#$&* course Mth 158 11/10 4 023. `* 23
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Given Solution: * * The function decreases until reaching the local min at (-8, -4), then increases until reaching the local max at (-2, 6). The function then decreases to its local min at (5, 0), after which it continues increasing. So the graph is decreasing on (-infinity, -8), on (-2, 0) and on (2, 5). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 3.3.22 (was 3.2.12). Piecewise linear (-3,3) to (-1,0) to (0,2) to (1,0) to (3,3).Give the intercepts of the function.Give the domain and range of the function.Give the intervals on which the function is increasing, decreasing, and constant.Tell whether the function is even, odd or neither. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (a) intercept is (0,2) and on the x axis, (-1,0) and (1,0) (b) domain is [-3, 3] and range is [3, 3] (c) decreases on intervals (-3, -1) and (0, 1) increases on intervals (-1, 0) and (1,3) (d) Even and symmetric about the y-axis confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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: (a) x-int (-2.3, 0) and (3,0) y-int (0,1) (b) domain [-3,3] range [-2,0] (c) intervals decreasing (2,3) constant (-2, 0) increasing (-3,2) and (0,2) (d) neither, not symmetric confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * The function intersects the x axis at (-2.25, 0) and (3, 0), and the y axis at (0, 1). The function decreases from (2,2) to (3,0) so it is decreasing on the interval (2,3). The function increases from (-3,-2) to (-2, 1) so it is increasing on the interval (-3, -2). The function ioncreases from (0, 1) to (2, 2) so it is increasing on the interval (0, 2). The function value does not change between (-2, 1) and (0, 1) so the function is constant on (-2, 0). The domain of the function is the set of possible x values, so the domain is -3 <= x <= 3, written as the interval [-3, 3]. The range of the function is the set of possible y values, so the range is -2 <= y <= 2, written as the interval [-2, 2]. The function is not symmetric about the y axis, with f(-x) not equal to f(x) for many values of x; e.g., f(-3) is -2 and f(3) is 0. So the function is not even. x = 3 shows us that the function is not odd either, since for an odd function f(-x) = -f(x) and for x = 3 this is not the case. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 3.3.32 (was 3.2.24). sine-type fn (-pi,-1) to (0, 2) to (pi, -1).At what numbers does the function have a local max and what are these local maxima?At what numbers does the function have a local min and what are these local minima? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (a) local maximum (0,1) (b) local minimum (-pi,-1) and (pi, -1) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Local maximum is (0,1) Local minimum are (-pi,-1) and (pi,-1) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: * 3.3.76 / 46 (was 3.2.36) (f(x) - f(1) ) / (x - 1) for f(x) = x - 2 x^2 What 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: Here I go.... start with f(x)=2x^2+x but we have to plug in [f(x)-f(1)]/(x-1) so we get something pretty ugly like this [f(x)-f(1)]/(x-1) = [(2x^2+x)- (-1)]/(x-1) luckily it's a little more simpler than it looks. So start with simplifying the top part [(2x^2+x)- (-1)] by making the double negatives a positive and you get 2x^2+x +1 which factors into (2x+1)(-x+1) and the equation is now [(2x+1)(-x+1)]/(x-1) (-x+1) can become -(x-1) so the denominator is canceled out with part of the factor leaving [f(x)-f(1)]/(x-1) = -(2x+1) = -2x-1 in the lecture it said something about thinking of the secant line as the slope equation? It still kind of confuses me, but I guess the equation -(2x+1) would be the slope of a line anyway. If x=1 then I guess you'd say [f(1)-f(1)]/(1-1)= 0 and x=2 you get [f(2)-f(1)]/(2-1)= 1/1=1 so there should be a line running from points (1,0) to (2,1) with -(2x+1) as the slope between them confidence rating #$&*: 3? ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** STUDENT QUESTION 100724 I'm OK up to the point where ( f(x) - f(1) ) / (x - 1) = - ( 2 x + 1 ). Beyond that, I don't know where to go and can't follow the given solution. INSTRUCTOR RESPONSE You appear to understand the expression (f(x) - f(1) ) / (x - 1), and you appear to know how to simplify the resulting numerator, factor, and simplify the resulting expression. That's a very good start. Optional series of exercises: Here's a numerical exercise, using the same function. What is numerical value of the expression (f(3) - f(1) ) / (3 - 1)? &&&& What are the x = 1 and x = 3 points of a graph of f(x) vs. x? &&&& Sketch a set of coordinate axes, and plot these two points. Connect the two points with a straight line segment. Describe your sketch. &&&& What is the rise from the first point to the second? &&&& What is the run from the first point to the second? &&&& What therefore is the slope of your line segment? &&&& Between x = 1 and x = 3, do you think the graph of the actual function f(x) is a straight line? Why or why not? &&&& Your line segment went from one point of the f(x) graph to another. A line segment from one point on a curve to another is called a secant. So you have just calculated the slope of a secant line. Restate this in your own words: &&&& Your secant is a straight line segment, and you have just found its slope. What are the coordinates of the x = 1 point? &&&& You have found the slope of the secant, and you have just given the coordinates of a point on that secant. How do you find the equation of a straight line when you know its slope and the coordinates of a point on the line? &&&& What therefore is the equation of the secant line? &&&& Now repeat the same reasoning to do the following: Find the x = 1 and x = 2 points of the graph. The x = 1 point will be the same as before. Then find the slope of the secant line between the x = 1 and x = 2 points. Finally find the equation of the secant line and simplify it to the form y = m x + b. Having gone through the process for these points, you should obtain the equation y = -5 x + 4. Now, since the present function is f(x) = x - 2 x^2, the point (x, f(x)) has coordinates ( x, x - 2 x^2). What were the coordinates of your x = 1 point again? &&&& Between your x = 1 point and the general point (x, x - 2 x^2): What are the two x coordinates of your points? (The coordinates of the second point both contain x's). &&&& What therefore is the expression for the run between these points? (there will be x's in this expression also) &&&& What are the two y coordinates of your points? (yes, there will be x's in this expression .. ) &&&& What therefore is the expression for the slope between these points? (and it will be no surprise to find x's in this expression) &&&& A line segment from your x = 1 point to the point (x, x - 2 x^2) is a secant of the graph of f(x) = x - 2 x^2. You have just found the slope of the secant. Explain this in your own words: &&&& Your expression for the secant included the variable x. If you replace x by 2 and evaluate your expression for the secant, what do you get? &&&& What were the coordinates of your x = 1 point again? &&&& What was the expression you got for the slope of the secant line when you replaced x by 2? &&&& What therefore is the equation of the secant line through your x = 1 point, when x has been replaced by 2? &&&& What does all this tell you about the given solution to the problem? &&&& If you had trouble understanding the second half of the given solution, you should consider submitting the optional exercise. If you do, insert each answer in the line that contains the &&&&, but before the &&&& mark, so that your insertion ends with &&&&. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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: start with h(x) = 3 x^3 + 5 h(-x) = 3 (-x)^3 + 5 = -3 x^3 + 5 not even -h(x) = -(3 x^3 + 5) = -3 x^3 - 5 not odd function is neither confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):ok ------------------------------------------------ Self-critique Rating:ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!