Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 20 By Herbert I. Gross and Richard A. Medeiros next

Problem #1a © 2007 Herbert I. Gross next The functions f, g, and h are defined by… f(x) = 2x + 3, g(x) = 4x + 5 and h(x) = fog; that is, h(x) = f(g(x)). Express f -1 (x) in the mx + b form. Answer: f -1 (x) = 1 / 2 x + -3 / 2 next

Answer: f -1 (x) = 1 / 2 x + -3 / 2 Solution for #1a: Method 1 In terms of the “recipe” format, the best situation for finding the inverse function is if each step in the process is reversible. next © 2007 Herbert I. Gross This is always the case when we are dealing with a linear function, especially if it’s already in the mx + b form. next

Solution for # 1a: With respect to this problem, this leads to… next © 2007 Herbert I. Gross next That is… Output x – 3 2 Subtract 3x – 3InputxDivide by 2 x – 3 2 f – program Input Multiply by 2 Add 3 Output f -1 – program f -1 (x) = x – 3 2

Solution for # 1a: To convert our answer to the mx + b form we rewrite next © 2007 Herbert I. Gross next = x – 3 2 in the form… x – 3 2 (x – 3) 2 1 (x + - 3) 2 1 = = x + ( - 3) x =

Solution for #1a: Method 2 One way to obtain the inverse of a function is to paraphrase the relationship into a form that expresses the input in terms of the output. In this problem, if we use y to represent f(x), we start with… next © 2007 Herbert I. Gross y = 2x + 3 next

Solution for #1a: Method 2 Subtracting 3 from both sides of the equation y = 2x + 3 we obtain… next © 2007 Herbert I. Gross y – 3 = 2x next And if we now divide by 2, we obtain… y – 3 2 = x or 1212 x = y +

Solution for #1a: Method 2 If we now denote next © 2007 Herbert I. Gross next by q(y), q = f -1. And if we now denote the input by x, we obtain… That is… next y + f -1 ( ) = q( ) = ( ) + f -1 ( ) = q( ) = ( ) + …which agrees with the result we obtained using Method 1. x xx

Solution for #1a: Method 3 This method, in a way, is a refinement of Method 2. Namely, in the case of y = f(x) = 2x + 3, it was possible (in fact it was relatively easy) to paraphrase the equation so that x was expressed as a function of y. next © 2007 Herbert I. Gross next This is always possible when f is a linear function.

Solution for #1a: Method 3 However, if f is not linear, it might not be possible to express x explicitly as a function of y. next © 2007 Herbert I. Gross next For example, if f is defined by y = f(x) = x 5 + x, it is well beyond the scope of this course to solve for x in terms of y.

Solution for #1a: Method 3 In such cases, we may view the inverse of a function as the function we obtain when we interchange the input with the output. In this example, if we use y to represent f(x), we would start with… next © 2007 Herbert I. Gross next y = 2x + 3 and then interchange x and y to obtain… x = 2y + 3 Subtracting 3 and then divide by 2, we obtain the same result as before, namely… x – 3 2 y = next

Notes on Method 3 This approach is valid even when f is not linear. In general, if y = f(x) and f -1 exists, we find f -1 in the following way… © 2007 Herbert I. Gross ►Start with… next y = f(x) ►Interchange x and y to obtain… x = f(y) ►Next, if possible, paraphrase x = f(y) so that y is expressed as a function of x. If we denote this function by g (that is, y = g(x)), then f -1 = g. next

Caution #1 The existence of f doesn’t guarantee that f -1 exists. Remember that our definition of a function requires that for each given input there can be only one output. next © 2007 Herbert I. Gross Thus, if f(x) = x 2, f -1 does not exist. Namely, if c is any positive number, there are 2 values of x for which f(x) = c. For example, if x 2 = 9, then x can be either 3 or - 3.

next Caution #2 When we interchanged x and y in the equation y = f(x) to obtain the equation x = f(y); the y in y = f(x) was not the same as the y in x = f(y). next © 2007 Herbert I. Gross Namely, once we interchange x and y, the original y now becomes the new input; and by our general agreement the input is always the x-coordinate.

next Caution #3 As we saw in the case of y = f(x) = x 5 + x, when we interchange x and y it might not be possible to paraphrase the “new” y explicitly in terms of the “new” x. next © 2007 Herbert I. Gross For example, if we start with the function f where y = f(x) = x 5 + x and interchange x and y, we obtain the equation x = y 5 + y; in which case we cannot solve for y explicitly in terms of x. Further discussion of what we do in this case is presented in our note on graphing.

next Note on the Composition of Functions © 2007 Herbert I. Gross In terms of composition of functions, when f -1 exists the relationship between f and f -1 is f o f -1 = f -1 o f = I, where I(x) = x. next In this exercise… f( ) = 2( ) + 3 and f -1 ( ) = ( ) – 3 2

next Note on the Composition of Functions © 2007 Herbert I. Gross As a check, we see that… next and f(f -1 (x)) = … f -1 (f(x)) = f -1 (2x + 3) f -1 (f(2x + 3 )) 2x 2 = = x = (2x + 3) – 3 2 x – 3 2 f( ) x – 3 2 = 2( ) + 3 = (x – 3) + 3= x= x f -1 (f(x)) x f(f -1 (x)) x

Enrichment Note on Introducing y to Represent f(x) © 2007 Herbert I. Gross In Methods 2 and 3, we introduced the notation y = f(x) when we were trying to find the inverse of f. While there is nothing wrong with doing this, the notation y = f(x) is usually reserved for when we are discussing the graph of the function. That is, in terms of the graph of f, the y-axis represents f(x). next

Enrichment Note © 2007 Herbert I. Gross If we prefer not to introduce y and work directly with f, we may start with… next f(x) = 2x + 3 …and knowing that f -1 f(x) = x, we see that … x = f -1 (f(x)) = f -1 (2x + 3) next

Enrichment Note © 2007 Herbert I. Gross An expression such as f -1 (2x + 3) may seem more “ominous” than, say, f -1 (u). If this is the case, simply let u = 2x + 3 to obtain… x = f -1 (f(x)) = f -1 (2x + 3) = f -1 (u) next However, since the input of f -1 is u we would like the output to also be expressed in terms of u (which is not the case if we write x = f -1 (u)).

Enrichment Note © 2007 Herbert I. Gross To this end we may use the equation u = 2x + 3 to express x in terms of u. next More specifically… u = 2x + 3 → u – 3 = 2x →= x u – 3 2 And if we now replace x by = f -1 (u) u – 3 2 f -1 (u) = u – 3 2 or by symmetry next u – 3 2 in the equation x = f -1 (u) we see that…

Enrichment Note © 2007 Herbert I. Gross The use of u in the equation next f -1 (u) = u – 3 2 is not important. What is important is that whatever we replace u by on the left side of our equation is what we have to replace u by on the right side of our equation. In fact, that’s why we suggested the use of the notation f( ) in place of f(x) when dealing with such expressions as f(2x + 3). So we may replace u by x in our equation to obtain the more traditional form… f -1 ( u ) = u – 3 2 next x x

Graphing f -1 © 2007 Herbert I. Gross To obtain the inverse of a function we simply interchange the input and the output. next In terms of a graph this means that the point P(a,b) becomes the point Q(b,a). And from the geometric point of view, these two points are mirror images of one another with respect to the line y = x. This is illustrated in the following slides.

Graphical Interpretation © 2007 Herbert I. Gross next In terms of a graph, the function f defined by f(x) = 2x + 3 is represented by the line shown on the graph… ( - 4, - 5) ( - 3, - 3) ( - 2, - 1) ( - 1,1) (0,3) (1,5) f(x) = 2x + 3

Graphical Interpretation © 2007 Herbert I. Gross next To find the graph of f -1 we start with the graph of f. ( - 4, - 5) ( - 3, - 3) ( - 2, - 1) ( - 1,1) (0,3) (1,5) f(x) = 2x + 3 We then draw the line y = x y = x

Graphical Interpretation © 2007 Herbert I. Gross next ( - 4, - 5) ( - 3, - 3) ( - 2, - 1) ( - 1,1) (0,3) (1,5) f(x) = 2x + 3 y = x We then reflect each point on the graph of f about the line y = x to obtain… ( - 5, - 4) ( - 1, - 2) (1, - 1) (3,0) (5,1)

Graphical Interpretation © 2007 Herbert I. Gross next The line which passes through the reflected points is the graph of f -1. ( - 4, - 5) ( - 3, - 3) ( - 2, - 1) ( - 1,1) (0,3) (1,5) f(x) = 2x + 3 y = x ( - 5, - 4) ( - 1, - 2) (1, - 1) (3,0) (5,1) y = f -1 (x) x – 3 2 =

Graphical Interpretation © 2007 Herbert I. Gross next ( - 4, - 5) ( - 3, - 3) ( - 2, - 1) ( - 1,1) (0,3) (1,5) y = f(x) Notice that a short cut for obtaining the graph of f -1 is to replace each point (a,b) in the graph of f by (b,a). ( - 5, - 4) ( - 1, - 2) (1, - 1) (3,0) (5,1) y = f -1 (x)

Graphical Interpretation © 2007 Herbert I. Gross next The algebraic definition of the inverse function is much more abstract than the geometric interpretation. To illustrate the geometric interpretation, the curve C on the following slide represents the graph of the function f Keep in mind that for f -1 to exist, the curve C must either be always rising or always falling.

Graphical Interpretation © 2007 Herbert I. Gross next If d is any input, to compute f(d), we start at the point P(d,0) and move parallel to the y-axis until we get to the to the point Q(d,f(d)) on C. P (d,0) Q(d,f(d)) next C: y = f(x) O

Graphical Interpretation © 2007 Herbert I. Gross next Through Q, we then move parallel to the x-axis until we get to the point R(0,f(d)) on the y- axis. P (d,0) Q(d,f(d)) C: y = f(x) O R(0,f(d)

Graphical Interpretation © 2007 Herbert I. Gross next f(d) is the directed distance from O to R (or from P to Q). P (d,0) Q(d,f(d)) C: y = f(x) O R(0,f(d) next

Graphical Interpretation © 2007 Herbert I. Gross next Note that it would have been no less logical to start with the point (0,e) on the y-axis and look for the corresponding point on the x-axis. Suppose, for example, that we didn’t know how the point R was obtained in the previous graph, and we labeled it (0,e). Finding f -1 is simply a matter of reversing the steps we used in going from P to Q to R. More specifically…

Graphical Interpretation © 2007 Herbert I. Gross next To find f -1 (e), we start at the point R(0,e) and move parallel to the x-axis until we get to the to the point Q on C. R(0,e) Q

Graphical Interpretation © 2007 Herbert I. Gross next Through Q, we then move parallel to the y-axis until we get to the point P on the x-axis. Q C: y = f(x) O R(0,e) P next

Graphical Interpretation © 2007 Herbert I. Gross next f -1 (e) is the directed distance from O to P (or from R to Q). P (f -1 (e),0) C: y = f(x) O R(0,e) Q next

Graphical Interpretation © 2007 Herbert I. Gross next In summary, the point P can be labeled either as (d,0) or (f -1 (e),0) and the point R as either (0,e) or (0,f(d)). This is shown more visually in the following slide.

Graphical Interpretation © 2007 Herbert I. Gross next To find f(d)… P Q C: y = f(x) O R To find f -1 (e)…

next © 2007 Herbert I. Gross Notice that the process we used previously to find f -1 didn’t depend on f being linear. next Graphical Interpretation More specifically, If we wanted to graph the inverse of the given function f, we would once again draw the line y = x and then reflect each point on C about the line y = x. The resulting curve would then be the graph of f -1.

next © 2007 Herbert I. Gross next Graphical Interpretation This is illustrated in the following graph. Q P R y = x C: y = f(x) C': y = f -1 (x) next P' R' Q' next

Problem #1b © 2007 Herbert I. Gross next The functions f, g, and h are defined by… f(x) = 2x + 3, g(x) = 4x + 5 and h(x) = fog; that is, h(x) = f(g(x)). Express g -1 (x) in the mx + b form. Answer: g -1 (x) = 1 / 4 x + -5 / 4 next

Answer: g -1 (x) = 1 / 4 x + -5 / 2 Solution for #1b: Problem #1a is a specific illustration of the more general case in which f is defined by y = f(x) = mx + b, where m ≠ 0. next © 2007 Herbert I. Gross Namely, starting with y = mx + b we begin by subtracting b from both sides of the equation to obtain… next y – b = mx

Solution for #1b: We then divide both sides of this equation by m (that is, we multiply both sides by 1 / m ) to obtain… next © 2007 Herbert I. Gross or next (y – b) = x 1 m y – b = x 1 m 1 m and if we now interchange x and y we see that… y = x – b 1 m 1 m ( = f -1 (x)) next

Solution for #1b: Problem #1b is another specific illustration of the equation y = mx + b with m = 4 and b = 5. In this case, the equation next © 2007 Herbert I. Gross becomes… next y = x – b 1 m 1 m y = x

next Caution In the equation y = mx + b, it’s important that m ≠ 0. In particular, if m = 0, we cannot divide by it. Moreover, if m = 0, the equation y = mx + b becomes y ( = f(x) ) = b. next © 2007 Herbert I. Gross In this case, f -1 does not exist because f(x) = b for all values of x.

Problem #2a © 2007 Herbert I. Gross next Suppose f(x) = 2x + 3, g(x) = 4x + 5 and h(x) = fog; that is, h(x) = f(g(x)). Express h(x) in the mx + b form. Answer: h(x) = 8x + 13 next

Answer: h(x) = 8x + 13 Solution for #2a: Note: This problem is a review of the composition of functions as well as the first step in solving problem #2b next © 2007 Herbert I. Gross h is defined by… next h(x) = f(g(x)) and since… We may replace g(x) in h(x) = f(g(x)) by its value 4x + 5 to obtain… (4x + 5) g(x) = 4x + 5 next

Solution for #2a: To compute f(4x + 5) we view f in the form… next © 2007 Herbert I. Gross and then “fill in” both sets of parentheses with the expression 4x + 5 to obtain… next f( ) = 2 ( ) + 3 next f( ) = 2 ( ) + 34x + 5 = (8x + 10) + 3 = 8x + 13

Problem #2b © 2007 Herbert I. Gross next Suppose f(x) = 2x + 3, g(x) = 4x + 5 and h(x) = fog; that is, h(x) = f(g(x)). Express h -1 (x) in the mx + b form. Answer: h(x) = 1 / 8 x / 8 next

Answer: h(x) = 1 / 8 x / 8 Solution for #2b: From our solution to part (a) of this problem we know that… next © 2007 Herbert I. Gross Therefore… next h(x) = 8x + 13 By definition… next h -1 (h(x)) = x h -1 (h(x)) = h -1 (8x + 13)

Solution for #2b: We may rewrite h -1 h(x) = h -1 (8x + 13) in the equivalent form… next © 2007 Herbert I. Gross If we now replace 8x + 13 by u in x = h -1 (8x + 13), we see that… next x = h -1 (8x + 13) x = h -1 (u)

Solution for #2b: And since u = 8x + 13, we see that… next © 2007 Herbert I. Gross or next u – 13 = 8x next = x u – 13 8 If we now replace x in x = h -1 (u) by its value in, we obtain… = x u – 13 8 = h -1 (u) u – 13 8 h -1 (u) = u – 13 8 or

Solution for #2b: The equation next © 2007 Herbert I. Gross next is a special form of and if we fill in both sets of parentheses with “x”, we obtain the more traditional form… h -1 ( ) = ( ) – 13 8 h -1 (u) = u – 13 8 h -1 (x) = x – 13 8

next Notes on Problem #2b If you are more comfortable using the y = h(x) format, you can replace the equation h(x) = 8x + 13 by… © 2007 Herbert I. Gross next y = 8x + 13 and then interchange x and y to obtain… next x = 8y + 13 or equivalently… 8y + 13 = x

next Notes on Problem #2b Then subtract 13 from both sides of the equation 8y + 13 = x to obtain… © 2007 Herbert I. Gross next 8y = x – 13 and then divide both sides of the equation above by 8 to obtain… next y = x – 13 8 = h -1 (x)

next Notes on Problem #2b If you prefer to write the recipe in “plain English”, you may represent h and h -1 by… © 2007 Herbert I. Gross next h-programh-program Input x Multiply by 8 8x Add 13 8x + 13 h -1 -program Divide by 8 x – 13 8 Subtract 13 x – 13 Output 8x + 13 Input x Output x – 13 8

Problem #3a © 2007 Herbert I. Gross next Suppose f(x) = 2x + 3, and g(x) = 4x + 5. Express f -1 (g -1 (x)) in the mx + b form. Answer: f -1 (g -1 (x)) = 1 / 8 x / 8 next

Answer: f -1 (g -1 (x) = 1 / 8 x / 8 Solution for #3a: From Problem #1a we know that… next © 2007 Herbert I. Gross and from Problem #1b we know that… next Therefore… 1212 f -1 (x) = x + g -1 (x) = x f -1 (g -1 (x) = f -1 ( x + )

Solution for #3a: Rewriting next © 2007 Herbert I. Gross in the form… next 1212 f -1 (x) = x + f -1 ( ) = ( ) +x we see that… f -1 ( ) = ( ) +

Solution for #3a: and if we now replace next © 2007 Herbert I. Gross next we see that… f -1 ( x + ) x f -1 ( ) = ( ) + x with its value… f -1 (g -1 (x))

Solution for #3a: Using our rules, we see that… next © 2007 Herbert I. Gross next = x ( ) + and… x ( ) x = x and… x = x x = next

Problem #3b © 2007 Herbert I. Gross next True or false: f -1 og -1 = (fog) -1. Answer: false next

Answer: false Solution for #3b: In Problem #2b, we saw that… next © 2007 Herbert I. Gross However, in Problem #3a, we saw that… next And since… (fog) -1 (x) = 1 / 8 x / 8 f -1 og -1 (x) = 1 / 8 x / 8 1 / 8 x / 8 ≠ 1 / 8 x / 8, f -1 og -1 ≠ (fog) -1

next Notes on Problem #3b Often in mathematics things that might seem to be true are actually false. © 2007 Herbert I. Gross next For example, when we want to compute 1 / 2 × 1 / 3, it seems “natural” to accept the “fact” that to find the product we simply “multiply across” to obtain 1 / 6 as the product; that is… × = 1 × 1 2 × 3 = 1 6

next Notes on Problem #3b In a similar way, when we want to compute 1 / / 3, it seems “natural” to accept the “fact” that to find the sum we simply “add across” to obtain 2 / 5 as the sum; that is… © 2007 Herbert I. Gross next However, as natural as this might seem the fact is that the correct answer is 5 / 6 not 2 / = = 2 5

next Notes on Problem #3b In the above vein, it might seem “natural” to accept the “fact” that (fog) -1 = f -1 og -1. However, this problem shows us that this is not the case. © 2007 Herbert I. Gross next What is true, however, is that (fog) -1 = g -1 of -1. This follows from the fact that even though we write f first in the expression fog(x), we compute g(x) first. And when we undo an operation, the last step that we did is the first step that we undo.

© 2007 Herbert I. Gross and from Problem #1b we know that… next Therefore… 1212 f -1 (x) = x + g -1 (x) = x g -1 (f -1 (x)) = g -1 ( x + ) Notes on Problem #3b To check this in terms of our specific problem, from Problem #1a we know that…

next © 2007 Herbert I. Gross in the form… next g -1 ( ) = ( ) + we see that… x g -1 ( ) = ( ) + Notes on Problem #3b Rewriting g -1 (x) = x

next © 2007 Herbert I. Gross next we see that… g -1 ( x + ) x g -1 ( ) = ( ) + x and if we now replace… g -1 (f -1 (x)) Notes on Problem #3b with its value…

next © 2007 Herbert I. Gross next = x ( ) + and… x ( ) x = x and… x = x x = next Notes on Problem #3b Using our rules we see that…

next © 2007 Herbert I. Gross next In summary, if f and g both posses inverses, our rules imply that… + x (fog) -1 = Notes on Problem #3b and this is the same answer we got in Problem #2b, namely… (fog) -1 = g -1 of -1

Problem #4 next Suppose that the function f is defined by: -- the domain of f is [0,4] and -- f(x) = Express f -1 (x) in the mx + b form. Answer: f -1 (x) = next 2x, if 0 ≤ x ≤ 2 3x – 2, if 2 ≤ x ≤ 4 1 / 2 x, if 0 ≤ x ≤ 4 1 / 3 x + 2 / 3, if 4 ≤ x ≤ 10 © 2007 Herbert I. Gross

Answer: Solution for #3b: In essence, f is the union of the two functions g an h, where… next © 2007 Herbert I. Gross next g(x) = 2x, dom g = [0,2] 1 / 2 (x), if 0 ≤ x ≤ 4 1 / 3 x + 2 / 3, if 4 ≤ x ≤ 10 h(x) = 3x – 2, dom h = [2,4]

© 2007 Herbert I. Gross next Solution for #4 For reference, the graph of f is shown to the right. next (0,0) (2,4) (4,10) y = g(x) = 2x y = h(x) = 3x – 2

Solution for #4: Letting y = g(x) = 2x and interchanging x and y we see that x = 2y. Therefore, next © 2007 Herbert I. Gross And letting y = h(x) = 3x – 2 we may interchange x and y to obtain x = 3y – 2; and solving for y in terms of x we see that… next y = 1 / 2 x = g -1 (x), where 0 ≤ x ≤ 4 y = = h -1 (x), where 4 ≤ x ≤ 10 x + 2 3

Solution for #4: In summary… next © 2007 Herbert I. Gross next f -1 (x) = h -1 (x) = if 4 ≤ x ≤ 10 x g -1 (x) = x, if 0 ≤ x ≤

© 2007 Herbert I. Gross next Solution for #4 The graph of f -1 is shown to the right. next (0,0) (2,4) (4,10) (4,2) (10,4) y = g(x) = 2x y = h (x) = 3x – 2 y = g -1 (x) = x 1212 y = h -1 (x) = x + 2 3

© 2007 Herbert I. Gross next Solution for #4 next (0,0) (2,4) (4,10) (4,2) (10,4) y = f(x) y = f -1 (x) y = x Notice that y = f(x) and y = f -1 (x) are symmetric with respect to the line y = x

next © 2007 Herbert I. Gross next Notice also how the domain of f becomes the image of f -1. That is, the domain of f is [0,4}, but the domain of f -1 is [0,10]. (0,0) (2,4) (4,10) (4,2) (10,4) y = f(x) y = f -1 (x) y = x Solution for #4

next © 2007 Herbert I. Gross next …or… Notes on Problem #4 Since there can be no more than one value of f(x) for each value of x ordinarily we would have written either… g(x) = 2x if 0 ≤ x ≤ 2 h(x) = 3x – 2, if 2 < x ≤ 4 f(x) = g(x) = 2x if 0 ≤ x < 2 h(x) = 3x – 2, if 2 ≤ x ≤ 4 f(x)

next © 2007 Herbert I. Gross next Notes on Problem #4 However, since g(2) = h(2) = 4, it was okay to allow 2 to be in the domain of both g and h.… g(x) = 2x if 0 ≤ x ≤ 2 k(x) = 3x + 2, if 2 ≤ x ≤ 4 f(x) = On the other hand, we would have been in “trouble” if the definition of f had been, say… …in which case f(2) = g(2) = 4, and at the same time f(2) = k(2) = 8. Since f(2) has two values, f is not a function. next

Problem #5a next Define the function j by j(x) = |x|, dom j = [0, ∞). Express j -1 (x) in the mx + b form. Answer: j -1 (x) = x, dom j -1 = [0, ∞) next © 2007 Herbert I. Gross

Answer: j -1 (x) = x, dom j -1 = [0, ∞) Solution for #5a: Since x is non-negative |x| = x. Hence, the graph of j is y = x; and if we interchange x and y, we obtain x = y. That is, y = x = j -1 (x). next © 2007 Herbert I. Gross next In other words, j is its own inverse, and its graph is shown on the following slide.

© 2007 Herbert I. Gross next (0,0) (5,5) Solution for #5a y = j -1 (x) = |x| = x dom j -1 = [ 0, ∞) next

Problem #5b next Define the function k by k(x) = |x|, dom k = ( - ∞, 0]. Express k -1 (x) in the mx + b form. Answer: k -1 (x) = - x, dom k -1 = [ - ∞ 0, ) next © 2007 Herbert I. Gross

Answer: k -1 (x) = - x, dom k -1 = (∞,0] Solution for #5b: Since x is negative |x| = - x. Hence, the graph of k is y = - x( = - 1x); and if we interchange x and y, we obtain - x = y. That is, y = - x = k -1 (x). next © 2007 Herbert I. Gross next The graphs of k and k -1 are shown on the following slide.

© 2007 Herbert I. Gross next Solution for #5b next y = k(x) = |x| = - x dom k = ( - ∞, 0] P ( - 4,4) Q ( - 2,2) (0,0) P' (4, - 4) Q' (2, - 2) y = k -1 (x) = |x| = - x dom k -1 = [ 0,∞) next By replacing (a,b) by (b,a), we obtain…

Problem #5c next Define the function p by p(x) = |x|, dom p = ( ∞, ∞).( - ∞, 0]. Does the function p -1 exist? Answer: No next © 2007 Herbert I. Gross

Answer: No Solution for #5c: If p -1 existed, it would mean that for each x in the domain of p -1, there would be one and only one value of p -1 (x). next © 2007 Herbert I. Gross next However, except for x = 0, there are two values of p -1 (x) for each value of x. For example, since p -1 (x) = |x| if p -1 (x) = 7, then x can be either 7 or - 7.

© 2007 Herbert I. Gross next Notes on Problem #4 Problems #5a and #5b are actually a preface to Problem #5c. Namely, while there are two values of x for which |x| = 7, one of them (namely, x = 7) belongs to the domain of j and the other (namely, x = - 7) belongs to the domain of k. More precisely, the function p is the union of the two functions, j and k, each of which has an inverse.

next © 2007 Herbert I. Gross next Notes on Problem #4 Hence, we can refer to the existence of p -1 if we restrict our discussion to either j -1 or k -1. For example, if |x| = 7 and we insist the x has to be positive, then x can only be 7; but if we insist that x has to be negative, then x can only be - 7. We refer to j and k as branches of p, and although p doesn’t have an inverse, each of its branches does. next

© 2007 Herbert I. Gross next Summary next y = j(x)y = k(x) y = p(x) Every horizontal line above the x-axis intersects y = p(x) at two points.

next © 2007 Herbert I. Gross next Summary next If we interchange x and y in the equation y = p(x), we obtain the following graph. y = j -1 (x) y = k -1 (x) For each positive value of x, there are two values of y