Introduction to Mathematical Analysis I - Second Edition

110 4.2 THE MEAN VALUE THEOREM Define h ( x ) = f ( x ) − g ( x ) = f ( x ) − f ( b ) − f ( a ) b − a ( x − a )+ f ( a ) for x ∈ [ a , b ] . Then h ( a ) = h ( b ) , and h satisfies the assumptions of Theorem 4.2.2 . Thus, there exists c ∈ ( a , b ) such that h 0 ( c ) = 0. Since h 0 ( x ) = f 0 ( x ) − f ( b ) − f ( a ) b − a , it follows that f 0 ( c ) − f ( b ) − f ( a ) b − a = 0 . Thus, ( 4.5 ) holds. Example 4.2.1 We show that | sin x | ≤ | x | for all x ∈ R . Let f ( x ) = sin x for all x ∈ R . Then f 0 ( x ) = cos x . Now, fix x ∈ R , x > 0. By the Mean Value Theorem applied to f on the interval [ 0 , x ] , there exists c ∈ ( 0 , x ) such that sin x − sin0 x − 0 = cos c . Therefore, | sin x | | x | = | cos c | . Since | cos c | ≤ 1 we conclude | sin x | ≤ | x | for all x > 0. Next suppose x < 0. Another application of the Mean Value Theorem shows there exists c ∈ ( x , 0 ) such that sin0 − sin x 0 − x = cos c . Then, again, | sin x | | x | = | cos c | ≤ 1. It follows that | sin x | ≤ | x | for x < 0. Since equality holds for x = 0, we conclude that | sin x | ≤ | x | for all x ∈ R . Example 4.2.2 We show that √ 1 + 4 x < ( 5 + 2 x ) / 3 for all x > 2. Let f ( x ) = √ 1 + 4 x for all x ≥ 2. Then f 0 ( x ) = 4 2 √ 1 + 4 x = 2 √ 1 + 4 x . Now, fix x ∈ R such that x > 2. We apply the Mean Value Theorem to f on the interval [ 2 , x ] . Then, since f ( 2 ) = 3, there exists c ∈ ( 2 , x ) such that √ 1 + 4 x − 3 = f 0 ( c )( x − 2 ) . Since f 0 ( 2 ) = 2 / 3 and f 0 ( c ) < f 0 ( 2 ) for c > 2 we conclude that √ 1 + 4 x − 3 < 2 3 ( x − 2 ) . Rearranging terms provides the desired inequality.