Math 147 - Assignment 6 Solutions - Fall 2012 - BSU - Jaimos F Skriletz
1. Complete Factorization
1
(a) P (x) = x5 − 5x3 − 36x = x(x4 − 5x2 − 36) = x((x2 )2 − 5(x2 ) − 36) = x(x2 − 9)(x2 + 4) = x(x + 3)(x − 3)(x2 + 4)
Thus the complete factorization of P (x) over the reals is
x(x + 3)(x − 3)(x2 + 4)
√
The two non-real zeros of x2 + 4 = 0 or x2 = −4 is x = ± −4 = ±2i.
Thus the complete factorization of P (x) over the complex numbers is
x(x + 3)(x − 3)(x + 2i)(x − 2i)
And the five zeros of P (x) are {0, −3, 3, −2i, 2i}
(b) P (x) = x5 + x3 + 8x2 + 8 = x3 (x2 + 1) + 8(x2 + 1) = (x3 + 8)(x2 + 1) = (x + 2)(x2 − 2x + 4)(x2 + 1)
Note I used the sum of cubes formula to get the complete factorization over the reals,
(x + 2)(x2 − 2x + 4)(x2 + 1)
The non-real zeros of x2 + 1 = 0 are x2 = −1 or x = ±i. While the non-real zeros of x2 − 2x + 4 = 0 can be found
by the quadratic formula
p
√
√
√
√
2 ± 4 − 4(1)(4)
2 ± −12
2 ± 2i 3
−b ± b2 − 4ac
=
=
=
=1±i 3
x=
2a
2(1)
2
2
Thus the complete factorization over the complex numbers is
√
√
(x + 2)(x − (1 + i 3))(x − (1 − i 3))(x + i)(x − i)
√
√
And the five complex zeros are {−2, 1 + i 3, 1 − i 3, i, −i}
2. Complex Roots
There are three complex cube roots, only one of which is real. Find all three complex cube roots of 125.
If x3 = 125 then x3 − 125 = 0
By the difference of cubes we get that x3 − 125 = (x − 5)(x2 + 5x + 25)
Thus we see that x = 5 is the real cube root. The two non-real roots come from use the quadratic formula to solve
x2 + 5x + 25 = 0 so
p
√
√
√
−5 ± 25 − 4(1)(25)
−b ± b2 − 4ac
−5 ± −75
−5 ± 5i 3
x=
=
=
=
2a
2(1)
2
2
Thus the tree complex cube roots of 125 are
√
√
−5 + 5i 3 −5 − 5i 3
{5,
,
}
2
2
2
Math 147 - Assignment 6 Solutions - Fall 2012 - BSU - Jaimos F Skriletz
3. Rational Functions
2(x + 4)(x − 4)
2x2 − 32
=
Consider the rational function R(x) = 2
x +x−6
(x + 3)(x − 2)
(a) The domain of R(x) is all reals such that x2 + x − 6 = (x + 3)(x − 2) 6= 0. Thus the domain is
{x | x 6= −3 or x 6= 2} = (−∞, −3) ∪ (−3, 2) ∪ (2, ∞)
(b) The two vertical asymptotes occur at the division by zeros found in part (a). Thus the two vertical asymptotes
are
x = −3
and
x=2
(c) R(x) = 0 if 2x2 − 32 = 2(x + 4)(x − 4) = 0. Thus the two zeros are x = −4 and x = 4. So the two x-intercepts are
(−4, 0)
and
(4, 0)
(d) The sign table for R(x) is
2(x + 4)(x − 4)
(x + 3)(x − 2)
−4
(−)(−)
(−)(−)
= (+)
−3
(+)(−)
(−)(−)
= (−)
2
(+)(−)
(+)(−)
= (+)
4
(+)(−)
(+)(+)
= (−)
(+)(+)
(+)(+)
= (+)
(e) Since the degree of the numerator (top) equals the degree of the denominator (bottom) the horizontal asymptote
is
an
2
y=
= =2
bn
1
Thus as x → ±∞ then y = R(x) → 2.
(f) The y-intercept is (0, R(0)) = (0, (−32)/(−6)) = (0, 16/3)
(g) The graph of R(x) is
Note the local minimum (optional) is at the about (−0.835, 4.986)
3
Math 147 - Assignment 6 Solutions - Fall 2012 - BSU - Jaimos F Skriletz
4. Slant Asymptote
Consider the rational function R(x) =
x3 + 8
2x2 + x − 1
The domain of R(x) is all reals such that
2x2 + x − 1 = (x + 1)(2x − 1) 6= 0
So it is all reals except −1 and 1/2. Thus the domain is
(−∞, −1) ∪ (−1, 1/2) ∪ (1/2, ∞)
So the two vertical asymptotes are x = −1 and x = 1/2.
The zeros are found by solving x3 + 8 = 0 or x3 = −8. There is only one real cube root of -8, so x = −2.
The slant asymptote is found by polynomial division
1
2x
2x + x − 1
2
−
1
4
3
x
+8
− x3 − 21 x2 + 12 x
− 12 x2 + 12 x + 8
1
1
1 2
2x + 4x − 4
3
4x
+
31
4
+
3
31
4x + 4
2x2 + x −
Thus this gives that
x3 + 8
=
R(x) = 2
2x + x − 1
This shows us that the end behavior is like y =
The graph of R(x) is
1
1
x−
2
4
1
1
1
x − which is the slant asymptote.
2
4
Math 147 - Assignment 6 Solutions - Fall 2012 - BSU - Jaimos F Skriletz
5. Exponential Functions
4
Find an exponential function of the form f (x) = C · ax that passes though the two points (0, 3) and (3, 375).
Passing though (0, 3) means that f (0) = C · a0 = C = 3.
Thus the function is of the form f (x) = 3ax .
Passing though (3, 375) means that
f (3) = 3a3 = 375
a3 = 125
√
3
a = 125 = 5
Thus the function is f (x) = 3 · 5x .
This function is a vertical scale by a factor of 3 of the exponential function p(x) = 5x . The graph of this function is
6. Compound Interest vs Compound Continuous Interest
Suppose that $6500 is invested at 5.2% interest for 10 years.
(a) The future value of the investment is
10n
r nt
0.052
A=P 1+
= 6500 1 +
n
n
(b) Use part (a) to create the following table
Compounded
Annually
Semiannually
Quarterly
Monthly
Weekly
Daily (bank year)
Daily (real year)
Hourly
n
1
2
4
12
52
360
365
8760
A
$10791.23
$10860.77
$10896.60
$10920.90
$10930.34
$10932.75
$10932.78
$10933.09
(c) If the interest is compounded continuously the future value is
A = P ert = 6500e0.052(10) ≈ $10933.18
(d) As the number of times you compound increases (n → ∞) the value approaches compound continuous interest
(the cap you can get from increasing the times you compound). As you see the difference between compound daily
and compound continuously is a bout 20 cents.