100 4.2 The Mean Value Theorem Figure 4.4: Right derivative. Then g is differentiable on[a,b] andg′+(a) <0<g′ − (b). Thus, lim x→a+ g(x)−g(a) x−a <0. 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 andg 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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