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Inverse Relations in Two-Space
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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
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g 1[ g(x)] = g 1(1/x) = 1/(1/x) = x so therefore g [g 1(x)] = g (1/x) = 1/(1/x) = x
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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)
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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
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Finding an Inverse Function
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y 6 (0, 3) 4 2 x 6 4 2 2 4 6 Horizontal line representing y=3 2 4 6
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Figure 12-10
Cartesian graph of the function h (x) = 3
y Original function 6 4 2 x 6 4 2 2 4 6 2 4 6
Point reflector line
Inverse relation
Figure 12-11
Cartesian graph of the inverse of h (x) = 3 It s obviously not a function!
Inverse Relations in Two-Space
Solution
The inverse of g is not a function If we call the dependent variable y, we get y = x2/4 The domain is the entire set of reals, and the range is the set of non-negative reals If we swap the names of the independent and dependent variables, we get x = y2/4 which is the same as y2 = 4x Taking the complete square root of both sides gives us y = 2x1/2 The plus-or-minus symbol indicates that for every nonzero value of the independent variable x that we input to this relation, we get two values of the dependent variable y, one positive and the other negative We can also write g 1(x) = 2x1/2 The original function g is two-to-one (except when x = 0) That s okay But the inverse relation is one-to-two except when x = 0 That prevents g 1 from qualifying as a true function The other two functions, f and h, have inverses that are also functions Both f and h are one-toone, so their inverses are also one-to-one We have f (x) = x 11 and f 1(x) = x + 11 We also have h(x) = 32x5 and h 1(x) = ( x)1/5/2
Practice Exercises
Practice Exercises
This is an open-book quiz You may (and should) refer to the text as you solve these problems Don t hurry! You ll find worked-out answers in App B The solutions in the appendix may not represent the only way a problem can be figured out If you think you can solve a particular problem in a quicker or better way than you see there, by all means try it! 1 Use algebra to find the inverse of the relation f (x) = 2x + 4 2 Use algebra to find the inverse of the relation g(x) = x2 4x + 4 3 Use algebra to find the inverse of the relation h(x) = x3 5 4 Determine the real-number domains and ranges of the relations and inverses from the statements and solutions of Problems 1, 2, and 3 5 Consider the two-space relation x2/4 y2/9 = 1 Figure 12-12 is a graph of this relation in Cartesian coordinates It s a hyperbola centered at the origin, opening to the right and left, and crossing the x axis at (2,0) and ( 2,0)
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