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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