100 4.2 The Mean Value Theorem Figure 4.4: Right derivative. Then g is differentiable on [a,b] and g ′ (a) < 0 < g ′ − (b). Thus, + g(x) − g(a) lim < 0. x→a+ x − a It follows that there exists δ1 > 0 such that g(x) < g(a) for all x ∈ (a,a + δ1) ∩ [a,b]. Similarly, there exists δ2 > 0 such that g(x) < g(b) for all x ∈ (b− δ2,b) ∩ [a,b]. Since g is continuous on [a,b], it attains its minimum at a point c ∈ [a,b]. From the observations above, it follows that c ∈ (a,b). This implies g ′ (c)= 0 or, equivalently, that f ′ (c)= λ . □ ′ ′ Remark 4.2.1 The same conclusion follows if f (a) > λ > f − (b). + Exercises 4.2.1 ▷ Let f and g be differentiable at x0. Suppose f (x0)= g(x0) and f (x) ≤ g(x) for all x ∈ R. Prove that f ′ (x0)= g ′ (x0). 4.2.2 Prove the following inequalities using the Mean Value Theorem.
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