1061ᵬᵬᵬ03-04ஹஹஹ06-10IIIʟʟʟὃὃὃYYYȤȤȤᑖᑖᑖ 1. (20 points) Compute each of the following limits if it exists or explain why it doesn’t exist. (a) (5 points) lim

1061微微微甲甲甲03-04、、、06-10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. (20 points) Compute each of the following limits if it exists or explain why it doesn’t exist.

(a) (5 points) lim

x→0sin(_x¹²) sin x.

(b) (5 points) lim

x→0 tan x

√

1−cos 3x. (c) (5 points) lim

x→0(cos x)^x22 . (d) (5 points) lim

x→+∞

√x+√ x−√

x−√ x.

Solution:

(a) Method 1

(2 points) −1 ≤ sinx¹² ≤ 1 ⇒ 0 ≤ ∣ sinx¹²∣ ≤ 1 ⇒ 0 ≤ ∣ sin x sinx¹²∣ ≤ ∣ sin x∣

(2 points) lim

x→0sin x= 0 ⇒ lim_x→0∣ sin x∣ = 0 = lim_x→00.

(1 point) By Squeeze Theorem, lim

x→0∣ sin x sin_x¹²∣ = 0 ⇒ lim_x→0sin x sin_x¹2 = 0.

Method 2 To find the limit as x approaching 0, only to consider x∈ (−π, π).

(1 point) When x> 0, −1 ≤ sinx¹² ≤ 1 ⇒ − sin x ≤ sin x sinx¹² ≤ sin x.

(1 point) lim

x→0⁺− sin x = 0 = lim_x→0+sin x.

(1 point) When x< 0, −1 ≤ sinx¹² ≤ 1 ⇒ sin x ≤ sin x sinx¹² ≤ − sin x.

(1 point) lim

x→0⁻sin x= 0 = lim_x→0−− sin x.

(1 point) By Squeeze Theorem, lim

x→0⁺sin x sin_x¹2 = 0 = lim_x→0−sin x sin_x¹2. Hence, lim

x→0sin x sin_x¹2 = 0.

(b) (2 points) ^{√tan x}

1−cos 3x = ^{tan x}x

√ x

1−cos 3x = (cos x¹ x sin x)(3∣x∣^x

√ _(3x)2

1−cos(3x)) = ¹3 1 cos x

sin x x

x

∣x∣

√ _(3x)2

1−cos(3x)

for x≠ 0.

(1 point) lim

x→0⁺ tan x

√

1−cos 3x = lim_x→0+(¹_{3cos x}^{1 sin x}_{x ∣x∣}^x√ _(3x)2

1−cos(3x)) = ¹₃⋅ 1 ⋅ 1 ⋅ 1 ⋅√

2=^√₃². (1 point) lim

x→0⁻ tan x

√

1−cos 3x = lim_x→0−(¹3 1 cos x

sin x x

x

∣x∣

√ _(3x)2

1−cos(3x)) = ¹3⋅ 1 ⋅ 1 ⋅ (−1) ⋅√

2= −^√3². (1 point) lim

x→0⁺ tan x

√

1−cos 3x = ^√₃² ≠ −^√₃² = lim

x→0⁻ tan x

√

1−cos 3x ⇒ lim_x→0^√_{1−cos 3x}^{tan x} doesn’t exist.

(c) Observe that

limx→0(cos(x))^2/x2 = lim_x→0e2 ln(cos(x))/x² = e^x→0lim2 ln(cos(x))/x²

(1 pt).

Now since

limx→0

2 ln(cos(x)) x²

L= lim_x→02× − sin(x)/ cos(x)

2x (2 pts)

= lim_x→0− sin(x)

x × 1

cos(x)

= −1 (1 pt) × 1

= −1 (1 pt), we conclude that lim

x→0(cos(x))^2/x2 = e^x→0lim2 ln(cos(x))/x²

= e⁻¹.

(d)

x→+∞lim

√x+√ x−√

x−√

x = lim_x→+∞(x +√

x) − (x −√ x)

√x+√ x+√

x−√ x

(2 pts)

= lim_x→+∞ 2√

√ x x+√

x+√ x−√

x = lim_x→+∞ 2

√ 1+√

x x +

√ 1−√

x x

= lim_x→+∞ 2

√ 1+ 1

√x+

√ 1− 1

√x

(2 pts) = 2

√1+ 0 +√ 1− 0

= 2

1+ 1 = 1 (1 pt)

2. (15 points) Differentiate the following functions.

(a) (5 points) f(x) =1+cos x^{sin x} . (b) (5 points) f(x) = log2

√x+ tan⁻¹(x³).

(c) (5 points) f(x) = x^{cos x}.

Solution:

(a) (Method 1) Use quotient rule

f^′(x) =

³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ1pt cos x(cos x + 1) −

³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ1pt sin x(− sin x) (1 + cos x)²

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

1pt

= 1+ cos x

(1 + cos x)² = 1 1+ cos x

(All correct +2pts.)

(Method 2) Use product rule

f^′(x) = cos x

1+ cos x + (sin x) ⋅ (− (1+ cos x)^′ (1 + cos x)²) =

³¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µcos x1pt 1+ cos x +

³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ2pts sin²x (1 + cos x)²) (All correct +2pts.)

(b) (Method 1) Simplify f(x) as

f(x) =

­1pt ln x

2 ln 2 + tan⁻¹(x³) Then

f^′(x) =

³¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ1pt 1 (2 ln 2)x +

«1pt 3x² ⋅

³¹¹¹¹·¹¹¹¹µ1pt 1 1+ x⁶ (All correct +1pt.)

(Method 2) Differentiate f(x) directly

f^′(x) =

³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ1pt 1 (ln 2)√

x⋅

¬1pt 1 2√

x+

«1pt 3x² ⋅

³¹¹¹¹·¹¹¹¹µ1pt 1 1+ x⁶ (All correct +1pt.)

(c) (Method 1) Write f(x) as

³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ1pt f(x) = e^{cos x ln x} Then differentiate f(x)

f^′(x) =

³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ1pt

e^{cos x ln x}(cos x ln x)^′= e^{cos x ln x}

³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ1pt (− sin x ln x +cos x

x )

(All correct +2pts.)

(Method 2) Use logarithmic differentiation. Write f(x) as

³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ1pt ln f(x) = cos x ln x Differentiate

­1pt f^′(x)

f(x) =

³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ1pt (− sin x ln x +cos x

x ) Thus,

f^′(x) = x^{cos x}(− sin x ln x +cos x x ) (All correct +2pts.)

Remark 計算錯誤至少扣2分，答案正確但沒有計算過程或說明扣1分，關鍵過程大致正確但抄寫錯

誤扣1分。

3. (12 points) Let f(x) =⎧⎪⎪⎪

⎨⎪⎪⎪⎩

x⁴³ cos(1

x) , for x ≠ 0

0, for x= 0.

(a) (3 points) Is f(x) continuous at x = 0?

(b) (6 points) Compute f^′(x) for x ≠ 0 and f^′(0).

(c) (3 points) Is f^′(x) continuous at x = 0?

Solution:

(a) Because the cosine of any number lies between −1 and 1, we can write.

−1 ≤ cos (1 x) ≤ 1

Any inequality remains true when multiplied by a positive number. We know that x⁴³ ≥ 0 for all x and so, multiplying each side of the inequalities by x⁴³, we get

−x⁴³ ≤ x⁴³ cos(1 x) ≤ x⁴³ We know that

limx→0x⁴³ = lim_x→0−x⁴³ = 0 By Squeeze Theorem, we obtain

limx→0x⁴³ cos(1 x) = 0 Therefore, f(x) is continuous at x = 0.

(b) By definition of differential and pinching theorem,

f^′(0) = lim_x→0x⁴³ cos(¹x) − f(0)

x− 0 = lim_x→0x⁴³ cos(¹x)

x = lim_x→0x¹³cos(1 x) = 0

On the other hand, we consider x /= 0. By the Product Rule, f^′(x) = (x⁴³)^′cos(_x¹) + x⁴³{cos (¹x)}^′. Now by Chain Rule, {cos (¹x)}^′= sin (¹x) ⋅ _x¹². Therefore, we obtain

f^′(x) =4

3x¹³cos(1

x) + x⁻²³ sin(1 x) (c) Because lim sup

x→0

f^′(x) = ∞ and lim inf_x→0 f^′(x) = −∞, lim_x→0f^′(x) does not exist. Therefore, we can deduce that f^′(x) is not continuous at x = 0.

[Remark]

In question (b), suppose you know the definition of f^′(0) = lim_x→0f (x)−f (0)

x−0 . You can get 2 points.

Suppose you write f^′(0) = lim_x→0f^′(x). You can not get any point in (b).

4. (10 points) The figure shows a lamp located 4 units to the right of the y−axis and a shadow created by the elliptical region x²+5y²≤ 6. If the point (−6, 0) is on the edge of the shadow, how far above the x−axis is the lamp located?

? x y

0 3 _5

≈+4¥=5

x

²

+ 5y

²

= 6

. 6 4

Solution:

1. (Method 1)

By implicit differentiation, we have 2x+ 10yy^′= 0, or, y^′= −5y^x (5%)

Suppose the point of tangency is (x^o, y_o), then the tangent line is given by y = −5y^xoo(x − x_o) + y^o

Plug in (−6, 0), we have 0 = −5y^xoo(−6 − x^o) + y^o, or, x²_o+ 5yo²= −6x^o. Also, x²_o+ 5y²o = 6, so xo= −1, yo= 1(3%)

Then the tangent line is y =¹5x+⁶5, so y∣

x=4= 2(2%) 2. (Method 2)

Suppose the lamp is located at (4, h). Then the tangent line is given by y = 10^h(x+6)(4%) Since it’s tangent to the ellipse, the equation

⎧⎪⎪⎨⎪⎪

⎩

y= 10^h(x + 6) x²+ 5y²= 6

should have only one zero(repeated roots), or equivalently, the discriminant of x² + 5(10^h(x + 6))²= 6 sholud be zero.(4%)

Thus, 36h⁴− (20 + h²)(36h²− 120) = 0, we have h = 2(2%) 3. (Method 3) Suppose the point of tangency is(xo,

√6−x²_o

5 ), and the lamp is located at (4, h) Then we have

√

6−x2o 5 −0

xo−(−6) = 4−(−6)^h−0 , or h= ²

√ 30−5x²_o xo+6 (4%) We can find h by using the condition ^dh_dx = 0(why?)(4%) Thus, 0=

−10xo

√30−5x2o

⋅(xo+6)−2√ 30−5x²_o⋅1

(xo+6)² = ^−10x_(xo^2o₊₆₎^−60x2√^o−60+10x2o 30−5x²_o

Hence, x_o= −1, y^o= 1, and h = 2(2%)

5. (10 points)

(a) (3 points) Find the linearization of f(x) = sin x at ^π₆.

(b) (7 points) Explain why f satisfies the hypotheses of the Mean Value Theorem on [^π₆,^π₂]. Use the theorem to prove that

sin x< 1 2+

√3 2 (x −π

6) for x ∈ (π 6,π

2]

Solution:

5. (10 points)

(a) (3 points) Find the linearization of ( ) sin at 6

(b) (7 points) Explain why satisfies the hypotheses of the Mean Value Theorem on , . Use the theorem to prove that

6 2

f x x

f



 



 

 

 

1 3

sin ( ) for ( , ]

2 2 6 6 2

x_{ } x_



x_

 

0 0 0 0

6

Solution:

(a)

The linearization of ( ) is

( ) ( ) ( )( ), 6

1 3

Note that ( ) and ( ) (sin ) cos

6 2 6 6 2

Therefore, the linearization o

(2

f ( ) at i

pt

6 s

f

s)

x

f x

L x f x f x x x x

f f x

f x





  





    

 

   

1 3

( ( ) ) ( )

2 2 6

(b)

( ) sin is differentiable on .

So it is continuous on the close interval [ , ], and is dif (1 pt for cor

ferentiable on the open 6 2

interval

rect equation)

( , ) sa 6 2

L x

f

x

f x x



 

 

  

 R

tisfying the hypotheses of the Mean Value Theorem on [ , ].

6 2

If is in the interval ( , ], we can still apply MVT on [ , ] to obtain a in

6 2 6

( , ) such that:

6

(2 pts for required hypoth

es s) e

x x c

x

 

  



sin sin ( ) cos 6

6

Since cos is strictly decreasing on ( , ) (

(2 pts for applying MVT correctly)

(2

(cos ) 0), we get the inequality 6 2

cos 3 for c ( , ) pts

2 6 2

x

f c c

x

x x

c





 

 

   



 

 

 )

Then

sin sin 6 3

( ) cos for ( , ]

2 6 2

6 Therefore,

1 3

sin ( ) for ( , ]

(

2 2 6 6 2

1 pt for deducing the inequality)

x

f c c x

x

x x x

  



  

     



   

: using the racetrack principle Set

1 3

( ) sin and ( ) ( )

2 2 6

( ) 3 2

( ) cos and ( ) 3

6 2

Since (cos ) 0 on ( , ], cos is decreasing on ( , ]

6 2 6 2

Thus

( ) ( ) f

f x x L x x

L x

f x x f

x x

f x L x





   

   

 

   

 

   Method 2

or ( , ] 6 2 Note that

( ) ( ) 1

6 6 2

By the racetrack principle, ( ) ( ) for ( , ]

6 2 Therefore, we get the inequality

1 3

sin ( ) for ( , ]

2 2 6 6 2

x

f L

f x L x x

x x x

 

 

 

  



 

 

   

6. (10 points) In the engine, a 7-inch connecting rod is fastened to a crank of radius 3 inches. The crankshaft(機軸) rotates counterclockwise at a constant rate of 100 revolutions per minute. Find the velocity of the piston(活塞) when θ = ^π₃. (Reminder: the angular velocity of a circular motion at a constant speed of 1 revolution per minute is 2π rad/min.)

Solution:

First, we know ^dθ_dt = 100 ⋅ 2π = 200π (1 point) (Idea I)

∵7²= 3²+ x²− 2 ⋅ 3 ⋅ x cos θ (3 points)

∴0 = 2x^dxdt − 6(^dxdt cos θ− x sin θ^dθdt)

⇒ (6 cos θ − 2x)^dx_dt = 6x sin θ^dθ_dt ⇒ ^dx_dt = _{6 cos θ−2x}^{6x sin θ dθ}_dt (3 points) When θ= ^π3, then x= 8. ⇒ ^dxdt = −⁴⁸⁰⁰13^√3π. (2 points) (Idea II)

∵7²= 3²+ x²− 2 ⋅ 3 ⋅ x cos θ ⇒ x²− 6 cos θ − 40 = 0 (3 points)

∴x = 3 cos θ +√

9 cos²θ+ 40 (for x > 0) ⇒ ^dx_dt = −3 sin θ^dθ_dt +¹₂⋅ −18 cos θ sin θ√

9 cos²θ+40 ⋅^dθ_dt (3 points) when θ= ^π₃, we have ^dx_dt = [−3 ⋅ (^√₂³) −¹₂^18⋅12⋅

√3

√ 2 9

4+40] ⋅ 200π = −⁴⁸⁰⁰₁₃^√3π (2 points) (Idea III)

∵ cos θ = ^32+x_2⋅3⋅x²⁻⁷²(⇒ x = 3 cos θ +√

9 cos²θ+ 40 (for x > 0)) (3 points)

∴ − sin θ^dθdt = ^{2x⋅6xdxdt−6(x}36x^22−40)dxdt ⇒ − sin θ^dθdt =^x26x⁺⁴⁰² dx

dt (3 points) When θ= ^π3, then x= 8. ⇒ ^dxdt = − sin θ^dθdt ⋅x^6x2+40² ∣

x=8,θ=^π₃ = −⁴⁸⁰⁰13^√3π. (2 points) (Idea IV)

∵x = 3 cos θ +√

49− (3 sin θ)²(= 3 cos θ +√

40+ 9(1 − sin²θ) = 3 cos θ +√

9 cos²θ+ 40) (3 points)

∴^dxdt = −3 sin θ^dθdt +¹2⋅−18 sin θ cos θ√

49−9 sin²θ ⋅^dθdt (3 points) when θ= ^π3, we have ^dx_dt = [−3 ⋅ (^√2³) −¹2

18⋅

√3 2 ⋅¹

√ 2

49−²⁷₄ ] ⋅ 200π = −⁴⁸⁰⁰13^√3π (2 points) (Idea V)

7 sin φ= 3 sin θ ⇒ 7 cos φ^dφ_dt = 3 cos θ^dθ_dt

⇒ ^dφdt =7 cos φ^{3 cos θ} dθ

dt = ^√_{49−49 sin}^{3 cos θ} 2φ dθ

dt = ^√_{49−9 sin}^{3 cos θ}2θ dθ

dt (2 points)

∵x = 3 cos θ + 7 cos φ (= 3 cos θ + 7√

1−₄₉⁹ sin²θ= 3 cos θ + 7√

49− 9 sin²θ) (1 point)

∴^dx_dt = −3 sin θ^dθ_dt − 7 sin φ^dφ_dt = −3 sin θ^dθ_dt − 3 sin θ (^√_{49−9 sin}^{3 cos θ}2θ dθ

dt) (3 points) when θ= ^π₃, we have ^dx_dt = [−3 ⋅ (^√₂³) −¹₂^18⋅

√3 2 ⋅¹

√ 2

49−²⁷₄ ] ⋅ 200π = −⁴⁸⁰⁰₁₃^√3π (2 points) (Idea VI)

7 sin φ= 3 sin θ ⇒ 7 cos φ^dφdt = 3 cos θ^dθdt, when θ= ^π3 ⇒ sin φ = ³14^√3, cos φ=¹³14

⇒ ^dφ_dt =_{7 cos φ}^{3 cos θdθ}_dt = ^√_{49−49 sin}^{3 cos θ} 2φ dθ

dt = ^√_{49−9 sin}^{3 cos θ}2θ dθ

dt (2 points)

∵sin(π−φ−θ)^x = sin θ⁷ ⇒ x = 7^{sin(π−φ−θ)}sin θ = 7^sin(θ+φ)sin θ (1 point)

∴^dxdt = 7^{cos(θ+φ)⋅(dθdt+dφdt})sin θ−cos θ sin(θ+φ)^dθ_dt

sin²θ (3 points)

When θ= ^π₃, we have ^dx_dt = 7^[(

1 2)(¹³

14)−(

√3 2 )(³

√3 14 )]⋅[

3

√ 2 49− 274

+1]

√3 2 −¹₂[(

√3

2 )(¹³₁₄)+(¹

2)(³

√3 14 )]

3

4 ⋅ 200π

⇒ ^dxdt = 7 [(28⁴) (¹⁶13)^√2³ −¹⁶56^√3] ⋅⁸⁰⁰3 π= 7 ⋅^{−9⋅16⋅800}3⋅56⋅13^√3π = −⁴⁸⁰⁰13^√3π (2 points) That is the velocity of the piston is ⁴⁸⁰⁰

√ 3

13 π inch/min. with the direction to the left. (1 point)

7. (10 points) A right circular cone is inscribed in a larger right circular cone so that its vertex is at the center of the base of the larger one. Denote the height of the large cone by H and the height of the small one by h. When the large cone is fixed, find h that maximizes the volume of the small cone and find out this maximum volume in terms of the volume of the large cone. (Hint: The volume of a right circular cone with height h and base radius r is ¹₃πr²h.)

H

h

Solution:

We denote the radii of the bases of the larger and smaller cones as R and r, respectively. Then we have the relation

H− h

H = r

R. This gives us

r= R

H(H − h).

(2 points)

Hence the volume of the small cone is

V_small= 1 3πr²h

= 1

3πh⋅ R²

H²(H − h)²

for 0≤ h ≤ H

(2 points). In order to compute the maximum of V

_small, we compute dV_small

dh = πR²

3H²(H − h)(H − 3h).

(2 points)

By setting ^dVsmall_dh = 0 we have h = H or h = H/3

(2 points). Since

V(0) = 0, V(H) = 0, V(H

3) = 4

81πR²H> 0, we know that when h= H/3, the maximum of V^small is

V_small= 4

81πR²H= 4

27V_large,

(2 points)

where V_large= ¹3πR²H is the fixed volume of the larger cone.

註1: h= ^H3 and V_small =27⁴V_large should both be answered. Not answering both of the two will cost you 2 points.

註2: Assuming H = 2R will cost you 2 points.

8. (18 points) Let f(x) = ^{ln ∣x∣}_x , x≠ 0. Answer the following questions by filling each blank below and give your reasons (including computations). Put None in the blank if the item asked does not exist.

(a) (3 points) Find all asymptote(s) of the curve y= f(x).

Vertical asymptote(s): .

Horizontal asymptote(s): .

Slant saymptote(s): .

(b) (4 points) f(x) is increasing on the interval(s) .

f(x) is decreasing on the interval(s) .

(c) (2 points) Find all local extreme values of f(x).

Local maximum point(s): (x, f(x)) = .

Local minimum point(s): (x, f(x)) = .

(d) (4 points) f(x) is concave upward on the interval(s) .

f(x) is concave downward on the interval(s) .

(e) (2 points) List the inflection point(s) of the curve y= f(x) ∶ (x, f(x)) = . (f) (3 points) Sketch the graph of f , and indicate all asymptotes, extreme values, and inflection

points.

Solution:

(a) 1. Vertical asymptote:

Answer : x= 0 (y-axis). Correctness : 0.5 point ; Explanation : 0.5 point.

Solution:

x> 0: lim^x→0+f(x) = lim^x→0+(^ln(x)x ) = ^−∞0 = −∞

( means f(x) → −∞ when x → 0⁺).

x< 0: limx→0⁻f(x) = limx→0⁻(−^ln(−x)_−x ) = ^+∞₀ = +∞

( means f(x) → +∞ when x → 0⁻).

Note: We can’t use L’Hospital’s Rule to solve this question.

2. Horizontal asymptote:

Answer : y= 0 (x-axis). Correctness : 0.5 point ; Explanation : 0.5 point.

Solution:

x> 0: lim^x→+∞f(x) = lim^x→+∞(^ln(x)x ) = lim^x→+∞(^1x1) = ⁰1 = 0 ( ^+∞_+∞ type, use L’Hospital’s Rule.)

( means f(x) → 0 when x → +∞).

x< 0: lim^x→−∞f(x) = lim^x→−∞(−^ln(−x)−x ) = lim^x→−∞(^1x1) = ⁰1 = 0

x> 0: lim^{x→+∞f (x)}x = lim^x→+∞(^ln(x)x² ) = lim^x→+∞(2⋅x^x1 ) = +∞⁰ = 0 ( ^+∞_+∞ type, use L’Hospital’s Rule.)

(This result is in contradiction, therefore a slant asymptote doesn’t exist).

x< 0: lim^{x→−∞f (x)}_x = lim^x→−∞(^ln(−x)_x² ) = lim^x→−∞(_2⋅x^1x ) = −∞⁰ = 0 ( ^+∞_+∞ type, use L’Hospital’s Rule.)

(This result is in contradiction, therefore a slant asymptote doesn’t exist).

Solution2:

A slant asymptote of f(x) only occurs when x → ±∞, but lim^x→±∞f(x) = 0 from the above, therefore a slant asymptote doesn’t exist.

(b) Answer : increasing interval:(0, e) ∪ (−e, 0), decreasing interval:(e, +∞) ∪ (−∞, −e).

Correctness : 1 point per question;

Determine f^′(x) (1 point) + Illustrate f^′(x) > 0 and f^′(x) < 0 (1 point).

Solution:

x> 0: f(x) = ^ln(x)_x , f^′(x) = ^1−ln(x)_x² (from the Quotient Rule).

f^′(x) > 0: 1 − ln(x) > 0 ⇒ ln(x) < 1 ⇒ 0 < x < e (increasing interval).

f^′(x) < 0: 1 − ln(x) < 0 ⇒ ln(x) > 1 ⇒ x > e (decreasing interval).

x< 0: f(x) = −^ln(−x)−x , f^′(x) = ^1−ln(−x)x² (from the Quotient Rule).

f^′(x) > 0: 1 − ln(−x) > 0 ⇒ ln(−x) < 1 ⇒ −e < x < 0 (increasing interval).

f^′(x) < 0: 1 − ln(−x) < 0 ⇒ ln(−x) > 1 ⇒ x < −e (decreasing interval).

(c) Answer : local maximum point:(e,¹_e), local minimum point:(−e,⁻¹_e ).

Correctness : 0.5 point per question; Explanation : 1 point.

Solution:

Local maximum: Because f(x) is increasing in (0, e) and decreasing in (e, +∞), local maximum point occurs at (e, f(e)) = (e,¹e) (f^′(e) = 0).

Local minimum: Because f(x) is decreasing in (−∞, −e) and increasing in (−e, 0), local minimum point occurs at(−e, f(−e)) = (−e,⁻¹e ) (f^′(−e) = 0).

(d) (4 points) f(x) is concave upward on the interval(s)

(−e

³²

, 0 ) ⋃(e

³²

, ∞)

. f(x) is concave downward on the interval(s)

(0, e

³²

) ⋃(−∞,−e

³²

)

.

⎧⎪⎪⎨⎪⎪

⎩

f^′′(x) = x > 0, ⇒ ^−1xx2−(1−ln x)2x

x⁴ =x(−3+2 ln x) x⁴

f^′′(x) = x < 0, ⇒ ^−1xx2−(1−ln(−x))2x

x⁴ =x(−3+2 ln(−x)) x⁴

(1)

f^′′(x) > 0 ⇒ (concave upward)

⎧⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎩

x(−3+2 ln x)

x⁴ > 0, x > 0 ⇒ −3 + 2 ln x > 0 ⇒ ln x > ³2 ⇒ x > e³² ⇒

choose x > e

³²

x(−3+2 ln(−x))

x⁴ > 0, x < 0 ⇒ −3 + 2 ln(−x) < 0 ⇒ ln(−x) > ³2 ⇒ x > −e³²

⇒

choose − e

³²

< x < 0

(2)

f^′′(x) < 0 ⇒ (concave downward)

⎧⎪⎪⎪⎪

⎪⎪⎨⎪⎪⎪

⎪⎪⎪⎩

x(−3+2 ln x)

x⁴ > 0, x > 0 ⇒ −3 + 2 ln x > 0 ⇒ ln x > ³2 ⇒ x > e³²

⇒

choose 0 < x < e

³²

x(−3+2 ln(−x))

x⁴ > 0, x < 0 ⇒ −3 + 2 ln(−x) < 0 ⇒ ln(−x) > ³₂ ⇒ x > −e³²

⇒

choose x < −e

³²

(3)

score: answer 0.5 point separately, right concept 2 points. ( notify: if the graph of f lies above all of its tangents on an interval I, then called concave upward. ex: f^′′(x) > 0 ⇒ concave upward)

if ⋃ write ⋂ lose 0.5 point

(e) (2 points) List the inflection point(s) of the curve y= f(x) ∶ (x, f(x)) =

(e

³²

,

³

2e³²

), (−e

³²

, −

³

2e³²

)

f^′′(x) = 0 ⇒ x = e³² and x= −e³²

f^′′(e³²) = ³₂e⁻³² and f^′′(−e³²) = −³₂e⁻³²

score: answer 0.5 points separately, right concept 1 points. ( notify f^′′(x) = 0, x = −e³² and e⁻³² )

(f) (3 points) Plot the f(x) and indicate asymptote, local point and infection point.