Mathematics 1375/D579, Fall 2015
Instructor: Suman Ganguli
Exam #2
Wednesday, November 11
Name:
Question:
1
2
3
4
5
6
Total
Points:
10
20
15
15
15
25
100
Score:
In order to receive full credit, you must show all your work and simplify your answers.
1. (10 points) Recall the Factor Theorem:
Factor Theorem: (x − c) is a factor of a polynomial f (x) if and only if f (c) = 0
Which of the following are factors of f (x) = x15 + x7 − x − 1? Justify your answers in terms of the Factor Theorem.
(a) x − 1
Solution: f (1) = 1 + 1 − 1 − 1 = 0 so x − 1 is a factor of f (x)
(b) x + 1
Solution: f (1) = −1 − 1 + 1 − 1 = −2 so x + 1 is not a factor of f (x)
2. (20 points) Divide by long division. In each case write your answer in the form “quotient + (remainder/divisor)”.
(a)
2x3 + x2 + 3x + 5
x−1
2x2 + 3x + 6
Solution:
x−1
2x3 + x2 + 3x + 5
− 2x3 + 2x2
3x2 + 3x
− 3x2 + 3x
6x + 5
− 6x + 6
11
3
Hence:
(b)
2
2x + x + 3x + 5
11
= 2x2 + 3x + 6 +
x−1
x−1
x3 + 27
x+3
x2 − 3x + 9
Solution:
x+3
3
x
− x3 − 3x2
+ 27
− 3x2
3x2 + 9x
9x + 27
− 9x − 27
0
3
Hence:
x + 27
= x2 − 3x + 9
x+3
Math 1375/D579, Fall 2015
Exam #2
Page 2 of 5
3. (15 points) Consider the polynomial p(x) = x3 + x2 − 3x − 3.
(a) Find all roots of the polynomials algebraically in radical form. (Hint: Start by using a graph of p(x) to identify an
integer root, and use the root to factor the polynomial via long division.)
Solution: x = −1 is a root: p(−1) = −1 + 1 + 3 − 3 = 0
x2
x+1
3
−3
2
x + x − 3x − 3
− x3 − x2
− 3x − 3
3x + 3
0
√
So x3 + 2x2 − 10x − 20 = (x + 1)(x2 − 3), and hence the roots of p(x) are x = −1, x = ± 3
(b) Sketch a complete graph of p(x). Indicate the x- and y-intercepts and relative maxima and minima on the graph.
Math 1375/D579, Fall 2015
4. (15 points) Consider the rational function f (x) =
Exam #2
Page 3 of 5
−6
.
x+2
(a) What is the domain of f (x)?
Solution: x + 2 = 0 for x = −2. Hence the domain of f is R − {−2} = (−∞, −2) ∪ (−2, ∞)
(b) Algebraically solve for the the x− and y−intercepts of the graph of f (x).
Solution:
There are no x−intercepts since the numerator of f (x) is a non-zero constant (and hence f (x) 6= 0 for all x).
−6
The y−intercept occurs at f (0) =
= −3, i.e., at the point (0, −3).
2
(c) Algebraically find the vertical and horizontal asymptotes.
Solution: The vertical asymptote occurs when x + 2 = 0, i.e., the vertical line x = −2
Since this rational function consists of a constant over a linear factor (i.e., degree of the numerator > degree of
the denominator), the horizontal asymptote is y = 0.
(d) Sketch a complete graph of the function. Indicate and label the x- and y-intercepts, and the vertical and horizontal
asymptotes.
Math 1375/D579, Fall 2015
5. (15 points) Consider the rational function f (x) =
Exam #2
Page 4 of 5
6x − 18
.
2x − 3
(a) What is the domain of f (x)?
Solution: 2x − 3 = 0 for x = 3/2. Hence the domain of f is R − {3/2} = (−∞, 3/2) ∪ (3/2, ∞)
(b) Algebraically solve for the the x− and y−intercepts of the graph of f (x).
Solution:
The x−intercept occurs when 6x − 18 = 0, hence at x = 3, i.e., at the point (3, 0)
−18
The y−intercept occurs at f (0) =
= 6, i.e., at the point (0, 6)
−3
(c) Algebraically find the vertical and horizontal asymptotes.
Solution: The vertical asymptote occurs when 2x − 3 = 0, i.e., the vertical line x = 3/2.
6
To find the horizontal asymptote look at the ratio of the leading coefficients: the horizontal asymptote is y = ,
2
i.e., the horizontal line y = 3
(d) Sketch a complete graph of the function. Indicate and label on your graph the x- and y-intercepts, and the vertical
and horizontal asymptotes.
Math 1375/D579, Fall 2015
Exam #2
Page 5 of 5
6. (25 points) Solve the inequalities. In each case express your solution as an interval.
6x − 18
6x − 18
(a)
≤ 0 (Hint: use the graph of f (x) =
you sketched in the previous exercise.)
2x − 3
2x − 3
Solution:
We see that the graph of f (x) =
6x − 18
is at or below the x-axis for x-values in the interval
2x − 3
3
,3
2
(b) x2 − 6x + 5 > 0
Solution:
The graph of f (x) = x2 − 6x + 5 = (x − 5)(x − 1) is an parabola opening upwards with x-intercepts at x = 5
and x = 1. Hence solution set of the inequality is the interval
(−∞, 1) ∪ (5, ∞)
(c) 7x − x2 ≥ 0
Solution:
The graph of f (x) = 7x − x2 = x(7 − x) is an parabola opening downwards with x-intercepts at x = 0 and
x = 7. Hence solution set of the inequality is the interval
[0, 7]