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// Pad on right with spaces str = strPadRight(20); ConsoleWriteLine("|" + str + "|"); // Trim spaces str = strTrim(); ConsoleWriteLine("|" + str + "|"); // Pad on left with #s str = strPadLeft(10, '#'); ConsoleWriteLine("|" + str + "|"); // Pad on right with #s str = strPadRight(20, '#'); ConsoleWriteLine("|" + str + "|"); // Trim #s str = strTrim('#'); ConsoleWriteLine("|" + str + "|"); } }

SOLUTION By inspection of the graph we see that f is one-to-one (ie, takes different domain values to different range values) and onto (ie, takes on all values in the range) from S = [ 2, 3] to T = [1, 5] Therefore f has an inverse The graph of f 1 is exhibited in Fig 150

The output is shown here:

10 0

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You can insert a string into another using the Insert( ) method, shown here: public string Insert(int start, string str) Here, str is inserted into the invoking string at the index specified by start The resulting string is returned You can remove a portion of a string using Remove( ), shown next: public string Remove(int start) public string Remove(int start, int count) The first form begins at the index specified by start and removes all remaining characters in the string The second form begins at count and removes count number of characters In both cases, the resulting string is returned You can replace a portion of a string by using Replace( ) It has these forms: public string Replace(char ch1, char ch2) public string Replace(string str1, string str2) The first form replaces all occurrences of ch1 in the invoking string with ch2 The second form replaces all occurrences of str1 in the invoking string with str2 In both cases, the resulting string is returned

22:

However, for the physical device to perform well dynamically it will probably be necessary to control velocities, accelerations, etc, at the boundaries Also, from Eq (520), note that if the displacement is differentiated twice, it is apparent that the acceleration of the cam, Yc(2) is governed by the second, third, and fourth derivatives of the output motion Therefore, to ensure continuous rst and second derivatives of the cam motion, Yc, the following ten boundary constraints for the output motion must be imposed (Chew et al, 1983; Wiederrich, 1981): at t = 0, Y = 0, Y (1) = 0, Y (2 ) = 0, Y (3) = 0, Y (4 ) = 0 at t = 1, Y = h, Y (1) = 0, Y (2 ) = 0, Y (3) = 0, Y (4 ) = 0 Here t is the normalized time (0 t 1), as de ned earlier Continuous cam motion acceleration is needed to improve the vibrational behavior of the follower Boundary conditions for the cam pro le curve that correspond to the above boundary conditions of the output motion are: at t = 0, Yc = 0, Yc(1) = 0, Yc(2 ) = 0, at t = 1, Yc = h ( Ks + K f ) K f ), Yc(1) = 0, Yc(2 ) = 0

Here is an example that demonstrates Insert( ), Remove( ), and Replace( ):

You Try It: Sketch f (x) = x 3 + x and its inverse Another useful fact is this: Since an invertible function must be one-to-one, two different x values cannot correspond to (that is, be sent by the function to ) the same y value Looking at Figs 151 and 152, we see that this means In order for f to be invertible, no horizontal line can intersect the graph of f more than once

// Inserting, replacing, and removing using System; class InsRepRevDemo { static void Main() { string str = "This test"; ConsoleWriteLine("Original string: " + str);

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

// Insert str = strInsert(5, "is a "); ConsoleWriteLine(str); // Replace string str = strReplace("is", "was"); ConsoleWriteLine(str); // Replace characters str = strReplace('a', 'X'); ConsoleWriteLine(str); // Remove str = strRemove(4, 5); ConsoleWriteLine(str); } }

The output is shown here:

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