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Properties of Functions New Functions from Old Problem. Use the graph to answer these questions about the function y = f(x). 1. For what values of x is y = 0? 2. For what values of x is y≤ 0 ? 3. What are the maximum and minimum values of y and for what x do they occur? 4. For what values of x is y = 20? Problem. Use the equation y = x 2 −3x + 2 to answer these questions about the function y = f(x). 1. For what values of x is y = 0? 2. For what values of x is y≥ 0 ? 3. Does y have a maximum and minimum value? If so find them and the x for which they occur. 4. For what values of x is y = −5? Solution: x2 −3x + 2 = (x −1)(x − 2) so we see first that y = 0 when x = 1 or x = 2. Also, the product of two numbers is >0 when the numbers are both positive or both negative. In this case both factors are positive if x >2, and both are negative when x < 1. Thus the answer to 2 is x ≤ 1 or x ≥ 2. To answer 3, we perform a computation called completing the square. We know that (x − a)2 = x2 − 2ax + a2 . Thus we can write x2 −3x + 2 = (x2 −3x + 9) + (2− 9) = (x − 3)2 − 1 4 4 2 4 Since the first factor has a minimum value of 0, we see that the function is a minimum when x = 3/2, and this minimum value is –1/4. There is no maximum value, since y gets arbitrarily large when x does. Finally if we try to solve y = −5, or x2 −3x + 7 = 0 we get no real solutions. Thus this never happens. Functions from Applied Problem Often the form of a function must be found from information describing a practical problem. This problem may also effect the domain of the resulting function. Example 8. A box is made from a 6 inch square of cardboard by cutting out a square of side x from each corner, and folding up the sides. x x x x 6 x x x x V = x(6− 2x)2 x 0≤x≤3 6 −2x Here the natural domain is not appropriate. We use the domain that corresponds to the practically realizable values of the independent variable. One practical problem is to find the value of x so that the resulting box has the largest possible volume. We begin by graphing the function. It appears from this graph that the maximum volume occurs when x is about 1. Later in the course we will be able to prove that it occurs exactly at x = 1. Piecewise Defined Functions Sometimes we can combine the methods of specifying a function. One example that occurs frequently is the piecewise defined function. Here we describe the function by different formulas that are valid for different sets of input values. This is, in a way, a combination of the verbal description and the formula description. Example 7. Define a function with real inputs and outputs as follows: x2 f(x) = | x<–1 x+2 | – 1 ≤ x < 3 14- x 2 | x ≥ 3 Then we see that f (−2) = (−2)2 = 4 f (0) = 0 + 2 = 2 and f (4) =14 −(4)2 = −2 The graph of this function is: One of the most useful functions defined in this way is the absolute value function. We let |x| = x| x≥0 –x| x<0 The graph of the function whose formula is f(x) = |x| is : We see that the expression |x – a| will equal x – a when x − a ≥0 and will equal –(x – a) when (x – a)<0. In other words x–a| x≥a |x – a| = a–x| x<a Thus we can simplify the following functions: 5x – 2 | x ≥ 0 1. |x| + 4x – 2 = 3x – 2 | x < 0 3x + 1 | x ≥ 1/3 2. 2 + |3x – 1| = – 3x + 3 | x < 1/3 Since 3x – 1 ≥ 0 means that 3x ≥ 1, or x ≥ 1/3. 2x – 4| x ≥ 3 3. |x – 1| + |x – 3| = 2 | 1 ≤ x<3 4 – 2x | x< 1 Since the expression becomes respectively (x – 1) + (x – 3), (x – 1) + (3 – x), and (1 – x) + (3 – x) New Functions from Old Given any two numerical functions, we can combine these functions by addition, subtraction, multiplication and division to get new functions. Let f and g be two numerical functions. Then we define f + g, f − g, fg, and f/g to be the functions whose values are produced by the following formulas: 1. (f + g) (x) = f(x) + g(x) 2. (f − g) (x) = f(x) − g(x) 3. (fg)(x) = f(x)g(x) 4. (f/g)(x) = f(x)/g(x) The domains of the first three are taken to be the intersection of the domains of f and g (all numbers common to both), and for 4. We also exclude any place where g(x) = 0. x f g f+g f(x) g(x) y = (f + g)(x) = f(x) + g(x) Example 1. Let f and g be defined by the formulas f(x) = 2 + x −1 g(x) = x− 4 and let each have their natural domain. Then 1. (f + g) (x) = f(x) + g(x) = 2 + x−1 + (x − 4) = (x − 2) + x−1 2. (f − g) (x) = f(x) − g(x) = 2 + x−1 − (x − 4) = (−x + 6) + x−1 3. (fg)(x) = f(x)g(x) = (2 + x −1) (x − 4) 4. (f/g)(x) = f(x)/g(x) = (2 + x −1) x−4 The domain natural of the first three is all x≥1 In the case of number 4, the natural domain consists of all x≥1 except x = 4. 1 Example 2. Let f(x) = x and g(x) = 2 . x The natural domain of f is all non negative numbers, and that of g is all real numbers other than 0. The common domain is therefore all positive real numbers. For x positive, (f + g) (x) = f(x) + g(x) = shown below. x + 1 2 . The graphs are x x + 12 x x 1 x2 Stretches and Compressions If we multiply a function by a positive constant, we get a new function whose graph is the graph of the original function stretched or compressed vertically, depending on whether the constant is greater than or less than 1. If we multiply the independent variable by a positive constant, the new function has a graph that is the graph of the original function stretched or compressed horizontally, again depending on whether the constant is less than or greater than 1. Let f be the function whose graph is shown below: We combine this function with 2f and .5f on the same graph. f 2f .5f Now we combine the functions with formulas f (x), f(2x), and f (.5x) on the same graph. .5x f(2x) f (x) Warning. Multiplying the independent variable by numbers > 1 compresses the graph. Multiplying by numbers < 1 stretches it. Translations If we add a nonzero constant to a function the graph is translated up or down by that amount, depending on whether the constant is positive or negative. If we add a nonzero constant to the independent variable, the graph is translated right or left, depending on whether the constant is negative or positive. Warning. Note that in the last statement, the sign of the constant and the direction of the horizontal shift are opposite to one another. Let f(x) = |x|. We have the following graphs: |x − 2| |x| + 2| |x + 2| |x| −2 Symmetry and reflections If we multiply a function by –1, we reflect the graph about the x axis. If we multiply the independent variable by –1, we reflect the graph about the y axis. f(x) f(x) f(– x) – f(x)