Download MGF 1105: Exam 1 Solutions

MGF 1105: Exam 1 Solutions
1. (10 points) In the graph below, tick marks along each axis represent one unit each.
a) List the x-intercepts, if any, of this graph. Show these on the graph.
b) List the y-intercepts, if any, of this graph. Show these on the graph.
c) Is the point (1, 2) part of the graph? Show why or why not in the picture.
Solution:
a) The x-intercepts are -4, -2, 3
b) The y-intercept is -2
c) The point (1, 2) is not part of the graph.
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2. (10 points) Sketch the graph of y = x2 −4 by plotting points with x = −3, −2, −1, 0, 1, 2, 3.
Solution:
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3. (10 points) Solve for x:
5x − (2x + 2) = x + (3x − 5)
Solution:
5x − (2x + 2) = x + (3x − 5)
5x − 2x − 2 = x + 3x − 5
3x − 2 = 4x − 5
x = 3
4. (10 points) Solve for x:
6
11
+5=
x−1
x−1
Solution: Note that, right away, we see that the ‘domain’ of this equation is x 6= 1.
The idea is to multiply both sides by x − 1, to cancel the denominators. (Since x 6= 1, this
is legal.)
6
+5 =
x−1
(x − 1) ·
·
(x
−1)
11
x−1
6
11
+ 5
=
· (x − 1)
x−1
x−1
6
+ (x − 1) · 5 =
x−
1
11
·
(x
−1)
x−
1
6 + (x − 1) · 5 = 11
6 + 5x − 5 = 11
5x = 10
x = 2 ( 6= 1! so, okay)
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5. (10 points) Solve for x:
6
5
−23
−
= 2
x+3 x−2
x +x−6
Solution: We start by factoring all the denominators:
• x+3
• x−2
• x2 + x − 6 = (x + 3)(x − 2)
Thus, the LCD is (x + 3)(x − 2) and the domain of the equation is x 6= 2, −3. Again,
the idea is to multiply both sides by the LCD to do away with the denominators.
(x + 3)(x − 2) ·
− 2) ·
(x
+3)(x
5
6
−
x+3 x−2
=
−23
x2 + x − 6
6
5
−
x+3 x−2
=
−23
(x + 3)(x − 2)
5
6
−
x+3 x−2
=
6
· 5
(x
−2)
− (x + 3)
x+
3
x−
2
=
−23
(x + 3)(x − 2)
· (x + 3)(x − 2)
−23
((
((−
(x(
+(3)(x
2)
(( · (
(−
(
(
(x
+
3)(x
2)
((
6 · (x − 2) − 5 · (x + 3) = −23
6x − 12 − 5x − 15 = −23
x − 27 = −23
x = 4 ( 6= 2, −3! so, okay)
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6. (15 points total) Simplify each expression involving complex numbers.
a) (2 − 3i)(1 + i) − (3 − i)2
Solution: Recall that here, we work just as if i were x, but with the rule that i2 = −1.
• (2 − 3i)(1 + i) = 2 + 2i − 3i − 3i2 = 2 − i − 3(−1) = 5 − i
• (3 − i)2 = (3 − i)(3 − i) = 9 − 6i + i2 = 9 − 6i + (−1) = 8 − 6i
Thus, subtracting these:
(2 − 3i)(1 + i) − (3 − i)2 = (5 − i) − (8 − 6i) = 5 − i − 8 + 6i = −3 + 5i
b)
2 − 8i
1+i
Solution: To divide by a complex number, we multiply top and bottom by the complex
conjugate:
2 − 8i
2 − 8i 1 − i
2 − 2i − 8i + 8i2
2 − 10i + 8(−1)
−6 − 10i
=
·
=
=
=
= −3 − 5i
2
1+i
1+i 1−i
1−i
1 − (−1)
2
c) (−2 +
√
−9)(3 +
√
−4)
Solution: Remember, here, the first thing you do is rewrite:
√
√
−a = i a.
√
√
√
√
(−2+ −9)(3+ −4) = (−2+i 9)(3+i 4) = (−2+3i)(3+2i) = −6−4i+9i+6i2 = −12+5i
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7. (15 points) A new car worth $23, 000 depreciates in value by $2, 500 each year.
a) Write a formula for this model.
Solution: Letting y be the value of the car after x years,
y = 23, 000 − 2, 500x
b) According to the model, when does the car become worthless (value = 0)?
Solution: In other words, for which x is y = 0?
0 = 23, 000 − 2, 500x
=⇒
x=
23, 000
= 9.2
2, 500
According to the model, the car will be worth $0 after exactly 9.2 years.
c) What does the model predict for the value of the car after 100 years? Explain.
Solution: According to the model, after 100 years, i.e., for x = 100, the car is worth:
y = 23, 000 − 2, 500(100) = −227, 000
Since this is pretty much absurd, rather than believe the car will actually be worth negative
$227,00 dollars after 100 years, we should interpret x = 100 as being outside the range
where this model (”depreciates in value by $2, 500 each year”) is meaningful.
8. (10 points) When 7 is subtracted from 4 times a number, the result is 50% of the number.
What is the number?
Solution: Letting x be the number,
4x − 7 = .5x
=⇒
4x − .5x = 7
6
=⇒ 3.5x = 7
=⇒
x=2
9. (10 points) A = 12 h(a + b)
a) Solve this equation for h.
Solution:
1
2A
A = h(a + b) =⇒ 2A = h(a + b) =⇒ h =
2
a+b
b) Solve this equation for a.
Solution:
1
2A
2A
A = h(a + b) =⇒ 2A = h(a + b) =⇒ a + b =
=⇒ a =
−b
2
h
h
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