Introduction to Mathematical Analysis I - Second Edition

158 Solutions and Hints for Selected Exercises Then h has an absolute maximum at x 0 . Thus, h 0 ( x 0 ) = f 0 ( x 0 ) − g 0 ( x 0 ) = 0 , which implies f 0 ( x 0 ) = g 0 ( x 0 ) . Exercise 4.2.3 . The inequality holds obviously if a = b . In the case where a 6 = b , the equality can be rewritten as sin ( b ) − sin ( a ) b − a ≤ 1 . The quotient sin ( b ) − sin ( a ) b − a is the slope of the line connecting ( a , f ( a )) and ( b , f ( b )) . We need to show that the absolute value of the slope is always bounded by 1, which can also be seen from the figure. The quotient also reminds us of applying the Mean Value Theorem for the function f ( x ) = sin ( x ) . Figure 5.1: The function f ( x ) = sin ( x ) . Consider the case where a < b and define the function f : [ a , b ] → R by f ( x ) = sin ( x ) . Clearly, the function satisfies all assumptions of the Mean Value Theorem on this interval with f 0 ( x ) = cos ( x ) for all x ∈ ( a , b ) . By the Mean Value Theorem, there exists c ∈ ( a , b ) such that f ( b ) − f ( a ) b − a = f 0 ( c ) = cos ( c ) , which implies f ( b ) − f ( a ) b − a = | cos ( c ) | ≤ 1 . It follows that | f ( a ) − f ( b ) | ≤ | a − b | . The solution is similar for the case where a > b . It is essential to realize that the most important property required in solving this problem is the boundedness of the derivative of the function. Thus, it is possible to solve the following problems