Download Homework 8 Solutions

Math 115 Spring 2011
Written Homework 8
Due Friday, March 18
1. For each function, determine whether the function is a polynomial function or not. It it
is not a polynomial function, explain why not. If it is a polynomial function, determine the
leading coefficient, the leading term, the degree, and the constant term.
√
(a) f (x) := πx3 + 17x − 3.175
Solution: The definition of a polynomial is that the coefficients are real numbers and the
powers on the variable terms xn are always positive integers (n ∈ N). The function in (a) is
a polynomial. The leading coefficient is π, the leading term is πx3 , the degree is 3, and the
constant term is −3.175.
(b) f (x) := x−3 + x2 + 1
Solution: This is not a polynomial because of the x−3 term. The power is not a positive
integer.
(c) f (x) := 3xπ + 7
Solution: This is not a polynomial because of the xπ term. The power is not an integer.
(d) f (x) := (x + 3)(2x − 5)(2x2 − 1)( 12 x + 5)
Solution: This is a polynomial. The leading term is (x)(2x)(2x2 )( 21 x) = 2x5 . Thus the
leading coefficient is 2 and the degree is 5. The constant term is (3)(−5)(−1)(5) = 75.
2. Calculate lim f (x) and lim f (x) for each function.
x→−∞
3
x→∞
2
(a) f (x) := 2x − 11x + 4x + 5
Solution:
lim 2x3
h
i
= 2 lim x3
x→∞
h
i3
= 2 lim x
lim f (x) =
x→∞
x→∞
x→∞
= ∞
lim 2x3
3
= 2 lim x
x→−∞
3
= 2 lim x
lim f (x) =
x→−∞
x→−∞
x→−∞
= −∞ since (−‘ve number)3 < 0
(b) h(x) := x4 − 4x3 + 3x2 + 4x − 4
Solution:
lim x4
h
i4
=
lim x
lim h(x) =
x→∞
x→∞
x→∞
= ∞
lim x4
4
=
lim x
lim h(x) =
x→−∞
x→−∞
x→−∞
= ∞ since (−‘ve number)4 > 0
(c) g(x) := −3x4 + 5x3 + 6x5 + x2 + 3x − 9
Solution:
lim 6x5 since this is the highest degree term
h
i
= 6 lim x5
x→∞
h
i5
= 6 lim x
lim g(x) =
x→∞
x→∞
x→∞
= ∞
lim 6x5
5
= 6 lim x
x→−∞
5
= 6 lim x
lim g(x) =
x→−∞
x→−∞
x→−∞
= −∞ since (−‘ve number)5 < 0
(d) p(x) := (πx2 − 4)(3x + 2)(4x2 − 5)(2 − x)
Solution: Here, we need the leading term for p(x). Note that the highest degree term for
the polynomial is
(πx2 )(3x)(4x2 )(−x) = −12πx6 .
lim −12πx6
x→∞
h
i
6
= −12π lim x
x→∞
h
i6
= −12π lim x
lim p(x) =
x→∞
x→∞
= −∞ since the coefficient − 12π is negative
−12πx6
x→−∞
6
= −12π lim x
x→−∞
6
= −12π lim x
lim p(x) =
x→−∞
lim
x→−∞
= −∞ since − 12π < 0 and (−‘ve number)6 > 0
Theorem: The Rational Root Theorem
p
Let be a rational number written in fully reduced form. Consider the polynomial equation
q
cn xn + cn−1 xn−1 + . . . + c1 x + c0 = 0
p
is a zero (or
q
a solution) of the equation then p must divide the constant coefficient c0 and q must divide
where all of the coefficients ci are integers and cn 6= 0. If the rational number
the leading coefficient cn .
Example: Let f (x) := 6x4 + 7x3 − 12x2 − 3x + 2. For a rational number
p
q
to be a zero, p
must be a factor of c0 = 2 and q must be a factor of c4 = 6. Thus, p can be ±1 or ±2, and
q can be ±1, ±2, ±3, or ±6. The possible rational zeros, pq , are
1 1 1 2
±1, ±2, ± , ± , ± , ±
2 3 6 3
3. Find all the rational roots of f (x) := 6x4 + 7x3 − 12x2 − 3x + 2.
Solution: This is just plug-and-chug. Evaluate f at each of the possible rational roots.
f (1) = 6(1)4 + 7(1)3 − 12(1)2 − 3(1) + 2
= 6 + 7 − 12 − 3 + 2
= 0 hence x = 1 is a root of f
f (−1) = 6(−1)4 + 7(−1)3 − 12(−1)2 − 3(−1) + 2
= 6 − 7 − 12 + 3 + 2
= −8 hence x = −1 is not a root of f
f (2) = 6(2)4 + 7(2)3 − 12(2)2 − 3(2) + 2
= 100 hence x = 2 is not a root of f
f (−2) = 6(−2)4 + 7(−2)3 − 12(−2)2 − 3(−2) + 2
= 0 hence x = −2 is a root of f
f (1/2) 6= 0 hence x = 1/2 is not a root of f
f (−1/2) = 0 hence x = −1/2 is a root of f
f (1/3) = 0 hence x = 1/3 is a root of f
f (−1/3) 6= 0 hence x = −1/3 is not a root of f
f (1/6) 6= 0 hence x = 1/6 is not a root of f
f (−1/6) 6= 0 hence x = −1/6 is not a root of f
f (2/3) 6= 0 hence x = 2/3 is not a root of f
f (−2/3) 6= 0 hence x = −2/3 is not a root of f
4. (Some Algebra Review ) Let f (x) := 3x3 + x2 − 38x + 24.
(a) For a rational number
p
q
to be a zero, p must be a factor of c0 = 24 and q must be a
factor of c3 = 3. Thus, p can be ±1, ±2, ±3, ±4, ±6, ±8, ±12 or ±24, and q can be ±1 or
±3. The possible rational zeros, pq , are
1 2 3 4 6 8 12 24
±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24, ± , ± , ± , ± , ± , ± , ± , ± .
3 3 3 3 3 3
3
3
Removing the rational numbers that repeat when simplified, we get
1 2 4 8
±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24, ± , ± , ± , ± .
3 3 3 3
(b) Show that x = −4 is a zero (or root) of the function f (x) := 3x3 + x2 − 38x + 24.
What does this tell you about a factor of this polynomial function?
Solution: To show x = −4 is a zero, we just need to show that f (−4) = 0.
f (−4) = 3(−4)3 + (−4)2 − 38(−4) + 24
= 3(−64) + 16 + 152 + 24
= −192 + 192
= 0
The linear term x − (−4) is a factor of the polynomial function.
(c) Determine the quotient and remainder when 3x3 + x2 − 38x + 24 is divided by the factor
you found in part (a).
3x2 − 11x + 6
Solution:
x+4
3x3
+ x2 − 38x + 24
− 3x3 − 12x2
− 11x2 − 38x
11x2 + 44x
6x + 24
− 6x − 24
0
Hence, f (x) := 3x3 + x2 − 38x + 24 = (x + 4)(3x2 − 11x + 6).
(d) Use what you found in parts (b) and (c) to completely factor the polynomial function
f (x) := 3x3 + x2 − 38x + 24.
Solution: Continuing the factoring by factoring the quadratic term;
f (x) := (x + 4)(3x − 2)(x − 3)
5. (More Algebra Review ) Determine the zeros of the polynomial function g(x) := x4 +
4x3 − 10x2 − 28x − 15 and write it in factored form. (Hint: Use the Rational Root Theorem
to determine all possible rational roots and then use the division technique of problem 4 to
fully factor g(x).)
Solution: For a rational number
p
q
to be a zero, p must be a factor of c0 = 15 and q must be
a factor of c4 = 1. Thus, p can be ±1, ±3, ±5 or ±15, and q can be ±1 only. The possible
rational zeros, pq , are
±1, ±3, ±5, ±15.
Plugging these potential roots into g(x), the first one we see is that x = −1 is a root.
Hence, x − (−1) is a factor of g.
x3 + 3x2 − 13x − 15
x+1
x4 + 4x3 − 10x2 − 28x − 15
− x4 − x3
3x3 − 10x2
− 3x3 − 3x2
− 13x2 − 28x
13x2 + 13x
− 15x − 15
15x + 15
0
Hence, g(x) := x4 + 4x3 − 10x2 − 28x − 15 = (x + 1)(x3 + 3x2 − 13x − 15).
Continuing our use of the potential rational roots, we see that x = 3 is a root (since
32 + 3(32 ) − 13(3) − 15 = 0). Hence x − 3 is a factor of both g(x) (and of x3 + 3x2 − 13x − 15).
x2 + 6x + 5
x−3
x3 + 3x2 − 13x − 15
− x3 + 3x2
6x2 − 13x
− 6x2 + 18x
5x − 15
− 5x + 15
0
Hence, g(x) := x4 + 4x3 − 10x2 − 28x − 15 = (x + 1)(x − 3)(x2 + 6x + 5).
We can continue using our list of potential rational roots, but the quadratic term is easy
to factor.
g(x) := x4 + 4x3 − 10x2 − 28x − 15 = (x + 1)(x − 3)(x + 5)(x + 1)
The roots/zeros of g(x) are x = −5, x = −1, and x = 3.
6. Use the end behavior, the intercepts, and continuity to draw a rough sketch of each
polynomial function. (Note: you may use information you found in previous problems as
part of these solutions.)
(a) h(x) := (x − 6)(x2 − 3)(7 − 2x)
Solution:
1. end behavior
Need to examine the limits to infinity. Recall that long-run behavior of a polynomial
is dictated by the leading term. Here,
lim h(x) = lim (x)(x2 )(−2x) = lim −2x4 = −∞.
x→∞
x→∞
x→∞
Since h(x) is an even degree polynomial (degree is 4),
lim h(x) = lim h(x) = −∞.
x→−∞
x→∞
2. intercepts
(a) y-intercept
Since h(0) = (−6)(−3)(7) = 126. The graph of h(x) intersects the y-axis at
(0, 126).
(b) x-intercept(s)
Fortunately, h(x) is already factored. For the x-intercepts, we need to identify all
of the roots of the polynomial. Here, we see that the roots are
√
7
x = 6, x = ± 3, and x =
2
These correspond to the x-intercepts,
√
√
(6, 0), ( 3, 0), (− 3, 0) and
7
,0 .
2
3. graph
We use continuity to connect the end behavior and the intercepts. We need to test a
√
point in the interval ( 3, 72 ). I choose x = 2: h(2) = (2 − 6)(22 − 3)(7 − 2(2)) < 0.
We also need to test a point in the interval ( 72 , 6). I choose x = 5: h(5) = (5 − 6)(52 −
3)(7 − 2(5)) > 0. Putting all of this information together gives the following rough
sketch:
(b) f (x) := 3x3 + x2 − 38x + 24
Solution:
1. end behavior
Need to examine the limits to infinity.
lim f (x) = lim 3x3 = ∞.
x→∞
x→∞
Since f (x) is an odd degree polynomial (degree is 3),
lim f (x) = − lim f (x) = −∞.
x→−∞
x→∞
2. intercepts
(a) y-intercept
Since f (0) = 24. The graph of f (x) intersects the y-axis at (0, 24).
(b) x-intercept(s)
To determine the roots, we need to factor f (x). Fortunately, we factored f (x) in
problem 1; f (x) := (x + 4)(3x − 2)(x − 3). Here, we see that the roots are
x = −4, x = 3, and x =
2
3
These correspond to the x-intercepts,
(−4, 0), (3, 0), and
2
,0 .
3
3. graph
We use continuity to connect the end behavior and the intercepts. We need to test
a point in the interval ( 72 , 3). I choose x = 2: f (2) = (2 + 4)(3(2) − 2)(2 − 3) < 0.
Putting all of this information together gives the following rough sketch:
(c) g(x) := x4 + 4x3 − 10x2 − 28x − 15
Solution:
1. end behavior
Need to examine the limits to infinity.
lim g(x) = lim x4 = ∞.
x→∞
x→∞
Since g(x) is an even degree polynomial (degree is 4),
lim g(x) = lim g(x) = ∞.
x→−∞
x→∞
2. intercepts
(a) y-intercept
Since g(0) = −15. The graph of g(x) intersects the y-axis at (0, −15).
(b) x-intercept(s)
To determine the roots, we need to factor g(x). Fortunately, we factored g(x) in
problem 2; g(x) := (x + 1)2 (x − 3)(x + 5). Here, we see that the roots are
x = −1, x = 3, and x = −5
These correspond to the x-intercepts,
(−1, 0), (3, 0), and (−5, 0).
3. graph
We use continuity to connect the end behavior and the intercepts. We need to test a
point in the interval (−5, −1). I choose x = −3: g(−3) = (−3+1)2 (−3−3)(−3+5) < 0.
Putting all of this information together gives the following rough sketch:
7. In class we used a proof by contradiction to show that
√
2 was an irrational number.
Armed with the Rational Root Theorem, we have now have an alternative proof.
√
(a) Consider the function f (x) := x2 − 2. Algebraically show that ± 2 are the roots of
this polynomial function.
Solution:
f (x) = 0
x2 − 2 = 0
x2 = 2
√
x = ± 2
(b) Now, use the Rational Root Theorem to determine all of the the possible rational
roots of f .
Solution: For a rational number
p
q
to be a zero of f , p must be a factor of c0 = 2 and q
must be a factor of c2 = 1. Thus, p can be ±1 or ±2, and q can be ±1 only. The possible
rational zeros, pq , are
±1 or ± 2.
(c) Explain why your answers in (b) prove that
√
2 must be an irrational number.
√
2 is a root of f but is not contained in the list of possible rational
√
roots (±1 or ±2), x = 2 can not be a rational number. Hence, by the definition of irrational
√
numbers, 2 is an irrational number.
Solution: Since x =