BS206 Answers for Topic 1 2019.docx - Problem Set for Tutorial One(Week Two Revision Selectively review Chapters 1 – 5(useful discussion related to

BS206 Answers for Topic 1 2019.docx - Problem Set for...

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Problem Set for Tutorial One (Week Two) Revision: Selectively review Chapters 1 – 5 (useful discussion related to the topic of functions is contained in Sections 3.3, 3.5, 4.1, 4.8, 4.9, 5.1, 5.2, 5.3 (higher order polynomials), and 5.5) New material: Read Section 5.6, Chapter 6 and Sections 7.5, 7.7 and 7.14 The textbook contains progress exercises in each chapter, the answers to which are provided at the back of the book. Attempting these questions as part of your reading is likely to be a useful exercise. Question One Consider the function y = 3 + 2 x for x 0 a. Identify the dependent variable, independent variable, the argument, the parameter, the constant, the coefficient, the exponent, the power y = 3 + 2 x y = 3 + 2 x 1 / 2 y: dependent variable x: independent variable or argument 3: constant and parameter 2: constant, coefficient and parameter ½: power, exponent and parameter b. Provide a rough graph of this function in Cartesian space. Identify the range and the domain. Verbally describe the nature of this function. This function increases at a decreasing rate. It therefore has a positive slope, but this slope is becoming increasingly shallow. Let us now prove this. Note that the square root of x is x to a half: y = 3 + 2 x y = 3 + 2 x 1 / 2 Always start by finding the horizontal and vertical intercepts (if any) and then the turning points (if any). Think of it as a three step process: horizontal intercept, vertical intercept and turning points. Step One: Find the Horizontal intercepts (but there isn’t any) The horizontal intercept is found by asking what is x if y = 0? There is, however, no horizontal intercepts as you cannot square-root a negative number. An error term will appear on your calculator. Specifically: y = 3 + 2 x 0 = 3 + 2 x 3 = 2 x 1

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Y 0 x 2 3 Slope Δy/Δx (or rise over the run) is positive and falling as x increases Graph of the function increasing at a decreasing rate 3 x 4 16 7 11 3 / 2 = x Now x will have to be a negative number, but you cannot square root a negative number. There is consequently no horizontal intercept when y =0. Also note that since you cannot square root a negative number, x cannot take on any negative values in Cartesian space. This is why I have imposed the condition of x 0. Step Two: Find the Vertical Intercept The vertical intercept is found by asking what is y if x = 0? Thus, if x = 0, then y = 3 + 2 x = 3 + 2 0 = 3 Step Three: Find the Turning Point (but there isn’t one) You can find the turning point using calculus (where the first derivative is equal to zero) or using the vertex (-b/2a). But it is pretty obvious that there is no turning point, as y increases as x increases for all values of x. Thus, just provide a few rough coordinates. Take a few positive x values and see what the associated y value becomes. I would argue that x = 4 and x =16 are easy, since these numbers are easy to square root. They also give the sense of the function increasing at a diminishing rate, since moving from 4 to 16 quadruples x, but the resulting y will rise by less than quadruple (from 7 to 11): y = 3 + 2 x = 3 + 2 4 = 7 y = 3 + 2 x = 3 +
• One '19
• Greg Moore

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