 Home
 Products
 Integration
 Tutorial
 Barcode FAQ
 Purchase
 Company
vb.net generate qr code At A, Cartesian graph of the relation y = (x1/2) At B, Cartesian graph of the inverse relation in .NET
At A, Cartesian graph of the relation y = (x1/2) At B, Cartesian graph of the inverse relation Code 39 Reader In VS .NET Using Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in .NET framework applications. Printing USS Code 39 In .NET Using Barcode printer for VS .NET Control to generate, create Code 39 Full ASCII image in Visual Studio .NET applications. Finding an Inverse Relation
Scan USS Code 39 In VS .NET Using Barcode scanner for Visual Studio .NET Control to read, scan read, scan image in .NET applications. Print Barcode In .NET Framework Using Barcode encoder for Visual Studio .NET Control to generate, create bar code image in .NET applications. y 6 (1, 1) 4 (0, 0) 2 (4, 2) Barcode Reader In Visual Studio .NET Using Barcode recognizer for .NET framework Control to read, scan read, scan image in .NET applications. Encode USS Code 39 In C# Using Barcode creation for Visual Studio .NET Control to generate, create Code 3/9 image in Visual Studio .NET applications. 6 4 2 2 4 6 2 4 6 Code39 Generator In VS .NET Using Barcode creator for ASP.NET Control to generate, create ANSI/AIM Code 39 image in ASP.NET applications. Creating Code39 In VB.NET Using Barcode generator for Visual Studio .NET Control to generate, create USS Code 39 image in .NET applications. y 6 4 2 (2, 4) (1, 1) x 6 4 (0, 0) 4 6 2 2 2 4 6 Encoding Linear 1D Barcode In VS .NET Using Barcode maker for .NET framework Control to generate, create Linear Barcode image in Visual Studio .NET applications. Print EAN13 In VS .NET Using Barcode generation for .NET Control to generate, create EAN13 image in Visual Studio .NET applications. Figure 124 Paint Code128 In .NET Using Barcode generator for .NET Control to generate, create Code 128B image in .NET applications. Draw UPCE Supplement 5 In Visual Studio .NET Using Barcode encoder for Visual Studio .NET Control to generate, create UPCE Supplement 5 image in Visual Studio .NET applications. At A, Cartesian graph of the relation y = x1/2 At B, Cartesian graph of the inverse relation
Bar Code Maker In Visual C# Using Barcode generation for Visual Studio .NET Control to generate, create barcode image in VS .NET applications. UPC Code Maker In Java Using Barcode generation for BIRT Control to generate, create UPC Symbol image in BIRT applications. Inverse Relations in TwoSpace
ECC200 Drawer In None Using Barcode drawer for Word Control to generate, create Data Matrix 2d barcode image in Office Word applications. Code 128C Generation In Java Using Barcode encoder for Java Control to generate, create Code 128 image in Java applications. Are you confused
Draw Code39 In None Using Barcode encoder for Word Control to generate, create Code 3/9 image in Microsoft Word applications. Create EAN13 In .NET Using Barcode creation for ASP.NET Control to generate, create EAN13 image in ASP.NET applications. It s reasonable for you to wonder, Can any relation be its own inverse The answer is yes There are plenty of examples Consider the following equation: x2 + y2 = 25 The Cartesian graph of this equation is a circle centered at the origin and having a radius of 5 units (Fig 125) If we transpose the variables, we get y2 + x2 = 25 which is equivalent to the original relation If we perform the graphical transformation by mirroring the circle around the line y = x, we get another circle having the same radius and the same center Theoretically, all but two of points on the new circle are in different places than the points on the original circle, but the graph looks the same as the one shown in Fig 125 Creating Linear Barcode In Java Using Barcode generation for Java Control to generate, create Linear Barcode image in Java applications. Creating UCC.EAN  128 In ObjectiveC Using Barcode creation for iPhone Control to generate, create EAN / UCC  14 image in iPhone applications. Here s a challenge! Consider the following relation where the independent variable is x and the dependent variable is y: x 2/9 + y2/25 = 1 y Circle centered at the origin is symmetrical 6 4 2
Point reflector line
x 6 4 2 2 4 6 with respect to the point reflector line 2 4 6
Figure 125 Cartesian graph of the relation x2 + y2 = 25
This relation is its own inverse
Finding an Inverse Relation
y 6 4 2 x 6 4 2 2 4 6 2 4 6
Ellipse centered at origin
Figure 126 Cartesian graph of the relation x2/9 + y2/25 = 1 Figure 126 is a graph of this relation in Cartesian coordinates It s an ellipse centered at the origin The distance from the center to the extreme right or lefthand point on the ellipse measures 3 units, which is the square root of 9 The distance from the center to the uppermost or lowermost point on the ellipse measures 5 units, which is the square root of 25 Determine the inverse of this relation, and graph it Solution
We can obtain the inverse of this relation by swapping the variables That gives us the equation y 2/9 + x 2/25 = 1 which can be rewritten as x 2/25 + y 2/9 = 1 Figure 127 illustrates the graphs of the original relation and its inverse in Cartesian coordinates The new graph is another ellipse having the same shape as the original one, and centered at the origin just like the original one But the horizontal and vertical axes of the ellipse have been transposed The distance from the center to the extreme right or lefthand point on the inverse ellipse measures 5 units, which is the square root of 25 The distance from the center to the uppermost or lowermost point on the inverse ellipse measures 3 units, which is the square root of 9 Inverse Relations in TwoSpace
y Original relation 6 4 2
Point reflector line
x 6 4 2 2 4 6 Inverse relation 2 4 6
Figure 127 Cartesian graph of the relation x2/25 + y2/9 = 1, the inverse of the relation graphed in Fig 126 Finding an Inverse Function
If a function is a bijection (that is, a perfect onetoone correspondence) over a certain domain and range, then we can transpose the domain and range, and the resulting inverse relation will always be a function If a function is manytoone, then its inverse relation is onetomany, so it s not a function Undoing the work Suppose that f and f 1 are both true functions that are inverses of each other Then for all x in the domain of either function, we have f 1[ f (x)] = x and f [ f 1(x)] = x An inverse function undoes the work of the original function in an unambiguous manner when the domains and ranges are restricted so that the original function and the inverse are both bijections Finding an Inverse Function
Sometimes we can simply turn f insideout to get an inverse relation, and the inverse will be a true function for all the values in the domain and range of f But often, when we seek the inverse of a function f, we get a relation that s not a true function, because some elements in the range of f map from more than one element in the domain of f When this happens, we must restrict f to define an inverse f 1 that s a true function We can usually (but not always) find a way to force f 1 to behave as a true function by excluding all values of either variable that map to more than one value of the other variable Once we ve done that, we get a bijection, ensuring that there is no ambiguity or redundancy either way Making a relation behave as a function A little while ago, we looked at a relation whose graph is a circle with a radius of 5 units (Fig 125) The equation of that relation, once again, is x 2 + y 2 = 25 which can be rewritten as y 2 = 25 x 2 and then morphed to y = (25 x 2)1/2 If we use relation notation to express this equation and name the relation f, we have f (x) = (25 x 2)1/2 The verticalline test tells us that f is not a true function of x We can modify it so that it becomes a function of x if we restrict the range to nonnegative values Graphically, that eliminates the lower half of the circle, so that for every input value in the domain, we get only one output in the range Figure 128A is an illustration of this function, which we can call f+ and define as f+(x) = (25 x 2)1/2 Once again, we mustn t forget that when we take the 1/2 power of a quantity without including any sign, we mean, by default, the nonnegative square root of that quantity The solid dots indicate that the plotted points are part of the range of f+ Now suppose that we eliminate the top half of the circle including the points ( 5,0) and (5,0), getting the graph shown in Fig 128B The verticalline test indicates that this is a true function of x If we call this function f , we can write f (x) = (25 x2)1/2 The white dots (small open circles) tell us that the plotted points are not part of the range of f We can restrict the range further, say to values strictly larger than 1 or values smaller than or

