90 4.1 Defnition and Basic Properties of the Derivative ■ Example 4.1.1 (a) Let f : R → R be given by f (x)= x and let x0 ∈ R. Then f (x) − f (x0) x − x0 lim = lim = lim 1 = 1. x→ x0 x − x0 x→ x0 x − x0 x→ x0 It follows that f is differentiable at x0 and f ′ (x0)= 1. Since x0 was an arbitrary point in R, we have f ′ (x)= 1 for all x ∈ R. (b) Let f : R → R be given by f (x)= x2 and let x 0 ∈ R. Then 2 f (x) − f (x 0) x 2 − x (x − x 0)(x + x0) lim = lim 0 = lim = lim (x + x 0)= 2x0. x→ x0 x − x0 x→ x0 x − x0 x→ x0 x − x0 x→ x0 Thus, f is differentiable at every x ∈ R and f ′ (x)= 2x. (c) Let f : R → R be given by f (x)= |x| and let x0 = 0. Then f (x) − f (0) |x| x lim = lim = lim = 1, x→ 0+ x − 0 x→ 0+ x x→ 0+ x and f (x) − f (0) |x| −x lim = lim = lim = −1. x→ 0− x − 0 x→ 0− x x→ 0− x f (x)− f (0) Therefore, limx→ 0 x−0 does not exist and, hence, f is not differentiable at 0. Theorem 4.1.1 Let I be an open interval in R and let f be defned on I. If f is differentiable at x0 ∈ I, then f is continuous at this point. Proof: We have the following identity for x ∈ I \{x0}: f (x) − f (x0) f (x)= f (x) − f (x0)+ f (x0)= (x − x0)+ f (x0). x − x0 Thus, f (x) − f (x0) lim f (x)= lim (x − x0)+ f (x0) = f ′ (x0) · 0 + f (x0)= f (x0). x→x0 x→x0 x − x0 Therefore, f is continuous at x0 by Theorem 3.3.1. □ Remark 4.1.1 The converse of Theorem 4.1.1 is not true. For instance, the absolute value function f (x)= |x| is continuous at 0, but it is not differentiable at this point (as shown in Example 4.1.1(c)). Theorem 4.1.2 Let I be an open interval in R and let f,g: I → R. Suppose both f and g are differentiable at x0 ∈ I. Then the following hold. (i) The function f + g is differentiable at x0 and ( f + g) ′ (x0)= f ′ (x0)+ g ′ (x0). (ii) For any c ∈ R, the function cf is differentiable at x0 and (cf ) ′ (x0)= cf ′ (x0).
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