Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 18 part 2 By Herbert I. Gross and Richard A. Medeiros next

Problem #1a © 2007 Herbert I. Gross next f is defined by f(x) = 3x – 6. Compute |f( - 7)|. Answer: |f( - 7)|= 27 next

Answer: |f( - 7)|= 27 Solution for #1a: We begin by replacing x by - 7in the equation f(x) = 3x – 6 to obtain… next © 2007 Herbert I. Gross |f(x)| = |3x – 6| Hence… next |f( - 7)| = |3( - 7) – 6| |f( - 7)| = | - 21 – 6| |f( - 7)| = | - 27| = 27

Problem #1b © 2007 Herbert I. Gross next f is defined by f(x) = 3x – 6. Compute f (| - 7|). Answer: f |( - 7)| = 15 next

Answer: f(| - 7|) = 15 Solution for #1b: We know that | - 7| = 7. next © 2007 Herbert I. Gross Therefore, f(| - 7|) = f(7), next f(7) = 3(7) – 6 and since f(x) = 3x – 6 f(7) = 21 – 6 We see that f(| - 7|) = f(7) = 15.

Notes on Problem #1 This problem illustrates that while the expressions |f(x)| and f(|x|) may look alike, they are not the same. © 2007 Herbert I. Gross next In particular, we have just seen that when x = - 7. |f( - 7)| = 27, but f(| - 7|) = 15.

Notes on Problem #1 In terms of the “program” model, in Problem #1a, we first find the output of the program when the input is 7; and we then replace the output by its absolute value. © 2007 Herbert I. Gross next In Problem #1b, however, we first replace the input by its absolute value and then we find the output.

Notes on Problem #1 The fact that the input of a function is positive does not imply that the output also has to be positive. © 2007 Herbert I. Gross next For example, in terms of our present problem suppose x = 1. Then, f(1) = 3(1) – 6 = - 3. So while we usually associate the absolute value function with non-negative numbers, keep in mind that while |f(x)| can never be negative, f(|x|) can be!

Problem #2a © 2007 Herbert I. Gross next The function f is defined by f(x) = 3x – 6. In the same diagram, draw the graphs: y = f(x) and y = |f(x)|.

next © 2007 Herbert I. Gross (2,0) (1, - 3) (0, - 6) next We begin by graphing f(x) = 3x – 6 y intercept = (0, - 6) m (slope) = 3 We then draw the line that passes through these three points. thus obtaining the graph of f(x)

Solution for #2a: The “trick” to graphing y = |f(x)| lies in the fact that |f(x)| = f(x) when f(x) is non- negative; and when f(x) is negative, |f(x)| = - f(x). next © 2007 Herbert I. Gross next Notice that the points (x,f(x)) and (x, - f(x)) are mirror images of one another with respect to the x-axis. Thus, to obtain the graph of |f(x)| once we know the graph of f(x), we simply take the portion of the curve f(x) that’s below the x-axis and reflect it about the axis. More specifically…

© 2007 Herbert I. Gross And we then reflect about the x-axis the portion of the line that was below the x-axis… (1,3 ) (0,6) next (2,0) (1, - 3) (0, - 6) We start with the line y = 3x – 6. thus obtaining the graph of |f(x)| next

Note on #2a We drew the graph of f(x) = 3x – 6 by locating the three points (0, - 6), (1, - 3), and (2,0), even though only two points are needed to determine the line. © 2007 Herbert I. Gross next

Note on #2a Using the equation of the line to locate three points rather than two, guards against a computational (arithmetic) error in locating any of the points. Namely, if a computational error is made, it is highly unlikely that the three points will lie on the same line. © 2007 Herbert I. Gross next Therefore, a good rule to follow when using the equation is to locate three points and then draw the line.

next Note on #2a If we had preferred, we might have also solved this problem by using the algebraic definition of absolute value. Namely… © 2007 Herbert I. Gross next |3x – 6| = 3x – 6 if 3x – 6 ≥ 0 (i.e., x ≥ 2) - (3x – 6) = - 3x + 6 if x < 2 In terms of a picture…

© 2007 Herbert I. Gross (1,3 ) (0,6) next (2,0) We start with the line y = 3x – 6, where x ≥ 2 Then delete the portion for which x > 2. next We draw the line y = - 3x + 6, where the y intercept is (0,6) and the slope is - 3. This result agrees with our previous graph.

Problem #2b © 2007 Herbert I. Gross next The function f is defined by f(x) = 3x – 6. In the same diagram, draw the graphs y = f(x) and y = f(|x|).

Solution for #2b: If x is positive then, |x| = x; and if x is negative, |x| = - x. next © 2007 Herbert I. Gross next The points (x,f(x)) and ( - x,f(x)) are symmetric with respect to the y-axis. So if we already know what the curve y = f(x) looks like, we may obtain the curve y = f(|x|) in the following way…

© 2007 Herbert I. Gross Start with the graph of f(x) = 3x – 6 (2,0)( - 2,0) ( - 1, - 3) (1, - 3) (0, - 6) next If x < 0, replace the point (x,f(x)) by the point ( - x, f( - x)). And this is the graph that represents f(|x|)

Note on #2b Just as we did in Problem #2a, we may use the algebraic definition of absolute value to obtain the graph of y = 3|x| – 6. More specifically… © 2007 Herbert I. Gross next 3|x|– 6 = 3x – 6 if x ≥ 0 3( - x)– 6 = - 3x – 6 if x < 0 That is, the graph is y = 3x – 6 to the right of the y-axis and y = - 3x – 6 to the left of the y axis.

Note on #2b More generally: For any function f, if we know the graph of f(x) and want to find the graph of f(|x|), we simply replace the portion of the curve y = f(x) that is to the left of the y-axis by the mirror image of the portion of the curve y = f(x) that lies to the right of the y-axis. © 2007 Herbert I. Gross In other words, the graph of f(|x|) is always symmetric with respect to the y-axis. next

Problem #3a © 2007 Herbert I. Gross next f is defined by f(x) = 3x – 6 Compute |f -1 ( - 7)| Answer: |f -1 ( - 7)|= 1 / 3 next

Answer: |f( - 7) | = 1 / 3 Solution for #3a: To find the inverse of f, we interchange x and f(x). In the present problem, this leads to replacing y = 3x – 6 by x = 3y – 6 and obtaining… next © 2007 Herbert I. Gross next y = 1 / 3 (x + 6) 3y = x + 6 f -1 (x) = 1 / 3 (x) + 2 y = g(x) = f -1 (x) = 1 / 3 (x + 6)

Answer: |f( - 7) | = 1 / 3 Solution for #3a: Since g(x) = f -1 (x), g( - 7) = f -1 ( - 7) = 1 / 3 ( - 7) + 2 = -1 / 3 next © 2007 Herbert I. Gross next Finally, since f -1 ( - 7) = -1 / 3, |f -1 ( - 7)| = | -1 / 3 | = 1 / 3 As a check, we see that… f( -1 / 3 ) = 3( -1 / 3 ) – 6 = - 1 – 6 = - 7 next |f( -1 / 3 )| = | - 7| = 7

Problem #3b © 2007 Herbert I. Gross next f is defined by f(x) = 3x – 6. Draw the graph of y = |f -1 (x)|.

next © 2007 Herbert I. Gross (3,3 ) (6,4) next (0,2) This represents the graph of y = |f -1 (x)|. next We draw the line f -1 (x) = 1 / 3 x + 2 where the y intercept is (0,2) and the slope is 1 / 3. We simply take the portion of the line y = 1 / 3 x + 2 that is below the x-axis (i.e. where x < - 6) and reflect it about the x-axis. ( - 6,0) ( - 9, - 1) ( - 12, - 2) ( - 3,1) ( - 9,1) ( - 12,2)

Problem #4a © 2007 Herbert I. Gross next f is defined by f(x) = 3x – 6. Express f(f(x)) in the form f(f(x)) = mx + b. Answer: f(f(x)) = 9x – 24 next

Answer: f(f(x)) = 9x – 24 Solution for #4a: Writing f in the form f( ) = 3( ) – 6, we see that f(f(x)) = 3(f(x)) – 6. next © 2007 Herbert I. Gross next f( ) = 3( ) – 6 And since f(x) = 3x – 6, we may rewrite the above equation in the form… f(x) next 3x – 6 f(f(x)) = 9x – 18 – 6 f(f(x)) = 9x – 24

Problem #4b © 2007 Herbert I. Gross next f is defined by f(x) = 3x – 6. Compute |f(f( - 7))|. Answer: |f(f( - 7))| = 87 next

Answer: |f(f( - 7))| = 87 Solution for #4b: From Problem #4a, we know that f(f(x)) = 9x – 24. next © 2007 Herbert I. Gross next f( ) = 9( ) – 24 Therefore… -7-7 next -7-7 f( - 7) = - 63 – 24 f( - 7) = - 87 And if f( - 7) = - 87, then |f( - 7)| = 87.

next © 2007 Herbert I. Gross Sketch the graph y = |g(x)|. next (2,0) (0, - 4) ( - 7,0) The function g is defined by the curve below. Problem #5a

next Up to now, we have been given the rule that defines the function and asked to draw the graph that represents the function. In this problem, however, we are told what the graph of the function g(x) is and are asked to use the graph to find other properties that involve g(x). © 2007 Herbert I. Gross next Preface to Solution for #5a Determining the function from its graph is what usually happens in laboratory experiments.

next That is: we make various inputs and measure the corresponding outputs. We then obtain a graph of the data with the x-axis representing the input and the y-axis representing the output. © 2007 Herbert I. Gross next Preface to Solution for #5a We then sketch a smooth curve that passes through these points and this is the graph of the function that relates the input to the output.

Solution for #5a: No matter how g(x) is defined, the graph of |g(x)| is obtained from the graph of g(x) simply by reflecting about the x-axis the portion of the curve y = g(x) that is below the x-axis. next © 2007 Herbert I. Gross next This is illustrated in the following slide…

© 2007 Herbert I. Gross We reflect each point that’s below the x-axis about the x-axis. next (2,0) (0, - 4) ( - 7,0) We start with the given curve. next This represents the graph of y = |g(x)|. (0,4)

next © 2007 Herbert I. Gross Sketch the graph y = g(|x|). next (2,0) (0, - 4) ( - 7,0) The function g is defined by the curve below. Problem #5b

Solution for #5b: No matter how g(x) is defined, the graph of g(|x|) is obtained from the graph of g(x) simply by reflecting about the y-axis the portion of the curve y = g(x) that is to the right of the y-axis. next © 2007 Herbert I. Gross next This is illustrated in the following slide…

© 2007 Herbert I. Gross We reflect each point that’s to the right of the y-axis about the y-axis. next (2,0) (0, - 4) ( - 7,0) We start with the given curve. next This represents the graph of y = g(|x|). ( - 2,0) …from which we obtain…

next Notes on #5b From this problem, we see that the portion of the curve y = f(x) that lies to the left of the y-axis is irrelevant when we want to draw the curve y = f(|x|). © 2007 Herbert I. Gross Namely, no matter what the part of the curve y = f(x) that was to the left of the y-axis looks like; it gets replaced by the mirror image of the part of y = f(x) that was to the right of the y-axis. next

Notes on #5b In general given a function f, it is unlikely that f(|x|) = f(x) for all values of x in the domain of f. However, if it does happen that f(|x|) = f(x) for all values of x in the domain of f, it means that the graph of f (namely, the curve y = f(x)) is symmetric with respect to the y-axis. © 2007 Herbert I. Gross Namely, the portion of the graph of f(|x|) that lies to the left of the y-axis is always replaced by the reflection of the portion of y = f(x) that lies to the right of the y-axis. next

Notes on #5b We give a special name to a function for which f(|x|) = f(x) for all values of x in the domain of f. Namely, we call such a function an even function. © 2007 Herbert I. Gross One reason for this follows from the fact that the product of two negative numbers is positive. Thus, ( - x) 2 = x 2 = |x| 2. So, for example, if f(x) = x 2, then… f(|x|) = |x| 2 = x 2 = f(x). next

Notes on #5b © 2007 Herbert I. Gross More generally, if x appears only to even powers in the definition of f, the curve y = f(x) will always be symmetric with respect to the y-axis.