(b) Suppose that y = y(x) is implicitly defined by xy = yx

98學年度第1學期 微積分甲一組期中考解答

1. (18%) (a) Let f1(x) = x^aa + a^xa + a^ax, a > 0, x > 0, and f2(x) =

Z ^√ln x x^x

e^t2dt, x > 1.

Then f₁⁰(x) = ,

f₂⁰(x) = .

(b) Suppose that y = y(x) is implicitly defined by x^y = y^x. Then at (x, y) = (2, 4), dy

dx = .

Sol:

(a-1) First note that :

dx^b/dx = bx^b−1

&

db^x/dx = (ln b)b^x We use this and chain rule to solve the problem.

df1(x)/dx = d(x^aa)/dx + d(a^xa)/dx + d(a^ax)/dx

= a^ax^aa−1+ (ln a)a^xad(x^a)/dx + (ln a)a^axd(a^x)/dx

= a^ax^aa−1+ (ln a)a^xaa(x^a−1) + (ln a)a^ax(ln a)a^x

= a^ax^aa−1+ (ln a)a^xa+1(x^a−1) + (ln a)²a^ax+x (a-2)

df2(x)/dx = d Z

√ln x

x^x

f (t)dt/dx

= d Z a

x^x

e^t2dt/dx + d Z ^√ln x

a

e^t2dt/dx

= (d Z a

x^x

e^t2dt/dx^x)dx^x/dx + (d Z ^√ln x

a

e^t2dt/d√

ln x)d√

ln x/dx

= (−e^xx2)(x^x)(ln x + 1) + (e^{√ln x}

2

)(1/2)((ln x)^−1/2)(1/x)

= −e^x2xx^x(ln x + 1) + 1

2(ln x)^−1/2

 dx^x

x = de^{x ln x}

dx = e^{x ln x}dx ln x

dx = x^x(ln x + 1)



(b) Implicit function:

x^y = y^x First take ”ln” to both sides

y ln x = x ln y Differentiate both sides w.r.t x (take y as a function of x)

dy

dxln x + y

x = ln y + xdy/dx y

⇒ dy

dx = ln y − y/x lnx − x/y So

dy

dx|^(2,4)= ln 4 − 2 ln 2 − 1/2

= 2 ln 4 − 4

2 ln 2 − 1 or 4(ln 2 − 1)

ln 4 − 1 or ln 16 − 4 ln 4 − 1

2. (18%) Determine whether the following limits exist. If the limit exists, evaluate it. If the limit doesn’t exist, explain why.

(a) lim

x→0(1 + | sin x|)^x1 = ,

(b) lim

x→0

 2^x+ 3^x+ 5^x 3

x¹

= .

(c) First express

n

X

k=1

ln 2 + ln (n + k) − ln n

n + k as a Riemann sum for a function defined on [0, 2].

Then evaluate lim

n→∞

n

X

k=1

ln 2 + ln (n + k) − ln n

n + k .

Answer.

n

X

k=1

ln 2 + ln (n + k) − ln n

n + k = ,

n→∞lim

n

X

k=1

ln 2 + ln (n + k) − ln n

n + k = .

Sol:

(a) using the continuity of exponential function and L.H rule

1 1

= exp( lim

x→0⁺( cos(x)

1 + | sin(x)|)) = exp(1)

x→0lim⁻(1 + | sin(x)|)^1x = lim

x→0⁻exp(1

xln(1 + | sin(x)|))

= exp( lim

x→0⁻( − cos(x)

1 − sin(x))) = exp(−1) so the limit doesn’t exist .

(b) still using the continuity of exponential function and L.H rule

x→0lim⁺

 2^x+ 3^x+ 5^x 3

¹x

= lim

x→0⁺exp(1

xln(2^x+ 3^x+ 5^x

3 ))

= exp( lim

x→0⁺(

2^xln(2)+3^xln(x)+5^xln(x) 3

2^x+3^x+5^x 3

)) = exp(ln(2) + ln(3) + ln(5)

3 )

= exp(ln(30)

3 ) = 30¹³ the same argument for

x→0lim⁻

 2^x+ 3^x+ 5^x 3

¹_x

so the limit exists and the limit is 30¹³ (c)

n

X

k=1

ln 2 + ln(n + k) − ln(n)

n + k =

n

X

k=1

ln(2 +^2k_n) n + k

=

n

X

k=1

2 n

ln(2 + ²ⁿ_k ) 2 + ^2k_n or

n

X

k=1

1 n

ln 2(1 +^k_n) 1 + ^k_n

n→∞lim

n

X

k=1

ln 2 + ln(n + k) − ln(n)

n + k =

Z 2 0

ln(2 + x) 2 + x dx or

Z 1 0

ln(2(1 + x)) 1 + x dx

= Z 2

0

ln(2 + x)d(ln(2 + x)) = 3

2(ln(2))² 3. (8%) Let

H(x) =















e^1x , x < 0

m , x = 0

a sin x + b cos x + cx , x > 0 Find conditions of m, a, b, c, such that, respectively,

(a) H(x) is continuous everywhere. Answer: ,

(b) H(x) is differentiable everywhere. Answer: ,

(c) H(x) has an inflection point at x = 0. Answer: .

Sol:

(a) Since H(x) is continuous everywhere, thus we have

x→0lim⁺H(x) = lim

x→0⁻H(x) = H(0)

=⇒ lim

x→0⁻e^x1 = H(0) = lim

x→0⁺(a sin x + b cos x + cx) and lim

x→0⁻e^1x = 0, H(0) = m,

x→0lim⁺(a sin x + b cos x + cx) = b so b = m = 0.

(b) Since H(x) is differentiable everywhere, thus H(x) is continuous everythere, so b = m = 0 and lim

h→0⁻

H(h)−H(0)

h−0 = H⁰(0) = lim

h→0⁺

H(h)−H(0) h−0

so

h→0lim⁻ e^h1

h = lim

h→0⁻ 1 h

e^−h1

= lim

h→0⁻

−_h¹²

1 h²e^−h1

= − lim

h→0⁻e^h1 = 0 and lim

h→0⁺

a sin h+ch

h = a + c

thus the conditions are b = m = 0, a + c = 0.

(c) Since H(x) is continuous at x = 0, thus b = m = 0 For x  0,

H⁰(x) = −_x¹²e^1x and H^”(x) = ^(2x+1)e

1 x

x⁴ , so H^”(x) 0 for −¹2 < x < 0.

For x 0, H⁰(x) = a cos x + c and H^”(x) = −a sin x.

Since x = 0 is an inflection point and H^”(x) 0 for x ∈ (−1

2, 0), hence a 0

4. (12%) Let g(x) = a

x³+ 3x + 4 + b

x³+ x − 2, ab > 0. Show that g(x) = 0 has exactly one real solution.

Sol:

g(x) = ^a(x_(x³3^+x−2)+b(x+3x+4)(x³³+x−2)^+3x+4)

Define f (x) = a(x³+ x − 2) + b(x³+ 3x + 4).

x³ + x − 2 = (x − 1)(x²+ x + 2); x³+ 3x + 4 = (x + 1)(x² − x + 4) (x²+ x + 2 > 0, x² − x + 4 > 0, ∀x ∈ R)

f (1) = 8b 6= 0; f(−1) = −4a 6= 0

So f (x) = 0 and g(x) = 0 have the same roots.

Because f (1)f (−1) = −32ab < 0, there is at least one root of f(x) = 0 in (−1, 1) by IVT.

If f (α) = f (β) = 0 for some α < β.

By MVT, ∃ c ∈(α, β) such that

f⁰(c) = f (β) − f(α) β − α = 0.

But f⁰(c) = a(3c²+ 1) + b(3c²+ 3) > 0 (or < 0, depending on the signs of a and b), a contradiction.

So f (x) = 0 has exactly one root. =⇒ g(x) = 0 has exactly one root.

5. (12%) Car A at the lower level pulls car B, which is located on the upper level 9 meters higher, with constant velocity 5 m/min to the right while a pulley 6 meters above the upper level is used to connect the two cars. Suppose that the total length of the rope is 35 meters. Let O be the point right underneath pulley P on the lower level, and θ = ∠BP O. Find the changing rate of θ when car A is 20 meters away from point O. See figure below.

Answer: .

Sol:

Let OA = x(t), then we have

35 −p225 + x(t)² = 6 sec θ(t)

Differentiate both sides with respect to t and we get

− 2x(t) · x⁰(t)

2p225 + x(t)² = 6 sec θ(t) tan θ(t) · θ⁰(t) Now since x⁰(t) = 5 and when x(t⁰) = 20, we have sec θ(t⁰) = 5

3 and tan θ(t⁰) = 4

3, thus we get θ⁰(t⁰) = − 3

10 (rad/min).

6. (12%) A sector is cut off from a circle of radius R, R > 0. The remaining part (shaded region) is used to construct a right circular cone. Find the maximal possible volume of the cone. See figure below. ( Hint. The volume of a right circular cone = 1

3(base area)·(height). )

Answer: The maximal volume= .

Sol:

Assume that the height of the cone is equal to h > 0. Then the radius of the circle at the bottom is equal to √

R²− h². The volume of the cone is

V (h) = π 3

√

R²− h²2

h = π

3 R²h − h³
To find the maximum value of V (h), consider V⁰(h) = 0. This gives

π

3 R²− 3h²
= 0 So h² = R²/3. That is, h = R/√

3 since R > h > 0. Now we can check this is exactly the

trivially. And the maximum value is

V  R

√3



= 2√ 3πR³ 27

7. (20%) Let f (x) = x²(x − 2)

(x + 1)² . Answer the following questions.

(a) The domain of y = f (x) is .

(b) f⁰(x) = .

(c) y = f (x) has critical point(s) at .

(d) f⁰⁰(x) = .

(e) y = f (x) is increasing on interval(s) ,

y = f (x) is decreasing on interval(s) .

(f) y = f (x) is concave up on interval(s) ,

y = f (x) is concave down on interval(s) .

(g) Find the (x, y)-coordinates of the following points if exist.

Local maximum point(s) : ,

Local minimum point(s) : ,

Inflection point(s) : .

(h) Find the asymptotes of the graph of y = f (x) if exist.

Vertical asymptotes(s) : ,

Horizontal asymptotes(s) : as x → ,

Slant asymtotes(s) : as x → .

(i) Sketch the graph of y = f (x) on page 11.

Sol:

Let f (x) = x²(x − 2)

(x + 1)² . Answer the following questions.

(a) The domain of y = f (x) is R \ {−1}.

(b) f⁰(x) = x(x + 4)(x − 1) (x + 1)³ .

(c) y = f (x) has criticial point(s) at x = −4, 0, 1.

(d) f⁰⁰(x) = 2(7x − 2) (x + 1)⁴ .

(e) y = f (x) is increasing on interval(s) (−∞, 4) ∪ (−1, 0) ∪ (1, ∞).

y = f (x) is decreasing on intervals(s) (−4, −1) ∪ (0, 1).

(f) y = f (x) is concave up on interval(s)(2 7, ∞).

y = f (x) is concave down on interval(s) (−∞, −1) ∪ (−1,²7).

(g) Find the (x, y)−coordinates of the follwing points if exist.

Local maximal point(s): (−4,³²₃), (0, 0).

Local minimal point(s): (1, −1 4).

Inlfection point(s): (2 7, − 16

189).

(h) Find the asymptotes of the graph of y = f (x) if exist.

Vertical asymptotes(s): x = −1.

Horizontal asymptotes(s): none.

Slant asymptotes(s):y = x − 4 as x → ±∞.

(i)