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asp.net barcode reader free Illustration for the solution to Problem 2 in Chap 14 in VS .NET
14 Scan Code 39 In .NET Framework Using Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in .NET framework applications. USS Code 39 Creator In .NET Framework Using Barcode printer for Visual Studio .NET Control to generate, create Code 3 of 9 image in Visual Studio .NET applications. 1 Suppose that there s a real number x that satisfies the equation ex = 0 We know that our mystery number x can t be equal to 0, because we can plug x = 0 straightaway into the equation and get e0 = 1 based on the fact that any positive real number (including e) raised to the zeroth power is equal to 1 We ve assumed that x is real, and we ve discovered that x 0, so we can be Code 3 Of 9 Reader In .NET Using Barcode recognizer for .NET framework Control to read, scan read, scan image in .NET applications. Bar Code Encoder In .NET Using Barcode creation for Visual Studio .NET Control to generate, create bar code image in Visual Studio .NET applications. 14 539
Barcode Scanner In .NET Framework Using Barcode scanner for .NET Control to read, scan read, scan image in VS .NET applications. Encode Code 3 Of 9 In C#.NET Using Barcode encoder for VS .NET Control to generate, create Code 3 of 9 image in .NET framework applications. certain that 1/x is a nonzero real number Therefore, it s okay for us to take the (1/x)th power of both sides of the original equation to obtain (ex)(1/x) = 0(1/x) which we can rewrite using the algebraic rules for exponents as e[x(1/x)] = 0(1/x) When 0 is raised to any nonzero real power, the result is 0 That fact, along with another dose of the algebraic rules for exponents, allows us to streamline the above equation, getting e(x /x) = 0 which simplifies further to e1 = 0 and finally to e=0 This statement is patently untrue According to reductio ad absurdum, it follows that our original assumption must be false We must conclude that no realnumber power of e is equal to 0 2 The dashed gray curves in Fig B14 are the graphs of y = ex USS Code 39 Generation In VS .NET Using Barcode generation for ASP.NET Control to generate, create Code 39 Extended image in ASP.NET applications. Generating Code 3/9 In VB.NET Using Barcode maker for .NET framework Control to generate, create Code 3 of 9 image in .NET applications. Figure B14 Printing EAN 13 In Visual Studio .NET Using Barcode printer for .NET Control to generate, create EAN13 image in .NET applications. Creating Barcode In VS .NET Using Barcode creator for Visual Studio .NET Control to generate, create bar code image in Visual Studio .NET applications. Illustration for the solution to Problem 2 in Chap 14
Encode UPCA In .NET Framework Using Barcode encoder for .NET framework Control to generate, create UPCA Supplement 5 image in .NET applications. USPS Intelligent Mail Encoder In .NET Using Barcode creation for .NET Control to generate, create USPS Intelligent Mail image in Visual Studio .NET applications. y 10 Create European Article Number 13 In ObjectiveC Using Barcode generation for iPhone Control to generate, create EAN13 image in iPhone applications. Making Code 3 Of 9 In Java Using Barcode maker for Java Control to generate, create Code 3/9 image in Java applications. x 2 1 0 1 2 Code128 Generator In Java Using Barcode drawer for BIRT Control to generate, create Code 128 Code Set C image in Eclipse BIRT applications. Draw EAN13 In Java Using Barcode creation for Java Control to generate, create EAN13 Supplement 5 image in Java applications. 540 WorkedOut Solutions to Exercises: 1119 Generate EAN128 In ObjectiveC Using Barcode drawer for iPad Control to generate, create GS1 128 image in iPad applications. Decoding Barcode In Visual Basic .NET Using Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in Visual Studio .NET applications. and y = e x taken directly from Figs 141A and 142A Now let s consider the function y = ex e x Using the algebraic rules for exponents, we can rewrite this as y = e[x+( x)] which simplifies to y = e0 and further to y=1 The domain of this constant function encompasses all real numbers The range is the set containing the single element 1 The graph is a solid black horizontal line passing through the point (0,1) in Fig B14 3 The dashed gray curves in Fig B15 are the graphs of y = 10x and y = 10 x Drawing Bar Code In None Using Barcode creation for Software Control to generate, create bar code image in Software applications. Decoding UCC  12 In Visual Basic .NET Using Barcode decoder for Visual Studio .NET Control to read, scan read, scan image in .NET applications. Figure B15 Illustration for the solution to Problem 3 in Chap 14
y 10 x 1 0 1 14 541
taken directly from Figs 141B and 142B Now let s look at the ratio function y = 10x /10 x Using the algebraic rules for exponents, we can rewrite this as y = 10[x ( x)] which simplifies to y = 102x The domain encompasses all real numbers The range is the set of all positive real numbers The graph is the solid black curve in Fig B15 4 Figure B16 shows the same graphs as Fig B15 However, in this illustration, the y axis is logarithmic, spanning the three orders of magnitude from 01 to 100 The dashed gray lines are the graphs of y = 10x and y = 10 x Figure B16 Illustration for the solution to Problem 4 in Chap 14
x 1 0 1 542 WorkedOut Solutions to Exercises: 1119 The solid black line is the graph of y = 102x 5 Figure B17 is the graph of the function y = 10(1/x) for values of x ranging from 10 to 10 When we input x = 0, we get 101/0, which is undefined For any other real value of x, the output value y is a positive real, so the domain is the set of all nonzero reals No matter how large we want y to be when y > 1, we can always find some value of x that will produce it No matter how small we want y to be when 0 < y < 1, we can always find some value of x that will produce it However, we can t find any value for x that will give us y = 1 For that to happen, we must raise 10 to the zeroth power, meaning that we must find some x such that 1/x = 0 No such x exists, so the range of the function is the set of all positive reals except 1 The graph has a horizontal asymptote whose equation is y = 1, and a vertical asymptote corresponding to the y axis The open circle at (0,0) tells us that this point is not part of the graph 6 Suppose that there s a real number x that satisfies the equation ln 0 = x y 10

