第二章第二節

第二章第二節^, 平均值定理 2.2(1)

(tan⁻¹x)⁰ = 1 1 + x²

∫ 1

1 + x²dx = tan⁻¹x + C

2.2(2)

(sin⁻¹x)⁰ = 1

√1− x²

∫ 1

√1− x²dx = sin⁻¹x + C

2.2(3)

(− cos x)⁰ = sin x

∫

sin xdx = − cos x + C

2.3(1)

欲證^{(1 + x)r}≤ 1 + rx as 0 ≤ r ≤ 1, x > 0 即證^{(1 + x)r− 1}

(1 + x)− 1 ≤ r

令f (x) = (1 + x)^r f⁰(x) = (1 + x)^r∗ r ∗ 1 1 + x By平均值定理 可知

(1 + x)^r− 1

(1 + x)− 1 = (1 + y)^r∗ r ∗ 1

1 + y 0≤ y ≤ x

= (1 + y)^r−1∗ r ≤ r

2.3(2)

for r < 0, x > 0 (1 + x)^r− 1

(1 + x)− 1 = (1 + y)^r∗ r ∗ 1

1 + y 0≤ y ≤ x

= (1 + y)^r−1∗ r ≥ r 所以 ^{(1 + x)r} ≥ 1 + rx

1

2.6

Because v(t) ≥ 0 as 0 ≤ t ≤ π Hence 最遠的距離^{=∫ π}

0

sin xdt = 2

2.8

令^{F (x) = x}− tan⁻¹x

F (0) = 0 y = x與^{y = tan−1x}至少交一點 F (x)− F (0)

x− 0 = F⁰(b) = 1− 1

1 + b² > 0 0 < b < x as 0 < x≤ 1 F (x) = (1− 1

1 + b²)x > 0 (0,1]上無交點 同理^{F (x) = (1}− 1

1 + b²)x < 0 x < b < 0 as −1 ≤ x < 0 [-1,0)上無交點 故僅交一點

2.9

不妨設y = f (x) = 0有^k 個相異實根^a1 < a₂ < ... < a_k 因為^{f (a}i) = f (a_i+1) = 0 i=1,...,k-1

By Rolle’s thm 知 存在^ai < b_i < a_i+1使得^f0(bi) = 0 所以^{f0(x) = 0}至少有^b1, ..., b_k−1共 ^k-1個相異實根

2.10

from 2.9 可知

y = f (x) = 0有^k 個相異實根 所以^{f0(x) = 0}至少有 ^k-1個相異實根 thenf⁰⁰(x) = 0至少有^k-2個相異實根 then...重複此論調

sof^(k−1)(x) = 0至少有一實根

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