course Mth 158 023. `* 23
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Given Solution: * * 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: * 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Decreases at (-3,3) to (-1,0) and decreases at (-3,-1) Decreases at(0,2) to (1,0) and decreases at (0,1) Increases at (-1,0) to (0, 2) and increases at (-1,0) Increases at (1,0) to (3,3) and increases at (1,3) -3< = x = < 3 and 0 < = y = <3 Function is even Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: * 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Decreases at (2,2) to (3,0) and interval of (2,3) Increases at (-3,2) to (-2,1) and interval of (-3,-2) Increases (0,1) to (2,2) interval (0,2) Constant at (-2,0) -3 <= x <= 3 and -2 <= y <= 2 Function is not even and is not odd Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: * 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: So the local max is (0,1) and the local min (-pi, -1) and (pi,-1) Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Local maximum is (0,1) Local minimum are (-pi,-1) and (pi,-1) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: * 3.3.76 / 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: X = 1 so (x-2x^2) - (-1) = -2 x^2 + x + 1 when put in f(x) and then factor = (2x + 1) ( -x + 1) Solve for (f(x) - f(1) ) / ( x - 1) so - (2x + 1) x = 2 for -(2x + 1) giving ( f(2) - f(1) ) / ( 2 - 1) -(2 * 2 + 1) = -5 the slope. secant line is (y - (-1) ) = -5 * (x - 1) solving to get y = -5 x + 4 Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): It’s a long process to think about how to get the answer and keep everything straight but taking step by step is best way I found while working through these. Self-critique Rating: ********************************************* Question: * 3.3.36 / 50 (was 3.2.40). h(x) = 3 x^3 + 5. Is the function even, odd or neither? How did you determine algebraically that this is the case? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: h(x) = 3x^3 +5 h(-x) = 3-x^3 +5 = -3x^3 + 5 h(x) doesn’t equal h(-x) so function is not even and h(x) is not equal to –h(x) so the function is not odd Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: "