Math 115 Spring 2014
Written Homework 6-SOLUTIONS
Due Friday, March 14
Instructions: Write complete solutions on separate paper (not spiral bound). If
multiple pieces of paper are used, THEY MUST BE STAPLED with your name and
lecture written on each page. Please review the Course Information document for
more complete instructions.
1. For each function, determine whether the function is a polynomial function or not. If 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) := πx5 + 17x2 − 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 integers greater than or equal to 0. The function in (a) is a
polynomial. The leading coefficient is π, the leading term is πx5 , the degree is 5, and the constant
term is −3.175.
(b) f (x) := 7x−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)( 21 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.
(e) f (x) := (π)4
Solution: This is a polynomial - it could be written as f (x) := (π)4 x0 . The leading term, leading
coefficient, and constant term are all (π)4 and the degree is 0.
2. Determine the limit as x → −∞ and the limit as x → ∞ for each function.
(a) f (x) := −2x3 + 5x2 − 3x + 11
Solution:
lim −2x3
h
i
= −2 lim x3
lim f (x) =
x→∞
x→∞
x→∞
= −∞
lim −2x3
3
= −2 lim x
lim f (x) =
x→−∞
x→−∞
x→−∞
= ∞ since (negative number)3 < 0
(b) h(x) := x6 − 2x3 + x2 − 4x − 12
Solution:
lim h(x) = lim x6
x→∞
= ∞
x→∞
lim h(x) =
x→−∞
lim x6
x→−∞
= ∞ since (negative number)6 > 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→∞
= ∞
lim g(x) =
x→∞
x→∞
lim 6x5
5
= 6 lim x
lim g(x) =
x→−∞
x→−∞
x→−∞
= −∞ since (negative 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
= −12π lim x6
lim p(x) =
x→∞
x→∞
= −∞ since the coefficient − 12π is negative
−12πx6
x→−∞
6
= −12π lim x
lim p(x) =
x→−∞
lim
x→−∞
= −∞ since − 12π < 0 and (negative number)6 > 0
3. For each function below (i) determine the long run behavior as x → −∞ and x → ∞; (ii)
determine the intercepts of the graph of the function; (iii) draw a rough sketch of the graph of the
function.
(a) g(x) := (x − 1)(2x − 3)(3 − x)
Solution:
1. long-run behavior
Need to examine the limits to infinity. Recall that long-run behavior of a polynomial is
dictated by the leading term. Here,
lim g(x) = lim (x)(2x)(−x) = lim −2x3 = −∞.
x→∞
x→∞
x→∞
Since g(x) is an odd degree polynomial (degree is 3), lim g(x) is the opposite of lim g(x)
x→−∞
x→∞
so lim g(x) = ∞
x→−∞
2. intercepts
(a) y-intercept
Since g(0) = (−1)(−3)(3) = 9. The graph of g(x) intersects the y-axis at (0, 9).
(b) x-intercept(s)
Fortunately, g(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
3
x = 1, x = , and x = 3
2
These correspond to the x-intercepts,
3
, 0 , and (3, 0).
(1, 0),
2
3. graph
We use continuity to connect the end behavior and the intercepts. We need to test a point in
the interval (1, 32 ) and a point in the interval ( 23 , 3). First choose x = 54 : g( 54 ) = ( 54 − 1)(2( 54 ) −
3)(3 − 54 )) < 0. Now choose x = 2: g(2) = (2 − 1)(2(2) − 3)(3 − 2)) > 0. Putting all of this
information together should give a rough sketch similar to this:
(b) f (x) := (x − 5)(x2 − 1)(11 − 3x)
Solution:
1. long-run behavior
Need to examine the limits to infinity. Recall that long-run behavior of a polynomial is
dictated by the leading term. Here,
lim f (x) = lim (x)(x2 )(−3x) = lim −3x4 = −∞.
x→∞
x→∞
x→∞
Since f (x) is an even degree polynomial (degree is 4),
lim f (x) = lim f (x) = −∞.
x→−∞
x→∞
2. intercepts
(a) y-intercept
Since f (0) = (−5)(−1)(11) = 55. The graph of f (x) intersects the y-axis at (0, 55).
(b) x-intercept(s)
Fortunately, f (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
x = 5, x = ±1, and x =
11
3
These correspond to the x-intercepts,
(5, 0), (1, 0), (−1, 0) and
11
,0 .
3
3. graph
We use continuity to connect the end behavior and the intercepts. We need to test points in
the interval (1, 11
). I choose x = 2: f (2) = (2 − 5)(22 − 1)(11 − 3(2)) < 0. We also need to
3
, 5). I choose x = 4: f (4) = (4 − 5)(42 − 1)(11 − 3(4)) > 0.
test a point in the interval ( 11
3
Putting all of this information together should give a rough sketch similar to this:
4. Let f (x) := 3x3 + x2 − 38x + 24.
(a) Determine all of the possible rational zeros of f (x) (according to the Rational Root Theorem).
Solution: For a rational number pq 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 (i.e. show
that f (−4) = 0). 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) = 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 (b). (You may use either polynomial long division or synthetic division to do this.)
Solution:
3x2 − 11x + 6
x+4
3x3 + x2 − 38x + 24
− 3x3 − 12x2
− 11x2 − 38x
11x2 + 44x
6x + 24
− 6x − 24
0
3
2
Hence, f (x) := 3x + x − 38x + 24 = (x + 4)(3x2 − 11x + 6).
(d) Use what you found in parts (a) - (c) to completely factor the polynomial function f (x) :=
3x3 + x2 − 38x + 24.
Solution: Continue the factoring by factoring the quadratic term;
f (x) := (x + 4)(3x − 2)(x − 3)
5. Determine the zeros of the polynomial function g(x) := x4 + 4x3 − 10x2 − 28x − 15 and write
it in fully 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 pq 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) = 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
4
3
2
Hence, g(x) := x + 4x − 10x − 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 g(x) (and of x3 + 3x2 − 13x − 15).
x2 + 6x + 5
x3 + 3x2 − 13x − 15
− x3 + 3x2
6x2 − 13x
− 6x2 + 18x
5x − 15
− 5x + 15
0
4
3
2
Hence, g(x) := x + 4x − 10x − 28x − 15 = (x + 1)(x − 3)(x2 + 6x + 5).
x−3
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 long-run behavior, the intercepts, and continuity to draw a rough sketch of the graph
of each polynomial function. (Note: you may use information you found in previous problems as
part of these solutions.)
(a) 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
4; 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 ( 27 , 3). I choose x = 2: f (2) = (2 + 4)(3(2) − 2)(2 − 3) < 0. Putting all of this
information together gives a rough sketch similar to this:
(b) 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
5; 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 a rough sketch similar to this: