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vb.net generate qr code Inverse Relations in TwoSpace in Visual Studio .NET
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Bar Code Generator In Java Using Barcode creation for BIRT Control to generate, create barcode image in Eclipse BIRT applications. Barcode Decoder In C# Using Barcode reader for .NET framework Control to read, scan read, scan image in VS .NET applications. equal to 2, and we ll get more true functions You can doubtless imagine other restrictions we can impose on the range of the original relation f to get true functions of x Barcode Printer In None Using Barcode generation for Excel Control to generate, create barcode image in Office Excel applications. DataMatrix Drawer In Visual Basic .NET Using Barcode printer for Visual Studio .NET Control to generate, create ECC200 image in Visual Studio .NET applications. What about the inverse of f+ Let s manipulate f+ algebraically to find its inverse If we call the dependent variable y, then Barcode Encoder In None Using Barcode printer for Software Control to generate, create bar code image in Software applications. Reading GTIN  13 In Visual Studio .NET Using Barcode scanner for .NET Control to read, scan read, scan image in .NET applications. y = (25 x 2)1/2 Swapping the names of the variables, we get x = (25 y 2)1/2 Squaring both sides, we obtain x 2 = 25 y 2 Subtracting 25 from each side yields x 2 25 = y 2 When we multiply through by 1 and transpose the lefthand and righthand sides of the equation, we obtain y 2 = 25 x 2 Taking the complete square root of both sides gives us y = (25 x 2)1/2 Replacing y by f+ 1(x) to indicate the inverse of f+, we get f+ 1(x) = (25 x 2)1/2 Does this look like the same thing as the inverse of the original relation f Don t be fooled; it isn t the same! We haven t quite finished our work We must transpose the domain and range of f+ to get the domain and range of the inverse relation f+ 1 The domain of f+ is the closed interval [ 5,5], and the range of f+ is the closed interval [0,5] Therefore, the domain of f+ 1 is the closed interval [0,5], and the range of f+ 1 is the closed interval [ 5,5] Figure 129A is a graph of this inverse relation It s easy to see that f+ 1 fails the verticalline test, so it s not a true function of x What about the inverse of f Now let s go through the algebra to figure out the inverse of the function f This process is almost identical to the work we just finished, but it s a good practice to carry it out step by step anyway If we call the dependent variable y, then y = (25 x2)1/2 Inverse Relations in TwoSpace
y 6 (0, 5) is part of the graph 4 2
6 4 2 2 4 6 2 4 6 (0, 5) is part of the graph
y 6 (0, 5) is not part of the graph 4 2
6 4 2 2 4 6 2 4 6 (0, 5) is not part of the graph
Figure 129 At A, Cartesian graph of the inverse of the function y = (25 x 2)1/2 At B, Cartesian graph of the inverse of the function y = (25 x 2)1/2 Verticalline tests indicate that neither of these inverse relations is a function Finding an Inverse Function
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 righthand 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 128B The domain of f is the open interval ( 5,5), and the range of f is the halfopen interval [ 5,0) Transposing, we can see that the domain of f 1 is the halfopen interval [ 5,0), and the range of f 1 is the open interval ( 5,5) Figure 129B is a graph of f 1 This inverse relation fails the verticalline 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 125 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 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

