Solutions to Assignment #04 –MATH 1130
Precalculus
Section 2.3
(I) Complete Exercise #16 on p. 159. Remember that you will have a remainder and your …nal
terms will look like this:
???
x2 + 3
x
1
Here’s the long division.
x2
x3
1 )
x5
x5
+0x4 +0x3 +0x2 +0x
1x2
1x2
+7
+0x
+7
The remainder is x2 + 7 ; so the quotient is
x2 +
x2 + 7
:
x3 1
(II) Complete Exercises #28 & #35 on p. 159.
(#28) The divisor is (x + 3) ; so in synthetic division, we use x =
3 )
1
13
3
0
48
0
144
120
432
1
16
48
144
312 ( 856)
Thus, we know that f ( 3) =
Also, the quotient is x4
3:
856: You can check this.
16x3 + 48x2
(#35) The divisor (x + 1=2) ; so we use x =
144x + 132
856
:
x+3
1=2:
1=2 )
4
16
2
23
7
15
15
4
14
30
(0)
1
2
4x2 + 14x
So x = 1=2 is a zero of f (x) :
Thus, we can factor f (x).
f (x) =
x+
1
30
80
936
We note that the coe¢ cients of the quotient are all even, so we can factor out 2 and
then distribute that factor of 2 over the …rst factor.
1
f (x) =
x+
(2) 2x2 + 7x 15
2
= (2x + 1) 2x2 + 7x
= (2x + 1) (2x
1 3
; ;
2 2
The rational zeros are x =
15
3) (x + 5) :
5:
(III) Complete Exercises #47bc & #60 on p. 160.
(#47b) h (x) = 3x3 + 5x2
10x + 1
1=3 )
Thus, we have h (1=3) =
(#47c) Same.
3
5
1
3
6
10
2
1
8=3
8 ( 5=3)
5=3:
2 )
3
5
6
3
1
10
2
1
16
8 (17)
Thus, we have h ( 2) = 17:
Section 2.4
(IV) Complete Exercises #70 & #72 on p. 168.
(#70) Quadratic Formula. 16t2
4t + 3 = 0:
q
( 4)
( 4)2
t =
2 (16)
16
7 2
x
8
14x2
4
p
4 11i
1
=
32
The roots are nonreal and conjugate.
(#72) Multiply through by 16.
=
4 (16) (3)
3
5
x+
4
16
p
8
11i
=
4
p
176
32
:
= 0 16
12x + 5 = 0
x =
=
( 12)
q
( 12)2
4 (14) (5)
2 (14)
12
p
p
2 34i
6
34i
=
:
28
14
2
=
12
p
28
136
The roots are nonreal and conjugate.
(V) Complete Exercises #87 & #88 on p. 168. Be sure to justify your answer!
(#87) True or False? There is no complex number that is equal to its complex conjugate.
FALSE. Any purely real number: a + 0i is its own complex conjugate.
p
(#88)
6i is a solution of x4 x2 + 14 = 56:
TRUE. We can substitute in for x:
p
6i
p
4
6i
2
p
+ 14 =
6
4
= 36 (+1)
p
i4
6
2
i2 + 14
6 ( 1) + 14 = 56:X
We can also solve for ALL the zeros by factoring!
x4
x2 + 14 = 56 ) x4
x2
42 = 0:
Let u = x2 ; u2 = x4 :
u2
u
42 = (u + 6) (u
u =
Thus, we have u = x2 =
p
u = x2 = 7 ) x =
7:
6)x=
p
7) = 0
6; 7:
6i and
Section 2.5
(VI) Complete Exercises #9 & #18 on p. 179. Show your work using Synthetic Division and
FACTOR f (x) completely.
(#9) Use the Rational Zero Test to list all possible zeros of f:
f (x) = 2x4
17x3 + 35x2 + 9x
45:
The last coe¢ cient ( 45) gives us all the possible factors for the numerator:
1; 3; 5; 9; 15; 45:
The leading coe¢ cient (2) gives us all the possible factors for the denominator: 1; 2:
So here’s the list (in the “best”order –we start with the numerator factors and then go
down the line in the denominator factor list):
1;
1
;
2
3;
3
;
2
5;
5
;
2
9;
9
;
2
15,
15
;
2
3) (x
5) = 0
45;
wolframalpha gives us the factored form
f (x) = (x + 1) (2x
x =
1;
3
; 3; 5:
2
All four rational zeros appear on the list!
3
3) (x
45
:
2
(#18) Find all rational zeros.
f (x) = 3x3
Numerator factors ( 9): 1; 3;
Denominator factors (3): 1; 3:
The possible rational zeros are
19x2 + 33x
9
9:
1
;
3
1;
3;
9:
Most students tried x = 3 …rst. Synthetic:
3 )
3
19
9
33
30
9
9
3
10
3
(0)
It works. The resulting quadratic is easily factored.
3x2
10x + 3 = (3x
1) (x
3) = 0
1
; 3:
3
x =
So x = 3 has MULTIPLICITY 2.
That means that the curve y = f (x) will only touch the x-axis at x = 3:
The graph must look like this:
y
20
15
By the Leading Coe¢ cient Test, we have ODD degree and
the coe¢ cient is positive. Thus, the end behaviors of this
polynomial run o¤ into Quadrants I & III.
10
5
0
-1
0
1
2
3
4
5
x
-5
-10
(VII) Complete Exercises #83 & #84 on p. 181.
(#83) Use Descarte’s Rule of Signs to determine the possible number of positive and negative
zeros of the function.
f (x) = 2x3 3x2 3:
Since this is a cubic, we know there are 3 complex zeros. At least one of them must be
real (because we can have one real and two nonreal complex conjugate zeros).
The number of changes in sign is 1. Thus, it must have exactly one positive zero.
Now look at f ( x) :
f ( x) = 2 ( x)3
3
=
2x
4
3 ( x)2
2
3x
3:
3
There are NO changes in sign. Thus, f has NO negative real zeros.
From the previous part, there can only be one real zero. We can verify this by graphing.
y
10
8
Since the graph changes directions twice, we’re reasonably
sure that we have one positive real zero and two nonreal
zeros.
6
4
2
0
-1
0
1
2
3
4
The real zero is approximately x = 1:91 and the nonreal
zeros are approximately x = 0:205 0:862i:
x
-2
-4
-6
-8
-10
(#84)
f (x) = 4x3
3x2 + 2x
1
There are 3 sign changes. Thus, there are either 3 or 1 positive real zeros.
Look at f ( x) :
f ( x) = 4 ( x)3
=
3
3 ( x)2 + 2 ( x)
2
4x
3x
2x
1
1:
There are NO sign changes, so f has NO negative real zeros.
Since f is a third-degree polynomial, it must have 3 complex zeros.
Here’s the graph.
y
10
It turns out that the graph never changes direction,
so there is only one positive real zero.
The other two zeros must be nonreal complex conjugates.
5
0
-2
-1
0
1
2
x
-5
-10
(VIII) Complete Exercise #104 on p. 181.
The box has two square ends and its length is y:
Its girth is the perimeter of the square, 4x:
The sum of the girth and length is 4x + y
120 inches.
x2 y:
The volume of the box is V =
We want the maximum volume, so it makes sense that
we will use all 120 inches of the girth and length combination.
(a) Thus, we have
4x + y = 120 ) y = 120
2
V (x) = x (120
The physical domain for x is x 2 (0; 30) :
5
4x
4x) = 4x2 (30
x) :
(b) We see from V (x) that V has the zero x = 0 with multiplicity 2 and x = 30 with
multiplicity 1.
Thus, the graph must look like this:
y 2e+4
1e+4
We must have a relative minimum point at x = 0 and
there must be a relative maximum point somewhere between
x = 0 and x = 30:
0
-20
-10
0
10
20
30
40
x
-1e+4
If we told you that the derivative was this (and you could set it to zero)
12x2 + 240x = 0
12x (x
20) = 0 ) x = 0; 20
then we can solve this and …nd that the relative maximum must occur at x = 20:
This means that the maximum volume is achieved when x = 20 and
y = 120 4 (20) = 40 inches.
The maximum volume is V (20) = 4 202 (30 20) = 16000 cubic inches.
(c) Find values of x which solve
V (x) = 4x2 (30
x) = 13500
This is a cubic. Divide through by 4.
4x3
120x2 + 13500 = 0
x3
30x2 + 3375 = 0:
: (x 15) x2 15x 225 It turns out that the factors of 3375 are
1; 3; 5; 9; 15; 25; 27; 45; 75; 125; 135; 375; 675; 1125; 3375:
The zero we’re looking for is near x = 20 and wolframalpha gives x = 15 as the rational
zero.
By synthetic, we have
(x 15) x2 15x 225 = 0:
The other two zeros are not rational.
q
( 15)
( 15)2
x =
2 (1)
p
15 15 5 :
=
= 24:27;
2
4 (1) ( 225)
=
15
p
1125
2
9:27:
So this gives us one more solution. We reject the negative value of x:
The acceptable values of x are 15 and 24.27. Both of these values of x will give us a
box volume of 13500 cubic inches.
6
Section 2.6
(IX) Complete Exercises #36 & #38 on p. 194.
1
(#36) f (x) =
2)2
(x
(a) Domain: x 6= 2:
(b) If we set f (x) = 0; we have NO solution, since the numerator can never equal zero.
There are NO x-axis intercepts.
When x = 0; we have f (0) = 1=4: This is the y-axis intercept.
(c) Vertical asymptote at x = 2:
Since the degree of the numerator is zero and the degree of the denominator is 2,
the x-axis, y = 0; is the horizontal asymptote on both ends.
(d) Plot.
y
10
8
6
4
2
0
-2
-1
0
1
2
3
4
5
6
x
-2
-4
-6
-8
-10
(#38) f (x) =
x2
2x
x2
8
=
9
(x + 2) (x
(x + 3) (x
4)
:
3)
(a) Domain: x 6= 3; 3:
(b) If we set the numerator equal to zero, we have x = 2; 4: These are the x-axis
intercepts.
When x = 0; we have f (0) = 8=9: This is the y-axis intercept.
(c) Vertical asymptotes at x = 3 and x = 3:
Since the degree of the numerator and denominator are equal, we look at the ratio
of their leading terms.
Thus, y = 1 is the horizontal asymptotes on both ends.
(d) Plot.
y
20
15
10
5
0
-5
-4
-3
-2
-1
0
1
2
3
4
5
x
-5
-10
-15
-20
7
(X) Complete Exercises #76 & #77 on pp. 195-196.
(#76) A 1000-liter tank contains 40 liters of a 25% brine solution.
Think of the pure brine solution as pure alcohol.
So this 40 liter mixture contains (25%) (50 liters) = 12:5 liters of pure brine.
We make a table.
Mixture
Concentration
Pure Brine
50
25%
12.5
x
75%
0.75x
x + 50
0:75x + 12:5
The …rst line gives the information about the original 50-liter mixture. We then add
x liters of 75% brine mixture.
The total volume is (x + 50) and the total amount of pure brine in that mixture is
(0:75x + 12:5) liters.
(a) Thus, the concentation is
0:75x + 12:5
:
C (x) =
x+5
This is a percentage. We can rewrite this in the same form as the text by multiplying
top and bottom by 4.
C (x) =
3x + 50
3x + 50
=
:
4 (x + 50)
4x + 200
(b) The physical domain requires us to use the fact that the maximum volume in the
tank is 1000 liters.
Thus, we must have x 950 liters.
It makes no sense to have x 0 in this problem, so the physical domain must be
x 2 (0; 950]:
(c) Graph. We don’t care about the vertical asymptote at x = 50 since x cannot be
negative.
y
1
0. 75
0. 5
0. 25
0
0
100
200
300
400
500
600
700
800
900
x
8
(d) As the tank is …lled, we see that the slope of the tangent line begins to decrease.
Thus, the concentration is stabilizing as x increases.
If we had in…nite capacity, then the curve would approach the horizontal asymptote
y = 3=4 = 75%:
This should make sense. The total volume begins to resemble the 75% solution as
the original mixture becomes less signi…cant as x increases.
When we get to x = 950 liters, however, the …nal concentration is
C (950) =
(#77) The area of the blue print area is (x
Thus, we can solve for y:
(x
4) (y
3 (950) + 50
= 72:5%:
4 (950 + 50)
4) (y
2) and this must be equal to 30 sq. inches.
2) = 30 ) y
y = 2+
The total page area is A = xy = x 2 +
30
x
4
2=
30
x
4
30
x
4
:
:
(a) Rewrite this as
A (x) = 2x +
=
2x (x 4) + 30x
2x2 + 22x
30x
=
=
x 4
x 4
x 4
2x (x + 11)
:
x 4
(b) Physical domain.
Clearly, we must have x > 4; else there is no print area.
In theory, we could make x gigantic, and then y would approach 2 inches and height
of the print area would become tiny.
So the physical domain for x is x > 4; (4; +1) :
(c) We want the lowest possible area for the page area. We’re only interested in values
of x > 4:
y
100
75
wolframalpha gives us an approximate value for x at the
local minimum point: x ) 11:75 inches.
50
The associated minimum page area is approximately 69 sq. inches
25
0
0
5
10
15
20
x
9