(12 pts) Find the following limits

1081模模模組組組03-07班班班期期期中中中考考考解解解答答答和和和評評評分分分標標標準準準 1. (12 pts) Find the following limits.

(a) (5 pts) lim

x→∞

√

2x⁸+4x + 1 + 5x³

x⁴+1 . (b) (7 pts) lim

x→1tan⁻¹(4√ x − 4 x²−1 ).

Solution:

(a)

x→∞lim

√

2x⁸+4x + 1 + 5x³ x⁴+1 = lim

x→∞

√

2 +_x⁴7 +_x¹₈ +⁵_x 1 +_x¹4

4 pt

=

x→∞lim

√

2 +_x⁴7 +_x¹₈ + lim

x→∞

5 x

1 + lim

x→∞

1 x⁴

=

√

2. 1 pt (b) We first calculate

limx→1

4√ x − 4 x²−1 =

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎩ limx→1

4(√ x − 1) (x + 1)(√

x + 1)(√

x − 1) =lim

x→1

4 (x + 1)(√

x + 1) limx→1(

4(√ x − 1) x²−1 ⋅

√x + 1

√x + 1) =lim

x→1

4(x − 1) (x²−1)(√

x + 1) =lim

x→1

4 (x + 1)(√

x + 1) limx→1

√2 x

2x =lim

x→1

1 x√

x (by using L’Hospital rule) each is 4 pt

=1. 1 pt

Note that y = tan⁻¹x is a continuous function. Hence, limx→1tan⁻¹(

4√ x − 4

x²−1 ) =tan⁻¹(lim

x→1

4√ x − 4

x²−1 ) 1 pt

=tan⁻¹1

= π

4. 1 pt

2. (12 pts) Find the following limits.

(a) (5 pts) lim

x→0

tan⁻¹(a x)

tan⁻¹(b x), where a and b ≠ 0 are constants.

(b) (7 pts) lim

x→0(1 + sin 2x)^3x1 . Solution:

(a)

limx→0

tan⁻¹(ax)

tan⁻¹(bx) = lim

x→0

(tan⁻¹(ax))^′

(tan⁻¹(bx))^′( Type 0

0, using L’Hospital’s Rule)( 1 point)

= lim

x→0

1 1 + a²x² ⋅a

1 1 + b²x² ⋅b

(2 points)

= lim

x→0

a(1 + b²x²) b(1 + a²x²)

(1 point)

= a

b( 1 point).

(b) We have that ln(1 + sin 2x)^3x1 = 1

3xln(1 + sin 2x). (2 points)

limx→0

ln(1 + sin 2x)

3x = lim

x→0

2 cos 2x 1 + sin 2x

3 (Type0

0, Using L’Hospital’s Rule)( 2 points)

= 2

3( 1 point).

Hence

limx→0(1 + sin 2x)^3x1 = e^x→0lim

1

3xln(1+sin 2x)

= e^2/3( 2 points).

Another Method

limx→0(1 + sin 2x)^3x1 = lim

x→0(1 + sin 2x)^{sin 2x1 ⋅sin 2x3x}

= lim

x→0[(1 + sin 2x)^{sin 2x1} ]

sin 2x 3x

= e^2/3

3. (11 pts) Find the derivative of f (x).

(a) (5 pts) f (x) = x⁴³ +x ⋅ 2^(x2+1). (b) (6 pts) f (x) = x ⋅ sec⁻¹x −1

2ln (x²+1).

Solution:

(a)

f^′(x) = 4 3x¹³

±

1 pt

+ [2^(x2+1)+x ⋅ (2^(x2+1))^′]

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

1 pt(For product rule)

= 4

3x¹³ +2^(x2+1)+x ⋅ ln 2

°

1 pt

⋅2^(x2+1)

´¹¹¹¹¹¸¹¹¹¹¹¹¶

1 pt (derivative of 2^x)

⋅ 2x

¯

1 pt (chain rule)

= 4

3x^1/3+2^(x2+1)+2 ln 2 ⋅ x²⋅2^x2+1 (b)

f^′(x) = 1 ⋅ sec⁻¹x + x ⋅ 1 x√

x²−1

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

product rule 1 pt

(sec−1 x)′= 1 x

√ x2−1

2 pts

− 1 2

1 x²+1

´¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¶

1 pt (derivative of ln x)

⋅ 2x

¯

1 pt(chain rule)

=sec⁻¹x + 1

√

x²−1− x x²+1

4. (12 pts)

(a) (6 pts) f (x) = (x²+1)^{cos x}. Find f^′(x).

(b) (6 pts) Find the equation of the tangent line to the curve satisfying x²³ +y²³ +y = 6 at (8, 1).

Solution:

(a) ln f (x) = cos x ln(x²+1)

d dx

ÔÔ⇒

f^′(x)

f (x) = −sin x ln(x²+1) + cos x ⋅ 2x x²+1

⇒f^′(x) = f (x)[− sin x⋅ln(x²+1)+cos x⋅ 2x

x²+1] = (x²+1)^{cos x}[−sin x⋅ln(x²+1)+cos x⋅ 2x x²+1] 微分錯誤，一個錯誤扣1分。

(b) 2 3x⁻¹³ +

2

3y⁻¹³y^′+y^′=0 (x, y) = (8, 1) 帶入得 2

3 ⋅1 2+ (2

3+1)y^′=0, y^′= −1 y − 1 5

x − 8 = − 1 5

ˆ (3分)隱函數微分。

ˆ (1分)以y(x)和x寫出y^′(x)。

ˆ (1分)計算y^′(1).

ˆ (1分)寫出切線。

5. (16 pts) Consider the function

f (x) =

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

x³sin1

x+2x if x > 0, ax + b if x ≤ 0.

Suppose that f is differentiable everywhere.

(a) (8 pts) Find the values of a and b.

(b) (8 pts) Write down the linear approximation of f (x), at x = 0. Use the linear approximation to estimate f (0.01).

Solution:

(a) Step1: (1 point) By Squeeze Theorem we have that lim

x→0⁺f (x) = lim

x→0⁺2x + x³sin1 x =0.

Step2: ( 3 points) Since f is differentiable everywhere, f is continuous everywhere (1 point).

Thus we have that lim

x→0⁻f (x) = b = lim

x→0⁺f (x) = 0. (2 points)

Step3: (2 point) Since f is differentiable everywhere, f is differentiable at x = 0. So the following limit exists.

limh→0

f (0 + h) − f (0)

h = lim

h→0

f (h) − 0

h .

Step 4: So we have that

h→0lim⁺ f (h)

h = lim

h→0⁺

2h + h³sin1 h h

= 2(1 point)

by Squeeze Theorem, lim

h→0⁺h²sin1

h =0.(1 points)

h→0lim⁻ ah

h = lim

h→0⁻a.

Thus a = 2. Therefore if a = 2 and b = 0, then f (x) is differentiable everywhere.

(b) Since f (x) is differentiable at x = 0, the linear approximation of f (x) at x = 0 is L(x) = f (0) + f^′(0)(x − 0)( 4 points)

= 2x.(2 points) Thus f (0.01) ≈ 2 ∗ 0.01 = 0.02 (2 points).

6. (18 pts) f (x) = 3x²−x + 2 x − 2 .

(a) (6 pts) Compute f^′(x) and find interval(s) of increase of f (x) and interval(s) of decrease of f (x). Find local extreme values of f (x).

(b) (4 pts) Compute f^′′(x) and find concavity and inflection points of y = f (x).

(c) (4 pts) Find all vertical, horizontal and slant asymptotes of the curve y = f (x).

(d) (4 pts) Sketch the graph of f (x).

Solution:

(a) f^′(x) = 3(x − 4)x

(x − 2)² (2pts). f (x) is increasing for x > 4 or x < 0 (2pts). f (x) is decreasing for 0 < x < 4 (2pts).

(b) f^′′(x) = 24

(x − 2)³ (1pts). f (x) is concave upward for x > 2 (1pts). f (x) is concave downward for x < 2 (1pts). Inflection point: none (1pts).

(c) Vertical asymptotes: x = 2 (1pts). Horizontal asymptotes: none (1pts). Slant asymptotes:

f (x) = 3 x²−x + 2

x − 2 =5 + 3x + 12

x − 2 Ô⇒ y = 5 + 3 x (2pts).

(d) The curve y = f (x) (2pts). Vertical asymptotes x = 2 (1pts). Slant asymptotes y = 5 + 3 x (1pts).

-6 -4 -2 2 4 6

-40 -20 20 40 60

7. (19 pts) A firm finds that the total cost C(x)(in dollars) of manufacturing x tennis rackets/day is given by C(x) = 400 + 4x + 0.0001x². Each racket can be sold at a price of p dollars related to x by the equation p(x) = 12 − 0.0004x.

(a) (5 pts) Find the daily level of production, x1, that minimizes the average cost C(x)

x . (You need to check that the value you find is indeed the minimum value.)

(b) (4 pts) Show that the average cost, C(x)

x , equals the marginal cost C^′(x) when x = x₁. (c) (5 pts) Find the daily level of production, x₂, that maximizes the profit Π(x) = x ⋅ p(x) − C(x).

(You need to check that the value you find is indeed the maximum value.)

(d) (5 pts) Find the inverse function of p(x) = 12 − 0.0004x which is denoted by x = F (p). Find the point elasticity  = F^′(p) ⋅ p

F (p) . In the interval p ∈ (0, 12), find values of p such that −1 <  <

0(inelastic) and values of p such that  < −1(elastic).

Solution:

(a)

Let AC(x) = C(x)

x =

400

x +4 + 0.0001x (1 pt)

(AC(x))^′= − 400

x² +0.0001 (1 pt)

For x ∈ (0, 2000), (AC(x))^′<0 and (AC(x))^′>0 for x ∈ (2000, ∞) ( 2 pts

First derivative test

)

Hence AC(x) obtains minimum value at x1=2000 (1 pt)

(b)

sol 1 C(2000)

2000 =0.2 + 4 + 0.2 = 4.4 (1 pt)

C^′(x) = 4 + 0.0002x (1 pt)

C^′(2000) = 4.4 (1 pt)

Hence at x₁=2000, C(x)

x equals C^′(x) (1 pt) sol 2

(1 pt) ∵ AC(x) = C(x)

x is differentiable

∴ At the minimum value x = x1, AC^′(x1) =0 by Fermat’s Theorem (1 pt) However AC^′(x) = C^′(x) ⋅ x − C(x)

x²

(2 pts) AC^′(x₁) =0 ⇒ C^′(x₁) ⋅x₁−C(x₁) =0 ⇒ C^′(x₁) =

C(x₁) x₁ (c)

Π(x) = x ⋅ p(x) − c(x) = x ⋅ (12 − 0.0004x) − 400 − 4x − 0.0001x²

(d) (2 pts)(The inverse function of p(x)) p = 12 − 0.0004x ⇔ x = 1

0.0004(12 − p) = 2500(12 − p) i.e. F (p) = 2500(12 − p) (2 pts)(computation of )

(p) = F^′(p) ⋅ p F (p) =

−2500 ⋅ p 2500(12 − p) =

−p 12 − p For 0 < p < 12, 12 ⋅ p > 0 and

(2 pts) −1 <  < 0 ⇔ −1(12 − p) < −p < 0 ⇔ 0 < p < 6

 < −1 ⇔ −p < −(12 − p) ⇔ 12 > p > 6 Ans: −1 <  < 0 for 0 < p < 6.  < −1 for 6 < p < 12.