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Math 117 Midterm Review 2
First part, free responses.
I. Solve the following problems.
(f) Find the principle square root of −16.
II. Suppose f(x) = (x + 9)(x − 2)^{2}.
(a) Find all x and yintercpets.
Setting f(x) = 0, and solving for x, we obtain x intercepts at points (−9, 0),
(2, 0).
Also in order to obtain yintercpet, we need to solve f(0). f(0) = (0 +
9)(0 − 2)^{2 } = 36. Therefore yintercpet is at the point (0, 9).
(b) Determine whether graph touches or crosses xaxis at each xintercept.
The real zero −9 has a multiplicity of 1, an odd interger, meaning graph
there crosses xaxis.
The real zero 2 has a multiplicity of 2, an even integer, meaning graph there
touches xaxis.
(c) For large value of x, the graph will resemble the power function y = x^{3}.
III. Let f(x) = (a − 1)x^{2} + 4x + 1.
(a) When a−1 < 0, the parabola opens down, so (−∞, 1) should be the answer.
(b) In order for the quadratic function to have two distinct real zeros, the
discriminant need to be positive, i.e., Δ = 4^{2}−4·(a−1)·1 = 16−4(a−1) =
20 − 4a > 0. Solving this inequality, we obtain the solution in interval
notation, (−∞, 5).
(c) When a = 5, f(x) = 4x^{2} + 4x + 1 = (2x + 1)^{2} ≥ 0, so the
domain for this
quadratic function is (−∞,∞), and the range is [0,∞).
IV. Let f(x) = x^{2} + 4x − 5.
(a) Completing the squares, we have f(x) = (x^{2}+4x+4−4)−5 = (x+2)^{2}−9.
Hence the vertex should be (−2,−9).
(b) Factoring the polynomial function, f(x) = (x−1)(x+5). So setting f(x) =
0, we obtain the real zeros of f are 1 and −5.
(c) Following part (b), we divide the real number line into 3 sections : (−∞,−5),
(−5, 1) and (1,∞). Setting up a table might clear things up:
So, in order for f(x) > 0, the solution intervals are
(−∞,−5) ∪ (1,∞).
V. (a) R(p) = p · x = p(−5p + 100) = −5p^{2} + 100p.
(b) By completing the squares, R(p) = −5(p^{2} − 20p + 10^{2} − 10^{2}) = −5(p −
10)^{2} + 500. So the vertex is at (10, 500), meaning when price p equals 10,
the quadratic function R(p), revenue, reaches its local maximum of 500.
VI. Suppose .
(a) The domain excludes x = −3, 4, so domain of f(x) is (−∞,−3) ∪
(−3, 4) ∪
(4,∞).
(b) Notice that f(x) is not in lowest terms , so we reduce it into
.
Then setting the denominator to 0, we obtain the vertical asymptote x = 4.
(c) We can use long division to obtian the horizontal asymptote, which is y = 2.
VII. Solve inequalities.
(a) x^{2} − 5x + 4 ≥ 0
x^{2} −5x+4 = (x−1)(x−4) = 0 → x = 1, 4. So we separate the xaxis into
three parts. Setting up a table:
Hence the solution for f(x) > 0 is (−∞, 1) ∪ (4,∞). Also
notice that the
inequality is not strict , we need add solutions for f(x) = 0 into the result,
so we obtain the result as (−∞, 1]∪ [4,∞)
First of all we subtract both sides by 1 and simplify the ineequality, then
we obtain . So the result is (−∞, 1).
Second part, multiple choices .
1. C.
2. C. By looking at the discriminants.
3. C. Using synthetic division twice, we can obtain the answer.
4. A.
5. E.
6. B.
7. D.
8. B.
9. A.
10. C.
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