6.2(Kalo Ini Nemu Di Internet)

#2) (a) f (x)=1 - 1/x 2 for x = 0. f (x) ≥ 0 for x ≥ 1 and x ≤-1. f (x) ≤ 0 for x ∈ [-1, 0) ∪ (0, 1]. f is increasing on (-∞, -1] ∪ [1, ∞) and decreasing on [-1, 0) ∪ (0, 1]. It has a relative maximum at -1 and a relative minimum at 1. (b) g (x)= 1-x 2 (x 2 +1) 2 . g ≥ 0 on [-1, 1], and g ≤ 0 on (-∞, -1] ∪ [1, ∞). g is increasing on [-1, 1], and g is decreasing on (-∞, -1] ∪ [1, ∞). g has a relative minimum at -1 and a relative maximum at 1. (c) h (x)= 1 2 √ x - 1 √ x+2 for x> 0. 2 √ x ≤ √ x + 2 if and only if 0 <x ≤ 2/3. Hence, h ≥ 0 on (0, 2/3] and h ≤ 0 on [2/3, ∞). Therefore, h is increasing on (0, 2/3] and decreasing on [2/3, ∞). It has a relative maximum at 2/3. (d) k (x)=2 - 2/x 3 . k (x) ≥ 0 for x ≥ 1 and for x ≤ 0. k (x) ≤ 0 for x ∈ (0, 1]. k is increasing on (-∞, 0) ∪ [1, ∞) and decreasing on (0, 1]. It has a relative minimum at 1. #5) f (x)= x 1/n - (x - 1) 1/n for x ≥ 1. f (x)= 1 n ( x (1-n)/n - (x - 1) (1-n)/n ) . Since the exponent is negative, if x> 1, then x>x - 1 > 0, and it follows that x (1-n)/n < (x - 1) (1-n)/n . Hence, f (x) < 0 for x> 1. So f is strictly decreasing. Since a/b > 1, this implies that f (a/b) <f (1). But f (1) = 1. Hence, a 1/n /b 1/n - (a - b) 1/n /b 1/n < 1. So a 1/n - (a - b) 1/n <b 1/n and so a 1/n - b 1/n < (a - b) 1/n . #6) We can assume without loss of generality that x<y. By the Mean-Value Theorem, there exists c ∈ (x, y) such that sin y-sin x y-x = cos c. Therefore, | sin x - sin y| = | cos c||x - y|≤|x - y|. 1 #8) Let > 0. Choose δ> 0 such that if 0 < |x - a| <δ and x ∈ [a, b], then |f (x) - A| <δ. Suppose x ∈ [a, b] such that 0 < |x - a| <δ. Then x>a and so by the Mean-Value Theorem, there exists c ∈ (a, x) such that f (x)-f (a) x-a = f (c). a<c<x<a + δ, so 0 < |c - a| <δ. Hence, f (x) - f (a) x - a - A = |f (c) - A| < . Therefore, f (a) exists, and equals A. #11) f (x)= √ x is continuous on [0, 1], and so uniformly continuous. It is diﬀerentiable on (0, 1), with derivative 1 2 √ x , which is not bounded. #13) Suppose that a<b, a, b ∈ I . We must show that f (a) <f (b). But f (b)-f (a) b-a = f (c) > 0 for some c ∈ I . Since b - a> 0, it implies f (b) - f (a) > 0 as desired. 6.2

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Transcript of 6.2(Kalo Ini Nemu Di Internet)

#2) (a) f ′(x) = 1− 1/x2 for x 6= 0. f ′(x) ≥ 0 for x ≥ 1 and x ≤ −1. f ′(x) ≤ 0 for x ∈ [−1, 0) ∪ (0, 1]. fis increasing on (−∞,−1] ∪ [1,∞) and decreasing on [−1, 0) ∪ (0, 1]. It has a relative maximumat −1 and a relative minimum at 1.

(b) g′(x) = 1−x2

(x2+1)2 . g′ ≥ 0 on [−1, 1], and g′ ≤ 0 on (−∞,−1]∪ [1,∞). g is increasing on [−1, 1], andg is decreasing on (−∞,−1] ∪ [1,∞). g has a relative minimum at −1 and a relative maximumat 1.

x− 1√

x+2for x > 0. 2

√x ≤

√x + 2 if and only if 0 < x ≤ 2/3. Hence, h′ ≥ 0 on

(0, 2/3] and h′ ≤ 0 on [2/3,∞). Therefore, h is increasing on (0, 2/3] and decreasing on [2/3,∞).It has a relative maximum at 2/3.

(d) k′(x) = 2 − 2/x3. k′(x) ≥ 0 for x ≥ 1 and for x ≤ 0. k′(x) ≤ 0 for x ∈ (0, 1]. k is increasing on(−∞, 0) ∪ [1,∞) and decreasing on (0, 1]. It has a relative minimum at 1.

#5) f(x) = x1/n − (x − 1)1/n for x ≥ 1. f ′(x) = 1n

(x(1−n)/n − (x− 1)(1−n)/n

). Since the exponent is

negative, if x > 1, then x > x− 1 > 0, and it follows that x(1−n)/n < (x− 1)(1−n)/n. Hence, f ′(x) < 0for x > 1. So f is strictly decreasing. Since a/b > 1, this implies that f(a/b) < f(1). But f(1) = 1.Hence, a1/n/b1/n − (a− b)1/n/b1/n < 1. So a1/n − (a− b)1/n < b1/n and so a1/n − b1/n < (a− b)1/n.

#6) We can assume without loss of generality that x < y. By the Mean-Value Theorem, there existsc ∈ (x, y) such that sin y−sin x

y−x = cos c. Therefore, | sinx− sin y| = | cos c||x− y| ≤ |x− y|.

#8) Let ε > 0. Choose δ > 0 such that if 0 < |x − a| < δ and x ∈ [a, b], then |f ′(x) − A| < δ. Supposex ∈ [a, b] such that 0 < |x − a| < δ. Then x > a and so by the Mean-Value Theorem, there existsc ∈ (a, x) such that f(x)−f(a)

x−a = f ′(c). a < c < x < a + δ, so 0 < |c− a| < δ. Hence,∣∣∣∣f(x)− f(a)x− a

−A

∣∣∣∣ = |f ′(c)−A| < ε.

Therefore, f ′(a) exists, and equals A.

#11) f(x) =√

x is continuous on [0, 1], and so uniformly continuous. It is differentiable on (0, 1), withderivative 1

2√

x, which is not bounded.

#13) Suppose that a < b, a, b ∈ I. We must show that f(a) < f(b). But f(b)−f(a)b−a = f ′(c) > 0 for some

c ∈ I. Since b− a > 0, it implies f(b)− f(a) > 0 as desired.

6.2