98 4.2 The Mean Value Theorem such that h′(c) =0. Since h′(x) =f ′(x)− f (b)−f (a) b−a , it follows that f ′(c)− f (b)−f (a) b−a =0. Thus, (4.4) 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, fix 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 x0 −0 =cosc. Therefore, |sinx0| |x0| =|cosc|. Since |cosc| ≤1 we conclude |sinx0| ≤ |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 0−x0 =cosc. Then, again, |sinx0| |x0| =|cosc| ≤1. It follows that |sinx0| ≤ |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√1+4x = 2 √1+4x . Now, fix x0 ∈Rsuch 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 p1+4x0 −f (2) =p1+4x0 −3=f ′(c)(x0 −2). Since f ′(2) =2/3 and f ′(c) <f ′(2) for c >2 we conclude that p1+4x0 −3< 2 3 (x0 −2) = 2 3 x0 − 4 3 . Hence, p1+4x0 < 2 3 x0 − 4 3 +3= (5+2x0)/3. Since x0 >2 is arbitrary, the result follows for every x >2.
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