 = 122
9
100
5 (100) + 32 = 212
2. Graph the function
K(x) = 3|2x − 1| + 2
and solve the equation
K(x) = 12x − 5
Solution:
To evaluate K(x), you first evaluate 2x − 1; if 2x − 1 ≥ 0, you keep it,
and multiply by 3 and add 2, to get
K(x) = 6x + 1 if 2x − 1 ≥ 0
If 2x − 1 < 0, you take its opposite, and get
K(x) = 3(−2x + 1) + 2 = −6x + 4 if 2x − 1 < 0
Since 2x − 1 ≥ 0 if x ≥ 1/2, this gives us
6x + 1 if x ≥
K(x) =
−6x + 5 if x <
2
1
2
1
2
3. Draw a graph of a quadratic function Q(x) with vertex (1, 1) and yintercept 3. Then write a formula for Q(x).
Solution:
We can in general write a quadratic function as
Q(x) = a(x − h)2 + k = ax2 − 2ahx + ah2 + k
and we know that (h, k) is the vertex, so in this case h = 1, k = 1; and we
want
Q(0) = ah2 + k = 3
so a = 2 and we have Q(x) = 2(x − 1)2 + 1 = 2x2 − 4x + 3 looks like this:
4. Let
F (x) =
x2 − 1
x−2
(a) Draw a graph of F . Mark zeros and vertical asymptotes.
(b) Estimate the value of
F (3, 423, 221)
to the nearest whole number. (Hint: divide.)
Solution: F (x) = 0 when its numerator is 0 and its denominator isn’t,
so that’s when x = ±1; it’s undefined its denominator is 0 which happens
when x = 2.
To find the limiting values of F (this will answer part b, and will also help
us graph) we can divide:
x2 − 1
x2 − 2x + 2x − 1
=
x−2
x−2
2x − 4 + 3
=x+
x−2
3
=x+2+
x−2
3
So when x is large F (x) ≈ x + 2.
Then we check where F (x) is positive and negative, using the fact that it’s
equal to (x − 1)(x + 1)/(x − 2)
Region
x < −1
−1 < x < 1
1<x<2
2<x
Sign
(−)(−)/(−) = −
(+)(−)/(−) = +
(+)(+)/(−) = −
(+)(+)/(+) = −
To answer part (b), we use the approximation we found: F (x) ≈ x + 2
when x is large, so F (3, 423, 221) ≈ 3, 423, 223.
5. Estimate
(1.001)2
ln(1.001)
e0.001
to within 3 decimal places.
Solution:
e0.001 ≈ 1.001 and ln(1.001) ≈ e0.001 so this is about
0.001
1.0012
= 0.001(1.001) = 0.001001
1.001
6. The number of half-tones between two notes is given by the formula
H = 12 log2 (F1 /F2 )
where F1 is the frequency of the higher note and F2 is the frequency of
the lower note.
The lowest sounds most humans can hear are about 20 Hz, and the highest
notes most humans can hear are about 20, 000 Hz. Estimate the number
of half-tones between the lowest and highest notes humans can year. Use
the fact that 210 ≈ 103 .
4
Solution:
The highest notes humans can hear are about 1000 times as high as the
lowest, so they’re about 103 times as high, or about 210 times as high –
that is, F1 /F2 ≈ 210 . So the number of half-tones between them is
12 log2 (210 ) = 12(10) = 120
7. You take out $100 loan at 1% annual interest.
(a) Write a formula for the balance on the loan after t years if the interest
is simple – that is, every year the lender adds 1% of its original value.
(b) Write a formula for the balance on the loan after t years if the interest
is compounded annually – that is, every year the lender adds 1% of
its current value.
(c) Write a formula for the balance on the loan after t years if the interest
is compounded continuously – that is, over a very small interval the
rate of increase is very close to 1% of its current value.
(d) For each of (a), (b), (c), estimate how long it will take for the balance
on the loan to double.
Solution:
(a) 1% of its original value is $1 so we have
S(t) = 100 + t
(b) This means we multiply by 1.01 every year, so we have
A(t) = 100(1.01)t
(c) This means we use
C(t) = P ert = 100e0.01 t
(d) In each case we need to figure out how long it takes to reach 200. In
the simple case it’s
S(t) = 100 + t = 200
when t = 100. In the annual case it’s
100(1.01)t = 200 =⇒ 1.01t = 2 =⇒ t = ln(2)/ ln(1.01)
Since ln(1.01) ≈ 1.01 and ln(2) ≈ 0.7, that’s
ln(2)
≈ 70
0.01
And for continuous interest we have
100e0.01t = 200 =⇒ e0.01t = 2 =⇒ t = 100 ln(2) ≈ 70
In the last two cases, you could have used the doubling time formula
70/P instead.
5
8. Draw the radius of the unit circle corresponding to a 5-radian angle. Estimate the sine and cosine of 5.
Solution: Since 5 is pretty close to 3π/2 but a little more, the angle looks
about like this:
Its cosine looks like about 0.3 and its sine looks like about −0.9
9. Pizza 1 is a 14-inch diameter pizza cut into 6 pieces. Pizza 2 is a 16-inch
diameter pizza cut into 8 pieces.
(a) Find the arc length of a slice of each pizza (measuring along the
crust). Which pizza has slices with a longer crust arc?
(b) Find the area of a slice of each pizza. Which pizza has slices with
greater area?
(c) Find the radian measure of a slice of each pizza (i.e., of the angle
formed at the center). Which pizza has slices that form a larger
angle?
Solution:
(a) A slice of Pizza 1 has an arc length of 1/6 the circumference of Pizza
1, or
1
14π
(2π(7)) =
6
6
A slice of Pizza 2 has an arc length of 1/8 the circumference of Pizza
2, or
1
(2π(8)) = 2π
8
A slice of Pizza 1 has a greater arc length.
(b) A slice of Pizza 1 has an area of 1/6 the area of Pizza 1, or
1
49π
(π(7)2 ) =
6
6
6
A slice of Pizza 2 has an area of 1/8 the area of Pizza 2, or
1
(π(8)2 ) = 8π
8
A slice of Pizza 1 has a greater area.
(c) A slice of Pizza 1 has a radian measure of 1/6 the radian measure of
a whole circle, or
π
1
(2π) =
6
3
A slice of Pizza 2 has a radian measure of 1/8 the radian measure of
a whole circle, or
1
π
(2π) =
8
4
A slice of Pizza 1 has a greater radian measure.
√
√
10. Draw a triangle ABC with m∠A = 45◦ , AB = 3, and BC = 2. Find
all possible measures of ∠C, and all possible lengths for side AC, exactly.
(There could be only one possibility, or more than one possibility.) Draw
pictures showing the possibilities.
Solution:
Using the Law of Sines, we can find that sin ∠C =
measure of 60◦ or 120◦ .
√
3/2 so ∠C has a
If m∠C = 60◦ , then m∠B = 75◦ and we can use the law of cosines to find
side AC:
√
b2 = a2 + c2 + 2ac cos(75◦ ) = 5 − 2 6 cos(75◦ )
We can find cos 75◦ using an angle-sum formula
√ √
√
√
√
2 3
21
6− 2
◦
◦
◦
cos(75 ) = cos(45 + 30 ) =
−
=
2 2
2 2
2
so
s
√
√
√ 6− 2
b= 5−2 6
2
◦
◦
If m∠C = 120 , then m∠B = 15 and we can use the law of cosines to
find side AC:
√
b2 = a2 + c2 + 2ac cos(15◦ ) = 5 − 2 6 cos(15◦ )
We can find cos 75◦ using an angle-difference formula
√ √
√
√
√
2 3
21
6+ 2
cos(75◦ ) = cos(45◦ − 30◦ ) =
+
=
2 2
2 2
2
so
s
√
√
√ 6+ 2
b= 5−2 6
2
These can of course be simplified.
7
11. Find the area of 4ABC, 4ABD and 4CBD.
Solution: We multiply the base of each by √its height and take half the
result: first√ we notice that AD = BD = 2/2 and that AC = 1, so
DC = 1 − 2/2 and we get
√
√
1
2
2
=
Area(4ABC) = (1)
2
2
4
√ √
1
1 2 2
=
Area(4ABD) =
2 2 2
4
√ !√
√
1
2
2
2−1
Area(4ABD) =
1−
=
2
2
2
4
12. In this picture, DB is an angle bisector.
(a) Find sin(∠BDC) and cos(∠BDC).
(b) Find BC, AB, and BD.
8
Solution:
(a) We can find sin ∠BDC and cos ∠BDC using half-angle identities:
r
r
1 − cos ∠ADC
1
sin ∠BDC = ±
=±
2
10
r
r
1 + cos ∠ADC
9
cos ∠BDC = ±
=±
2
10
In fact in both cases we want the positive square root since this is a
positive acute angle.
(b) Since
cos(∠DBC) =
4
,
BD
we can get
4
BD = p
9/10
Since
cos(∠DBC) =
we can get
BD =
13. (a) θ is an angle with 0 < θ <
θ/2 and find cos(θ/2).
√
π
2
110 p
BC
,
BD
4
9/10
=
4
3
and cos(θ) = 0.25. Draw a picture of
(b) ϕ is an angle with 3π
2 < ϕ < 2π and cos(ϕ) = 0.25. Draw a picture
of ϕ/2 and find cos(ϕ/2).
Here’s the picture:
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In both cases we need to find the cosines of these angles, and in both cases
we’ll use the half-angle formula to get
r
p
1 + cos(x)
cos(x/2) = ±
= ± 5/8
2
But in the case of θ/2, its cosine
p is positive and in the
p case of ϕ/2 its
cosine is negative: cos(θ/2) = 5/8 and cos(ϕ/2) = − 5/8.
14. Each day, the temperature in City X reaches its minimum of 50◦ F at 6 am
and its maximum of 70◦ F at 6 pm. If a stopwatch is started at midnight
and
T (t) = the temperature t hours after the stopwatch is started
is a transformation of a trig function, write a formula for T (t).
Solution:
We want a trig function with period 24, midline 60, and amplitude 10.
One choice is
10 sin(2πt/24) + 60
But we need it to have a minimum at t = 6, and right now it has a
maximum. So we’ll use
T (t) = 10 sin(2πt/2t + π) + 60
15. θ is an obtuse angle and sin θ = 0.2. What is sin θ −
10
π
2
?
Probably the easiest way to do this is with an angle-difference formula:
π
sin θ −
= sin θ cos π/2 − cos θ sin π/2 = − cos θ
2
p
Then we need to find cos θ. We know it’sp± 1 − sin2 θ and since it’s
√
2
obtuse θ has a negative
√ cos θ = − 1 − sin θ = − 0.96. And
√ cosine:
sin(θ − π/2) = −(− 0.96) = 0.96
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