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Solutions to the math 111 quiz for week 7. It covers topics such as determining if one polynomial is a factor of another, finding horizontal and vertical asymptotes, analyzing the graph of a polynomial function, and solving inequalities.
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Solution: Recall that โx โ a is a factor of polynomial f (x)โ just means โf (a) = 0โ. To judge whether xโ1 is a factor of f (x), just plug 1 into f (x), we have f (1) = 1^3 +7ยท 12 +4ยท 1 โ12 = 0, so g(x) = x โ 1 is a factor of f (x). In this example, f (x) is relatively simple, you can also use polynomial divi- sion to verify that g(x) is a factor of f (x) (using brute force)
x(x โ 1) (x โ 2)^3 b) f (x) =
x(x โ 4) x(x โ 5) Is x = 0 a pole of function f in a)? Is x = 0 a pole of function f in b)? (30 points)
Solution: Plug x = 0 into f (x) in a), the numerator is 0 and the denominator is not 0, that is f (0) = 04 = 0. So x = 0 is not a pole in a) (because 0 is in the domain of f (x) in a) ). Plug x = 0 into f (x) in b), the numerator is 0 and the denominator is also 0, so x = 0 is not in the domain of f (x) in b), besides, for all the x 6 = 0, we can cancel x from the numerator and denominator to get f (x) = (^) (xxโโ^4 5) , x 6 = 0. So x = 0 is a pole of f (x) in b).
-4 (^) -2 2 4
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Solution: a) The two ends of the graph goes in opposite direction, so the degree of f (x) is odd. b) Positive. Because the right end of the graph goes upward. c) -1 (odd), 0 (odd), 2 (even), 4 (odd). d) The degree of f (x) is the sum of the multiplicity of every root. As the smallest odd number is 1 and the smallest even number is 2, the smallest possible degree of f (x) is 1 + 1 + 2 + 1 = 5.
Solution: a) โ 9 < 2 x + 3 < 9 โ 12 < 2 x < 6