dx . (b) (3 分) 求在點 (0, 4) 的 d 2 y

微乙小考三 (2018/10/25)

1. (6 分) 給定方程式 3x ² + xy + y ² = 16.

(a) (3 分) 求在點 (0, 4) 的 dy

dx . (b) (3 分) 求在點 (0, 4) 的 d ² y

dx ² . sol: (a) By implicit differentiation we have

6x + y + xy ⁰ + 2yy ⁰ = 0 (x + 2y)y ⁰ = −6x − y

y ⁰ = −6x − y x + 2y Hence

dy dx  

 _(0,4) = −4 8 = −1

2

(b) Differentiate first equation in (a) with respect to x we have

6 + y ⁰ + y ⁰ + yy ⁰⁰ + 2(y ⁰ y ⁰ + yy ⁰⁰ ) = 0 6 + 2y ⁰ + xy ⁰⁰ + 2(y ⁰ ) ² + 2yy ⁰⁰ = 0 In order to find ^d _dx

^y

 _(0,4) we can take ^dy _dx  

 _(0,4) = ⁻¹ ₂ , x = 0 and y = 4 into above equation 6 + 2 · −1

2 + 0 · y ⁰⁰ + 2 · ( −1

2 ) ² + 2 · 4 · y ⁰⁰ = 0 6 − 1 + 1

2 + 8y ⁰⁰ = 0 Hence

d ² y dx ²  

 _(0,4) = −11 16 2. (6 分)

(a) (3 分) 求出曲線 y = sin ⁻¹ x 在點 x = 1

2 的切線方程式。

(b) (3 分) 估計 sin ⁻¹ (0.501).

sol: (a) Let f (x) = sin ⁻¹ x then f ⁰ (x) = ^{√ 1}

1−x

. The linear approximation at x = ¹ ₂ is L(x) = f ⁰ ( 1

2 )(x − 1 2 ) + 1

2 Since

f ⁰ ( 1

2 ) = 1 q

1 − ¹ ₂

= 2

√ 3

and

f ( 1

2 ) = sin ⁻¹ ( 1 2 ) = π

6 We can now write the linear approximation as

L(x) = 2

√ 3 (x − 1 2 ) + π

6

1

(b)

sin ⁻¹ (0.501) ≈ L(0.501) = 2

√

3 (0.501 − 0.5) + π 6

= 2 √ 3

3 · 0.001 + π 6

=

√ 3 1500 + π

6

3. (8 分) 使用均值定理證明下列敘述。

(a) (3 分) 如果 f(0) = 0 且 f ⁰ (x) > 0, x > 0 ，則 f(x) > 0, x > 0.

(b) (3 分) 試證 e ^x > 1 + x 對所有 x > 0. ( 提示: 可以利用 e ^x > 1, x > 0.) (c) (2 分) 試證 e ^x > 1 + x + 1

2 x ² , x > 0.

sol: (a) For any x > 0, by MVT we have

f (x) − f (0)

x − 0 = f ⁰ (c) where c ∈ (0, x)

Since x and f ⁰ (x) are all positive if x is positive, we know that x · f ⁰ (c) > 0.

Hence f (x) − f (0) = x · f ⁰ (c) > 0 which imply f (x) > f (0) = 0 (b) There are two methods for this subproblem

1 : For all x > 0, let f (x) = e ^x and apply MVT we have f (x) − f (0)

x − 0 = f ⁰ (c) where c ∈ (0, x)

Since f ⁰ (c) = e ^c > 1 if x > 0, we know that e ^x − e ⁰

x − 0 = e ^c > 1 e ^x − 1

x > 1 which imply

e ^x > 1 + x

2 : For all x > 0, let f (x) = e ^x − 1 − x, one can easily verify that f (0) = e ⁰ − 1 − 0 = 1 − 1 = 0 and f ⁰ (x) = e ^x − 1 > 0 if x > 0

Then apply the result of (a) we have f (x) > 0 if x > 0 which imply e ^x > 1 + x if x > 0 (c) For all x > 0, let f (x) = e ^x − 1 − x − ¹ ₂ x ² . Then one can easily verify that

f (0) = e ⁰ − 1 − 0 − 0 = 0 and

f ⁰ (x) = e ^x − 1 − x > 0 by result of (b)

Then by the result of (a) we have f (x) > 0 if x > 0, which imply e ^x > 1 + x + ¹ ₂ x ² if x > 0

2