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The function y = -x2 - 2x + 8 has derivative y'= -2x - 2 = -2(x + I). Setting - 2(x + 1) = 0 produces the solution x = -1 and we already know y = 9 for x = -1. How does calculus help in graphing When the derivative of a parabola is zero, the curve has a A or U shape. Zero slope means the curve is flat and the only place where a parabola is flat is at a peak or a valley. The broader application of this approach is very helpful in higher (than 2) power curves such as the one in the next problem.
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= 5. We see that the required limit ( ) exists, and that it equals 5. Thus the function f (x) = x 2 + x is differentiable at x = 2, and the value of the derivative is 5. Math Note: When the derivative of a function f exists at a point c, then we denote the derivative either by f (c) or by (d/dx)f (c) = (df /dx)(c). In some contexts (e.g., physics) the notation f (c) is used. In the last example, we calculated that f (2) = 5. The importance of the derivative is two-fold: it can be interpreted as rate of change and it can be interpreted as the slope. Let us now consider both of these ideas. Suppose that (t) represents the position (in inches or feet or some other standard unit) of a moving body at time t. At time 0 the body is at (0), at time 3 the body is at (3), and so forth. Imagine that we want to determine the instantaneous velocity of the body at time t = c. What could this mean One reasonable interpretation is that we can calculate the average velocity over a small interval at c, and let the
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(1) 103 106 109 (8 bits) 210 210 220 106 220 230 109 240 1012 250 1015 Kbytes, KiB KiB/s Mbytes, MiB Mbytes, MB MiB/s Gbytes, GiB Gbytes, GB Tbytes, TiB Tbytes, TB Pbytes, PiB Pbytes, PB 2 KB per DVD sector 150 KB/s CD-ROM data rate 650 MB in CD-ROM 682 M bytes in CD-ROM 1.32 MB/s DVD data rate 4.37 GB in a DVD 4.7 G bytes in a DVD Kbps, kb/s, Kb/s 56 kbps modem mbps, mb/s, Mb/s 11.08 Mbps DVD data rate gbps, gb/s, Gb/s
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Unfortunately, this attempt won t work. Because T is a generic type, the compiler has no way to know precisely how two objects should be compared for equality. Should a bitwise comparison be done Should only certain fields be compared Should reference equality be used The compiler has no way to answer these questions. Fortunately, there is a solution. To enable two objects of a generic type parameter to be compared, they must implement the IComparable or IComparable<T>, and/or IEquatable<T> interfaces. Both versions of IComparable define the CompareTo( ) method and IEquatable<T> defines the Equals( ) method. The IComparable interfaces are intended for use when you need to determine the relative order of two objects. IEquatable is used for determining the equality of two objects. These interfaces are defined by the System namespace, and they are implemented by all of C# s built-in types, including int, string, and double. They are also easy to implement for classes that you create. Let s begin with IEquatable<T>. The IEquatable<T> interface is declared like this: public interface IEquatable<T> The type of data being compared is passed as a type argument to T. It defines the Equals( ) method, which is shown here: bool Equals(T other) It compares the invoking object to other. It returns true if the two objects are equal and false otherwise. When implementing IEquatable<T>, you will usually also need to override GetHashCode( ) and Equals(Object) defined by Object, so they act in a manner compatible with your implementation of Equals( ). The program that follows shows an example. Using IEquatable<T>, here is a corrected version of IsIn( ):
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Whether you re going to purchase your first digital camera or purchase a second digital camera with more powerful features, it s a good idea to do your homework first. If you rush out blindly and buy the first digital camera that looks cool or the one the friendly salesperson recommends, you may end up with a bad case of buyer s remorse when you actually use the camera.
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