981ʟ ᑖʟὃY

98學年度第1學期 微積分乙期中考解答

1. (10%) 令 f (x) = x^{ln x} , 求 f⁰(x) 。 Sol:

f(x) = x^{ln x} = e^{(ln x)2}

f⁰(x) = e^{(ln x)2} · ((ln x)²)⁰ = x^{ln x}· 2 ln x · 1

x = 2(ln x)x^{ln x−1} 2. (10%) 令 f (x) = ln (x +√

1 + x²) , 求 f⁰(x) 。 Sol:

f⁰(x) = 1 x+√

1 + x²(1 + x

√1 + x²) = 1

√1 + x²

3. (10%) 求 f (x) = 1

1 + tan x 在 x = π

4 處之切線 。 Sol:

Let f (x) = 1

(1 + tan x), then f⁰(x) = − sec²x (1 + tan x)². Because f (π

4) = 1

2, f⁰(π

4) = −1

2, we get y − 1 2 = −1

2(x − π 4).

4. (10%) 令 f (x) = 1

x , 1 ≤ x ≤ 2 。 求 ξ ∈ (1, 2) 使得 f⁰(ξ) = f(2) − f(1) 2 − 1 。 Sol:

f(x) = 1

x ⇒ f⁰(x) = −1 x² f(2) − f(1)

2 − 1 = −1 2

⇒ ξ² = 2 ⇒ ξ =√ 2

5. (10%) 用線性逼近求 tan⁻¹(1.02) 之近似值 。 (用弳度量) Sol:

Let f (x) = arctan x ⇒ f⁰(x) = 1

x²+ 1. We have

f(1.02) ≈ f(1) + (1.02 − 1) · f⁰(1)

= π

4 + 0.02 · 1 1²+ 1

= π

4 + 0.01

1

6. (15%) 求曲線 x³+ y³− 3x²y= 3 在點 (1, 2) 之 y⁰ 及 y⁰⁰ 。 (前者10% , 後者5%) Sol:

Differentiating both sides of x³+ y³− 3x²y= 3 with respect to x and regarding y as a function of x, we have

3x²+ 3y²y⁰− 6xy − 3x²y⁰ = 0. (∗) Substituting x = 1 and y = 2 into (∗), we get y⁰ = 1.

To find y⁰⁰, we differentiate (∗) implicitly with respect to x to obtain

6x + 6y(y⁰)²+ 3y²y⁰⁰− 6y − 6xy⁰− 6xy⁰− 3x²y⁰⁰ = 0.

Plugging x = 1, y = 2 and y⁰ = 1 into the above equation, we get y⁰⁰= 2 3. 7. (10%) 令 f (x) =√

1 + x + x² , x ≥ 0 。 求 (f⁻¹)⁰(√

3) =? (建議 : 不要直接求出反函數 f⁻¹(x) 之 表示式 。)

Sol:

Since x > 0, f⁻¹(x) exist. This will give:

f(1) = 3 ⇒ f⁻¹(√ 3) = 1

We have: (f⁻¹)⁰(x) = 1

f⁰(f⁻¹(x)) and f⁰(x) = 2x + 1 2√

1 + x + x² Hence,

(f⁻¹)⁰(√

3) = 1

f⁰(f⁻¹(√ 3))

= 1

f⁰(1)

= 1

3 2√

3

= 2√ 3

3 (or 2

√3)

2

8. (10%) 邊長為 12 公分之正方形紙板 。 四個角處 , 各截去大小相同之小正方形 。 剩餘部分折成無蓋之紙 盒 。 如圖 。 求最大容積 。 (不必測試所得值是否極大 。)

Sol:

Let the height of the box be x. Then both two lengthes of the box is 12 − 2x. The volume of the box is

f(x) = x(12 − 2x)², 0 ≤ x ≤ 6.

Since f (x) is differential everywhere,

the maximal value of f (x) can only happen at {x|f⁰(x) = 0, or x = 0, 6}.

f⁰(x) = x2(12 − 2x)(−2) + (12 − 2x)²(1)

= (12 − 2x)(−4x + 12 − 2x) = 12(6 − x)(2 − x)

f(2) = 128, f (6) = 0, f (0) = 0 Maximum value is f (2) = 128(cm³)

9. (15%) 令 f (x) = x⁵ − 5x + 1 。 回答下列問題 , 寫出計算過程 , 將答案填入空格 。

(a) 求出 f (x) 之極大發生在 x = , 極小發生在 x = 。

(b) 求出 f (x) 遞增之區間為 , 遞減之區間為 。

(c) 求出 f (x) 之反曲點在 x = 。

(d) 求出 f (x) 凹向上之區間為 , 凹向下之區間為 。

(e) 繪出 y = f (x) 之圖 。 Sol:

f(x) = x⁵− 5x + 1

3

⇒ f⁰(x) = 5x⁴− 5 = 5(x⁴− 1) = 5(x²+ 1)(x²− 1)

⇒ f⁰(x) = 0 ⇐⇒ x = ±1 When x < −1, f⁰(x) > 0, when −1 < x < 1, f⁰(x) < 0, when x > 1, f⁰(x) > 0.

(a) The relative maximal point is x = −1, the relative minimal point is x = 1.

(b) The increasing area is x < −1 or x > 1, the decreasing area is −1 < x < 1.

Since f⁰⁰(x) = 20x³, ⇒ f⁰⁰(x) = 0 ⇐⇒ x = 0.

When x < 0, f⁰⁰(x) < 0; when x > 0, f⁰⁰(x) > 0.

(c) The inflective point is x = 0.

(d) Concave upward area is x > 0, concave downward area is x < 0,

and f (−1) = −1 + 5 + 1 = 5, f(1) = 1 − 5 + 1 = −3, f(0) = 0 − 0 + 1 = 1.

(e) The graph is

4