course MTH 158 ܼ⯮M~assignment #023
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12:13:47 3.3.16 (was 3.2.6). Key pts and behavior: far left decr, far right incr, zeros at -10, 5, 0, 5, peaks at (-8,-4), (-2, 6), (2, 10). Local min, max among listed points.List the intervals on which the function is decreasing.
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RESPONSE --> Intervals (-infinity,-8), (-2,0) and (2,5) confidence assessment: 2
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12:14:30 ** The function decreases until reaching the local min at (-8, -4), then increases until reaching the local max at (-2, 6). The function then decreases to its local min at (5, 0), after which it continues increasing. So the graph is decreasing on (-infinity, -8), on (-2, 0) and on (2, 5). **
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RESPONSE --> ok self critique assessment: 3
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12:32:53 3.3.22 (was 3.2.12). Piecewise linear (-3,3) to (-1,0) to (0,2) to (1,0) to (3,3).Give the intercepts of the function.Give the domain and range of the function.Give the intervals on which the function is increasing, decreasing, and constant.Tell whether the function is even, odd or neither.
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RESPONSE --> Intercepts are (-1,0) (0,2) and (1,0) Domain is -3<= x <= 3 Range is 0 <= y <= 3 Increasing intervals are (-1,0) and (1,3) Decreasing intervals are (-3,-1) and (0,1) It's not constant anywhere The function is even. confidence assessment: 2
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12:33:54 ** The function intersects the x axis at (-1, 0) and (1, 0), and the y axis at (0, 2). The function decreases from (-3,3) to (-1,0) so it is decreasing on the interval (-3, -1). The function decreases from (0, 2) to (1, 0) so it is decreasing on the interval (0, 1). The function increases from (-1,0) to (0, 2) so it is increasing on the interval (-1, 0). The function ioncreases from (1,0) to (3, 3) so it is increasing on the interval (1, 3). The domain of the function is the set of possible x values, so the domain is -3 <= x <= 3, written as the interval [-3, 3]. The range of the function is the set of possible y values, so the range is 0 <= y <= 3, written as the interval [0, 3]. The function is symmetric about the y axis, with f(-x) = f(x) for every x (e.g., f(-3) = f(3) = 3; f(-1) = f(1) = 0; etc.). So the function is even. **
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RESPONSE --> ok self critique assessment: 3
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13:02:07 3.3.28 (was 3.2.18). Piecewise linear (-3,-2) to (-2, 1) to (0, 1) to (2, 2) to (3, 0)Give the intercepts of the function.Give the domain and range of the function.Give the intervals on which the function is increasing, decreasing, and constant.Tell whether the function is even, odd or neither.
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RESPONSE --> Intercepts on x are (-2.3, 0) and (3,0), y-intercept is (0,1) Domain is -3 <= x <= 3 Range is -2 <= y <= 2 Increasing intervals are from (-3, -2) and (0, 2) Decreasing intervals are from (2,3) Constant intervals are from (-2, 0) The function is neither even or odd because it is symmetrical to neither the y axis nor the x axis. confidence assessment: 2
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13:03:16 ** The function intersects the x axis at (-2.25, 0) and (3, 0), and the y axis at (0, 1). The function decreases from (2,2) to (3,0) so it is decreasing on the interval (2,3). The function increases from (-3,-2) to (-2, 1) so it is increasing on the interval (-3, -2). The function ioncreases from (0, 1) to (2, 2) so it is increasing on the interval (0, 2). The function value does not change between (-2, 1) and (0, 1) so the function is constant on (-2, 0). The domain of the function is the set of possible x values, so the domain is -3 <= x <= 3, written as the interval [-3, 3]. The range of the function is the set of possible y values, so the range is -2 <= y <= 2, written as the interval [-2, 2]. The function is not symmetric about the y axis, with f(-x) not equal to f(x) for many values of x; e.g., f(-3) is -2 and f(3) is 0. So the function is not even. x = 3 shows us that the function is not odd either, since for an odd function f(-x) = -f(x) and for x = 3 this is not the case. **
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RESPONSE --> ok self critique assessment: 3
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13:20:34 3.3.32 (was 3.2.24). sine-type fn (-pi,-1) to (0, 2) to (pi, -1).At what numbers does the function have a local max and what are these local maxima?At what numbers does the function have a local min and what are these local minima?
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RESPONSE --> The local maximum is at 0, with a maxima at 1 The local minimum is at -pi and pi, with a minima at -1 confidence assessment: 2
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13:21:03 ** Local maximum is (0,1) Local minimum are (-pi,-1) and (pi,-1) **
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RESPONSE --> Was that what I said? self critique assessment: 2
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14:52:03 3.3.46 (was 3.2.36) (f(x) - f(1) ) / (x - 1) for f(x) = x - 2 x^2What is your expression for (f(x) - f(1) ) / (x - 1) and how did you get this expression?How did you use your result to get the ave rate of change from x = 1 to x = 2, and what is your value?What is the equation of the secant line from the x = 1 point to the x = 2 point?
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RESPONSE --> With f(x) = x - 2x^2 {x - 2x^2 - F(1)} / x - 1 x - 2x^2 + 1 / x - 1 -2x^2 + x + 1 / x - 1 (-2x -1)(x-1) / x - 1 -2x - 1 is the average rate of change. Using the result to get the ave rate of change from x = 1 to x = 2, F(2) - F(1) / 2-1 We found out earlier that f(1) = -1, so F(2) - (-1) / 2-1 F(2) is going to be replacing x in x - 2x^2 with 2, which gives us (2) - 2(2)^2 = -6 -6 +1 / 2-1 -5 / 1 -5 is the slope or the average rate of change The equation of the secant line is y - y1 = m(x - x1) y - (-1) = -5(x - 1) y + 1 = -5x + 5 y = -5x + 4 confidence assessment: 2
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16:23:30 ** f(x) = f(1) = (x-2x^2) - (-1) = -2 x^2 + x + 1. This factors into (2x + 1) ( -x + 1). Since -x + 1 = - ( x - 1) we obtain (f(x) - f(1) ) / ( x - 1) = (2x + 1) ( -x + 1) / (x - 1) = (2x + 1) * -1 = - (2x + 1). A secant line runs from one point of the graph to another. The secant line here runs from the graph point (1, f(1) ) to the graph ponit ( x , f(x) ), and the expression we have just obtained is the slope of the secant line. For x = 2 the expression -(2x + 1) gives us ( f(2) - f(1) ) / ( 2 - 1), which is the slope of the line from (1, f(1) ) to (2, f(2)) . -(2 * 2 + 1) = -5, which is the desired slope. The secant line contains the point (1, -1) and has slope 5. So the equation of the secant line is (y - (-1) ) = -5 * (x - 1), which we solve to obtain y = -5 x + 4. **
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RESPONSE --> ok self critique assessment: 3
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16:34:01 3.3.50 (was 3.2.40). h(x) = 3 x^3 + 5Is the function even, odd or neither? How did you determine algebraically that this is the case?
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RESPONSE --> h(x) = 3x^3 + 5 3(-x)^3 + 5 = -3x + 5 this is not the same as the original, so it's not even -(3x^3 + 5) = -3x -5 this is not the same as the other, so it is not odd, either confidence assessment: 3
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16:34:12 ** h(x) = 3x^3 +5 h(-x) = 3-x^3 +5 = -3x^3 + 5 h(x) is not equal to h(-x), which means that the function is not even. h(x) is not equal to -h(-x), since 3 x^3 + 5 is not equal to - ( -3x^3 + 5) = 3 x^3 - 5. So the function is not odd either. **
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RESPONSE --> ok self critique assessment: 3
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x՚yЃ}{ assignment #024 024. `query 24 College Algebra 07-11-2008
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17:55:43 3.4.14 (was 3.3.6). Concave down then concave up. Does this graph represent a constant, linear, square, cube, square root, reciprocal, abs value or greatest integer function?
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RESPONSE --> Cube function confidence assessment: 3
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17:56:10 ** A linear function, represented most simply by y = x, has no curvature. A quadratic function, represented most simply by y = x^2, has a parabolic graph, which is either concave up or concave down. A cubic function, represented most simply by y = x^3, has a graph which changes concavity at the origin. This is the first power which can change concavity. A square root function, represented most simply by y = sqrt(x), have a graph which consists of half of a parabola opening to the right or left. Its concavity does not change. A reciprocal function, represented most simply by y = 1 / x, has a graph with a vertical asymptote at the origin and horizontal asymptotes at the x axis, and opposite concavity on either side of the asymptote. An absolute value function, represented most simply by y = | x |, has a graph which forms a 'V' shape. The greatest integer function [[ x ]] is a 'step' function whose graph is made up of horizontal 'steps'. The concavity of the reciprocal function changes, as does the concavity of the cubic function. The cubic function is the only one that matches the graph, which lacks the vertical and horizontal asymptotes. **
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RESPONSE --> ok self critique assessment: 3
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22:19:21 3.4.20 (was 3.3.12). Does your sketch of f(x) = sqrt(x) increase or decrease, and does it do so at an increasing or decreasing rate?
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RESPONSE --> It increases at decreasing rate confidence assessment: 2
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22:19:41 ** y = sqrt(x) takes y values sqrt(0), sqrt(2) and sqrt(4) at x = 0, 2 and 4. sqrt(0) = 0, sqrt(2) = 1.414 approx., and sqrt(4) = 2. The change in y from the x = 0 to the x = 2 point is about 1.414; the change from the x = 2 to the x = 4 point is about .586. The y values therefore increase, but by less with each 'jump' in x. So the graph contains points (0,0), (2, 1.414) and (4, 2) and is increasing at a decreasing rate. **
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RESPONSE --> ok self critique assessment: 3
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22:20:40 What three points did you label on your graph?
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RESPONSE --> (1,1), (4, 2) and (9, 3) self critique assessment: 3
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22:46:18 3.4.24 (was 3.3.24) f(x) = 2x+5 on (-3,0), -3 at 0, -5x for x>0. Given the intercepts, domain and range of the function
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RESPONSE --> x-intercept is at the point (0, 5/2) There are no y-intercepts Domain is x | x >= -3 Range is y | y = [-1. 3] and y = (0, -infinity) confidence assessment: 2
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22:49:19 ** From x = -3 to x = 0 the graph coincides with that of y = 2 x + 5. The graph of y = 2x + 5 has y intercept 5 and slope 2, and its x intecept occurs when y = 0. The x intercept therefore occurs at 2x + 5 = 0 or x = -5/2. Since this x value is in the interval from -3 to 0 it is part of the graph. At x = -3 we have y = 2 * (-3) + 5 = -1. So the straight line which depicted y = 2x + 5 passes through the points (-3, -1), (-5/2, 0) and (0, 5). The part of this graph for which -3 < x < 0 starts at (-3, -1) and goes right up to (0, 5), but does not include (0, 5). For x > 0 the value of the function is -5x. The graph of -5x passes through the origin and has slope -5. The part of the graph for which x > 0 includes all points of the y = -5x graph which lie to the right of the origin, but does not include the origin. For x = 0 the value of the function is given as -3. So the graph will include the single point (0, -3). **
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RESPONSE --> I see I forgot to put the negative sign in front of 5/2.. sorry! I also forgot to put the single point (0, -3), which is what I had indeed graphed, I just didn't see it in the instructions, so I didn't include it. And did the range and domain I gave match up? self critique assessment: 2
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