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Switching the names of the variables, we get x = (25 y2)1/2 Squaring both sides, we obtain x2 = 25 y2 Subtracting 25 from each side gives us x2 25 = y2 When we multiply through by 1 and transpose the left- and right-hand sides of the equation, we get y2 = 25 x2 Taking the complete square root of both sides, we have y = (25 x2)1/2 Replacing y by f 1(x) to indicate the inverse of f , we get f 1(x) = (25 x2)1/2 We transpose the domain and range of f to get the domain and range of f 1 Things get a little tricky here Refer again to Fig 12-8B The domain of f is the open interval ( 5,5), and the range of f is the half-open interval [ 5,0) Transposing, we can see that the domain of f 1 is the half-open interval [ 5,0), and the range of f 1 is the open interval ( 5,5) Figure 12-9B is a graph of f 1 This inverse relation fails the vertical-line test, so it s not a true function of x

Making an inverse behave as a function Do you get the idea that we can t make the relation graphed in Fig 12-5 behave as a function whose inverse is another function, no matter what limitations we impose on the domain and range Don t give up There are plenty of ways For example, we can restrict both the domain and the range of the original relation

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x2 + y2 = 25 to values that all show up in the first quadrant of the Cartesian plane When we do that, the domain and range are both narrowed down to the open interval (0,5) The relation becomes a true function of x, and its inverse also becomes a true function Similar things happen if we restrict both the domain and the range to values that show up entirely in the second quadrant, entirely in the third quadrant, or entirely in the fourth quadrant Feel free to draw the graphs, put a point reflector line to work, and see for yourself

You might ask, We ve seen an example of a relation that s its own inverse Can a function be its own inverse The answer is yes The function f (x) = x is its own inverse; the domain and range both span the entire set of real numbers It s the ultimate in simplicity The function s graph coincides with the point reflector line, so it s identical to its own reflection! We have f 1[ f (x)] = f 1(x) = x so therefore f [ f 1(x)] = f (x) = x

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Another example Consider g(x) = 1/x, with the restriction that the domain and range can attain any real-number value except zero This function is its own inverse We have

Now the time has come for you to do your project. Depending on what you chose, it may be just a few hours of work, or it may be many hours over days or weeks. See the project as your chance to learn the tracking tools for time, cost, quality, risk, and the other ve areas of project management knowledge that are covered in Part Three. Practice honest assessment and status reports. If you are working with others, ask them what you can do that would encourage them to be clear and honest about the status of their tasks. As you do this, ask yourself: What am I learning about how to keep a project on track Then take some time, write it down, and explain it to someone else. Or, if you can, arrange an opportunity to present what you are learning.

Still another example Consider the function h(x) = 3 for all real numbers x Figure 12-10 shows its graph When we transpose the variables, domain, and range, we must set x = 3 for h 1(x) to mean anything Then we end up with all the real numbers at the same time This relation fails the vertical-line function test in the worst possible way, because the graph is itself a vertical line (Fig 12-11)

Here s a challenge!

Consider the following three functions: f (x) = x 11 g(x) = x2/4 h(x) = 32x5 The inverse of one of these functions is not a function Which one