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Math 111–Exam 2 Solutions
- (3pts each) True or False:
(a) A polynomial function of degree 7 can have 6 real roots. True, for a polynomial function of degree 7 has at most 7 real roots, and 6 < 7. (b) If x + 1 is a factor of a polynomial function f (x), then x − 1 is also a factor of f (x). False, for example the polynomial functions with rules f (x) = x^2 + x, f (x) = (x + 1)(x − 2)^2 , and f (x) = x + 1 all have x + 1 as a factor but do not have x − 1 as a factor. (There are many more counterexamples). (c) x + 1 is a factor of x^4 − 15 x − 16. True, for (−1)^4 − 15(−1) − 16 = 0 so −1 is a root of x^4 − 15 x − 16. Thus x − (−1) = x + 1 is a factor of x^4 − 15 x − 16. (d) The range of some quadratic functions is (−∞, ∞). False, for all quadratic functions have either a minimum or a maximum value (the y-coordinate of the vertex). If the function has a minimum value, then no real number less than that value is in the range. Similarly if the function has a maximum value, then no real number greater than that value is in the range.
- (10pts) For what values of x is |x^2 − 1 | ≤ 8? Give your answer in interval notation.
Solution: First we consider the equation |x^2 − 1 | = 8. To solve for x we must solve the two equations x^2 − 1 = 8 and x^2 − 1 = − 8. Solving for x in the first equation we see
x^2 − 1 = 8 ⇒ x^2 = 9 ⇒ x = ± 3.
Solving for x in the second equation we see
x^2 − 1 = − 8 ⇒ x^2 = − 7 ⇒ x = ±
which is not a real number, thus there are no real solutions to the second equation. Now we must check one point in each of the intervals (−∞, −3), (− 3 , 3), and (3, ∞) in the original inequality. I will check − 4 , 0, and 4.
-4^0
Since
|(−4)^2 − 1 | = 15 6 ≤ 8 and |(0)^2 − 1 | = 1 ≤ 8 and |(4)^2 − 1 | = 15 6 ≤ 8
we see the solution is
or in interval notation [− 3 , 3].
- (10pts) Suppose that I invest $400 at 13% compounded quarterly and I do not withdraw any money. How much money will I have after 3 years? (You do NOT need to simplify your answer)
Solution: The rule f (t) = 400
)t
gives the amount of money in my account after t quarters. Since there are 12 quarters in 3 years we see that after 3 years I will have
f (12) = 400
dollars in my account.
- (10pts) Recall that the area A of a rectangle is given by A = lw where l is the length and w is the width. If a rectangle has the property that the sum of its length and 3 times its width is 12, what is the maximal area of the rectangle? (You must show work to justify your answer)
Solution: Since l + 3w = 12 we know l = 12 − 3 w, so we have
A = lw = (12 − 3 w)w = − 3 w^2 + 12w.
To find the maximal area, we must find the vertex of the parabola which is the graph of the equation A = − 3 w^2 + 12w. Notice
A = − 3 w^2 + 12w
A
= w^2 − 4 w
A
= w^2 − 4 w + 4
A
= (w − 2)^2
A
= (w − 2)^2 − 4
⇒ A = −3(w − 2)^2 + 12. Thus the vertex is (2, 12), which tells us the maximum area occurs when the width is 2, and the maximal area is 12.
- (3pts each) Let f be the rational function whose rule is given by
f (x) =
4(x + 2)(x − 3)^3 (x + 1)^2 8(x + 2)^2 (x − 3)^2 (x + 1)^3
Notice: f (x) =
4(x + 2)(x − 3)^3 (x + 1)^2 8(x + 2)^2 (x − 3)^2 (x + 1)^3
(x − 3) 2(x + 2)(x + 1)
(a) State the domain of f. Solution: Since we cannot divide by zero, the only real numbers which are not in the domain are the roots of the denominator, namely
x 6 = − 2 , x 6 = 3, x 6 = − 1.
(b) State the horizontal asymptote of the graph of f. Solution: Since the degree of the numerator is 6 and the degree of the denomi- nator is 7, we know the horizontal asymptote is
y = 0 (or the x-axis)
(c) Find all vertical asymptotes of the graph of f (if any exist). Solution: Since the only factors remaining in the denominator after simplifying are x + 2 and x + 1 we know the vertical asymptotes are
x = − 2 and x = − 1.
(d) Find all x-values of holes in the graph of f (if any exist). Solution: Since the factor x − 3 appears in the denominator of the rule, but does not appear in the denominator after simplifying the rule, we know there is a hole when x = 3. (e) Give the (x, y)-coordinates of all the holes of the the graph of f (if the graph has any holes). Solution: To find the y-value of the hole corresponding to x = 3, we evaluate the simplified rule at x = 3 and we get (^) 2(3+2)(3+1)(3−3) = 0 so that the hole in the graph of f is at the point (3, 0). (f) Find the y-intercept of the graph of f (if one exists). Solution: To find the y-intercept we set x = 0 (which is ok since 0 is in the domain of f ). Thus the y-intercept is
f (0) =
(g) Find the x-intercept(s) of the graph of f (if any exist). Solution: To find the y-intercept we set f (x) = 0. In doing so we only need to find the roots of the numerator. But none of the roots of the numerator are in the domain, so there are no x-intercepts.
- (3pts each)
y
x
Above is a complete graph of the polynomial function f. Circle the correct answer.
(a) The degree of f (x) is i. even ii. odd iii. neither even nor odd Answer: ii, because of the end behavior of the graph. (b) The leading coefficient of f (x) is i. positive ii. negative iii. zero Answer: i, because of the end behavior of the graph. (c) The degree of f (x) COULD be i. 4 ii. 5 iii. 8 iv. 11 v. none of the above Answer: iv, for the degree must be at least 7 and odd. (d) How many roots of odd multiplicity does f have? i. 0 ii. 1 iii. 2 iv. 3 v. 4 vi. none of the above Answer: iv, the only root of even multiplicity is root at x = 3, the other three are of odd multiplicity.
