course Mth 271 ›Ìä¼ÎTЙòtÓ˜ê¨}àÐðÉž„vÇ…qŽassignment #027
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16:18:08 Query 3.6.16 lim {x -> 2-} (1/(x+2)); graph shown. **** What is the desired the limit?
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RESPONSE --> - infinity confidence assessment: 2
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16:18:28 As you approach the vertical line x = -2 from the left (i.e., x -> -2-) y values drop asymptotically into unbounded negative values. So the limit is -infinity. **
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RESPONSE --> I got this one correct and understand how I got it. self critique assessment: 2
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16:19:26 Query 3.6.24 lim{x->infinity}( (5x^3+1) / (10x^3 - 3x^2 + 7) ) **** What is the desired limit and how did you obtain it?
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RESPONSE --> You take 5/10 (from num and den) and reduce it and get 1/2. confidence assessment: 2
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16:20:09 You can compare the leading terms of both numerator and denominator, which are both x^3 terms. The limit is therefore just the ratio of leading coefficiets: 5 / 10 = 1/2. A more rigorous algebraic treatment is sometimes called for if you are asked for a proof. See below: lim (x -> infinity) (5x^3 + 1)/(10x^3 - 3x^2 +7) = lim(x -> infinity) (5 x^3) / (10 x^3) = 5/10 or 1/2. You find this limit using the horizontal asymptote rules. Since the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is equal to a/b with a being the leading coefficient of the numerator and b being the leading coefficient of the denominator. You can also rearrange the expression by dividing numerator and denominator both by x^3 to get lim(x -> infinity) ( 5 + 1 / x^3) / ( 10 - 3 / x + 7 / x^3). Since the limits of 1/x^3, -3/x and 7 / x^3 are all zero you end up with just 5/10 = 1/2. **
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RESPONSE --> I got the correct answer and I understand how I got it. self critique assessment: 2
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16:28:38 ** Query 3.6.50 sketch f(x) = (x-2) / (x^2-4x+3). **** Describe your graph, including a description of all intercepts, extrema, asymptotes and concavity.
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RESPONSE --> Note: This is problem 52 in my book. f(x)= x-2/x^2-4x+3 f(x)= x-2/(x-3)(x-1) Vertical Asymptotes = 3 & 1 (Denominator) (- infinity, 1) (1,3) (3, infinity) Horizontal Asymptote y = 0 from the left and the right y intercept (0, 2/3) Numerator (- infinity, 2) ( 2, infinity) The graph starts out negative concaving downward, then becoming positive at 2 and (1,3) and concaving upward. self critique assessment: 2
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16:30:02 The function has zeros when the numerator is 0, at x = 2; only one x intercept which occurs at x = 2. The numerator is negative on (-inf,2) and positive on (2, inf). The y intercept is at x = 0; we get (0, -2/3). Vertical asymptotes when denominator is zero, which occurs at x = 1 and at x = 3. On (-inf,1), (1, 3) and (3, inf) the denominator is respectively positive, negative and positive. The function is always of one sign between zeros and asymptotes, i.e., on (-inf,1), (1, 2), (2, 3) and (3, inf). Putting together what we know about signs of the numerator and denominator we see that the respective signs on these intervals are negative, positive, negative and positive. For large pos and neg x values f(x) -> 0 so + and - x axis is an asymptote. First derivative is - (x^2 - 4·x + 5)/(x^2 - 4·x + 3)^2 and 2d derivative is 2·(x - 2)·(x^2 - 4·x + 7)/(x^2 - 4·x + 3)^3. The function is critical when 1st derivative is 0; numerator of 1st derivative is never 0 (quad formula, discriminant is negative) so there are no critical points. 2d derivative: x^2 - 4 x + 7 is always positive (discriminant negative, expression positive for x=0). Numerator zero only when x = 2. The denominator is zero at x=3, x=1. The 2d derivative is therefore of one sign on (-inf,1), on (1, 2), on (2,3) and on (3, inf). Substitution shows -, +, -, + on these respective intervals and concavity is therefore down, up, down and up. So the function starts with a horizontal asymptote below the negative x axis, remains negative and concave down to a vertical asymptote at x=1. It then becomes positive with upward concavity, descending from a vertical asymptote at x = 1 to a zero at x = 2, then becoming negative with downward concavity as it approaches a negative asymptote at x = 3. To the right of x = 3 the graph descends from a positive vertical asymptote to a horizontal asymptote at the positive x axis, remaining positive and concave up. **
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RESPONSE --> self critique assessment:
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16:34:47 SOLUTION TO PROBLEM #43:
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RESPONSE --> g(x) = x^2/x^2-16 (x+4)(x-4) Vertical Asymptote: x = 4 Horizontal Asymptote: y = 1 The intercept is at (0,0) - relative max The graph in quad I is concaving upward at the vertical asymptote of 4. The graph in quad II is concaving downward with a vertical asymptote at 4. In quads 3 & 4 the graph is concaving downward from the intercept (0,0) with the vertical asymptotes at 4. self critique assessment: 2
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16:35:27 The graph of x y^2 = 4 is not defined for either x = 0 or y = 0. The function has horizontal and vertical asymptotes at the axes. The graph of x y^2 = 4 is not defined for negative x because y^2 cannot be a negative number. However for any x value y can be positive or negative. So the first-quadrant graph is also reflected into the fourth quadrant and both are part of the graph of the relation. You therefore have the graphs of both y = 2 / sqrt(x) and y = -2 / sqrt(x). The graph of the first-quadrant function will be decreasing since its derivative is negative, and will as you say be asymptotic to the x axis. The graph of the fourth-quadrant function is increasing and also asymptotic to the x axis. **
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RESPONSE --> Yes. I think I got most of the problem correct. self critique assessment: 2
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