Download Past Midterm Exam 2 with Answers - College Algebra | MATH 111 and more Exams Algebra in PDF only on Docsity!
Midterm 2 for Math 111 (CRN 13936)
Note: Show your steps! No use of calculators!
Name
- (10 points) f (x) =
3 x + 2, g(x) = x^2 + 1, find the inverse function of f (x). Denote
this inverse function as h(x), what’s the domain of (
g h )(x)?
Solution: To find the inverse function of f (x), start from y =
3 x + 2 and solve for x. That is
y =
3 x + 2
Take the square of both sides, y^2 = 3x + 2
Subtract 2 on both sides, y^2 − 2 = 3x
Divide by 3 on both sides,
x =
y^2 − 2 3
So, h(y) = y^2 − 2 3
is the desired inverse function. Just use x to denote input, we have
h(x) = y^2 − 2 3
, and this h(x) is just the inverse function of f (x). To find the domain of (
g h )(x), we need to find the common part of the domain of g(x)
and h(x), and h(x) 6 = 0 because h(x) is the denominator. The domain of g(x) is all the real numbers, the domain of h(x) is NOT all the real numbers. As h(x) is the inverse function of f (x), the domain of h(x) is just the range of g(x), which is all the real numbers bigger than or equal to
2 (or x ≥
2). h(x) 6 = 0 just
means x^2 − 2 3
= 0, which means x^2 − 2 6 = 0, so x 6 =
2 and x 6 = −
- In conclusion, the
domain of ( g h )(x) is x >
- (10 points) Imagine you have a rope of length 12 meters and you use the rope to shape a rectangle. And one side of the rectangle is just the wall. How shall you shape
the rectangle (one side is the wall, and three sides are the rope) such that the area is the largest?
Solution: Let the length of the two parallel sides of ropes be x meters each, then the length of the wall shaping the rectangle is 12 − 2 x meters. So the area of the rectangle is x(12 − 2 x) meters. That is, the area is − 2 x^2 + 12x meters. The quadratic function f (x) = − 2 x^2 + 12x has an opening downward (because the coefficient for x^2 is negative). So, the vertex of the parabola just corresponds to the
maximal (that is, the maximal area). So, when x = − b 2 a
= 3, the area
reaches the maximal, that is 3 · (12 − 2 · 3) = 18 square meters.
- (15 points) Below is a complete graph of a polynomial f (x). a) Find all the extrema of the graph. Based on the number of all the extrema, what can you say about the degree of the polynomial? b) Write down all the roots of f (x) = 0, and decide whether the multiplicity of each root is even or odd? c) Is the degree of f (x) even or odd? Is the coefficient of the leading item of f (x) positive or negative?
-4 -2 2 4
20
40
Solution: a) 4 extrema, the degree of the polynomial is at least 4 + 1 = 5. b) -3 (odd), -2 (odd), 0 (even), 2 (odd) c) The degree of f (x) is odd and the coefficient of the leading item is positive. (Observe the end behavior of the graph!)
- (15 points) Solve the following inequalities:
As |x| is very big, according to the little-big principle, the value of the fraction will be very small (near to 0). So the horizontal asymptote is y = 0 (that is, the x-axis). To find the vertical asymptote, let the denominator be 0 and solve for x, we get x = 0 or x = 1. If x = 1, the numerator is 3 and the denominator is 0, so x = 1 is a vertical asymptote. If x = 0, both the numerator and the denominator are 0, and for all x 6 = 0, we can cancel x from the numerator and denominator, and after canceling, there is no factor x in the numerator and denominator, so x = 0 is a hole.
- (10 points) You have 1400 dollars and you save that amount in bank A. The annual compound interest rate for bank A is 4%. If the interest is compounded yearly, after two years, how much money will you have in the account? If the interest is compounded monthly, after two years, how much money will you have in the account?
Solution: If the interest rate is compounded yearly, after two years, the money in the account is 1400 · (1 + .04)^2. If the interest rate is compounded monthly, after two years, the money in the account
is 1400 · (1 +
)^24
- (10 points) Find the value of k such that x−1 is a factor of k^2 ·x^1000 − 8 k·x^95 +15 = 0.
Solution: Recall that if x − a is a factor of polynomial f (x), then f (a) = 0. As x − 1 is a factor of polynomial k^2 · x^1000 − 8 k · x^95 + 15 = 0, we have
k^2 · 11000 − 8 k · 195 + 15 = 0 that is
k^2 − 8 k + 15 = 0
(k − 3)(k − 5) = 0 So, k = 3 or k = 5.
- (10 points) Solve the inequality by considering the graph of the polynomial:
x^4 − 5 x^2 + 4 ≤ 0
Solution: Consider the graph of f (x) = x^4 − 5 x^2 + 4, it is the graph of polynomial. The degree of this polynomial is even and the coefficient of the leading item is positive, so the two ends of the graph goes upward. Now, consider the roots of this polynomial. Note that f (x) = 0 just means: x^4 − 5 x^2 + 4 = 0
that is (x^2 − 4)(x^2 − 1) = 0 (x + 2)(x − 2)(x + 1)(x − 1) = 0
So the roots of f (x) = 0 is −2, −1, 1 and 2, each root has the multiplicity 1 (kind of crossing). So far, we may sketch the graph of f (x) = x^4 − 5 x^2 + 4, it just as shown below:
