98 4.2 The Mean Value Theorem such that h′ (c)= 0. Since f (b) − f (a) h ′ (x)= f ′ (x) − , b − a it follows that f (b) − f (a) f ′ (c) − = 0. b− a Thus, (4.3) holds. □ ■ Example 4.2.1 We show that |sinx|≤|x| for all x ∈ R. Let f (x)= sinx for all x ∈ R. Then f ′ (x)= cosx. Now, fx x0 ∈ R, x0 > 0. By the Mean Value Theorem applied to f on the interval [0,x0], there exists c ∈ (0,x0) such that sinx0 − sin0 = cosc. x0 − 0 Therefore, |sin |x0 x | 0| = |cosc|. Since |cosc|≤ 1 we conclude |sinx 0|≤|x0|. Since x0 was an arbitrary positive real number, we can conclude that |sinx|≤|x| for all x > 0. Next suppose x0 < 0. Another application of the Mean Value Theorem shows there exists c ∈ (x0,0) such that sin0− sinx0 = cosc. 0− x0 Then, again, |sin |x0 x | 0| = |cosc|≤ 1. It follows that |sinx 0|≤|x0| for x0 < 0. Since x0 was an arbitrary negative real number, we have shown that |sinx|≤|x| for all x < 0. Since equality holds for x0 = 0, we conclude that |sinx|≤|x| for all x ∈ R. √ ■ Example 4.2.2 We show that 1 + 4x < (5 + 2x)/3 for all x > 2. We consider the function √ f (x)= 1 + 4x for all x ≥ 2. Then f ′ (x)= √ 4 = √ 2 . 2 1+ 4x 1+ 4x Now, fx x0 ∈ R such that x0 > 2. We apply the Mean Value Theorem to f on the interval [2,x0]. Then, since f (2)= 3, there exists c ∈ (2,x0) such that p p 1+ 4x0 − f (2)= 1 + 4x0 − 3 = f ′ (c)(x0 − 2). Since f ′ (2)= 2/3 and f ′ (c) < f ′ (2) for c > 2 we conclude that p 2 2 4 1 + 4x0 − 3 < (x0 − 2)= x0 − . 3 3 3 Hence, p 2 4 1+ 4x0 < x0 − + 3 =(5 + 2x0)/3. 3 3 Since x0 > 2 is arbitrary, the result follows for every x > 2.
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