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SOLUTION Graphs (a) and (b) in Fig 153 are the graphs of invertible functions since no horizontal line intersects each graph more than once Of course we must choose the domain and range appropriately For (a) we take S = [ 4, 4] and T = [ 2, 3]; for (b) we take S = ( 3, 4) and T = (0, 5) Graphs (a) and (b)

if(a || c) ConsoleWriteLine("a || c is true");

[w]1: [T]1 [w]2: [T]1

The true operator is first applied to a Since a is true in this situation, there is no need to use the | operator method However, if the statement were rewritten like this:

if(c || a) ConsoleWriteLine("c || a is true");

We next see that our new notion of exponentiation satis es certain familiar rules If a, d > 0 and b, c R then (i) a b+c = a b a c

[w]1: [T]2 [w]2: [T]2

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then the true operator would first be applied to c, which in this case is false Thus, the | operator method would be invoked to determine if a was true (which it is in this case) Although you might at first think that the technique used to enable the short-circuit operators is a bit convoluted, it makes sense if you think about it a bit By overloading true and false for a class, you enable the compiler to utilize the short-circuit operators without having to explicitly overload either Furthermore, you gain the ability to use objects in conditional expressions In general, unless you need a very narrow implementation of & and |, you are better off creating a full implementation

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In some situations, you will want to use an object of a class in an expression involving other types of data Sometimes, overloading one or more operators can provide the means of doing this However, in other cases, what you want is a simple type conversion from the class type to the target type To handle these cases, C# allows you to create a special type of operator method called a conversion operator A conversion operator converts an object of your class into another type Conversion operators help fully integrate class types into the C# programming environment by allowing objects of a class to be freely mixed with other data types as long as a conversion to those other types is defined There are two forms of conversion operators, implicit and explicit The general form for each is shown here: public static operator implicit target-type(source-type v) { return value; } public static operator explicit target-type(source-type v) { return value; } Here, target-type is the target type that you are converting to; source-type is the type you are converting from; and value is the value of the class after conversion The conversion operators return data of type target-type, and no other return type specifier is allowed If the conversion operator specifies implicit, then the conversion is invoked automatically, such as when an object is used in an expression with the target type When the conversion operator specifies explicit, the conversion is invoked when a cast is used You cannot define both an implicit and explicit conversion operator for the same target and source types To illustrate a conversion operator, we will create one for the ThreeD class Suppose you want to convert an object of type ThreeD into an integer so it can be used in an integer expression Further, the conversion will take place by using the product of the three dimensions To accomplish this, you will use an implicit conversion operator that looks like this:.

An alternative is to nd a numerical procedure that will approximate the solution to Eq (522), subject to the boundary conditions from the output motion These boundary conditions result from the ten conditions listed earlier plus any additional ones that come from the applications By considering the original ten constraints, the following conditions on the cam motion are obtained:

(ii) a b c = (iii) (iv) (v) (vi) (vii)

public static implicit operator int(ThreeD op1) { return op1x * op1y * op1z; }

Part I:

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