0) ●Clearly showing that they read the problem wrong, but work and answer is correct (3%) ●Minor simplification mistakes (-1%) ●L’Hospital’s Rule (-5%) (b) Formally: lim x→0+ln x

(1)

1091模模模組組組13-17班班班微微微積積積分分分1 期期期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. (15%) Find the following limits.

(a) (5%) lim

x→−∞

2e^x+5

3 − e^x . (b) (5%) lim

x→0⁺tan⁻¹(1 + ln x). (c) (5%) lim

x→∞( 1 e+

e x)

x

.

Solution:

(a) Formally:

x→−∞lim e^x=0

x→−∞lim 2e^x+5 = 5

x→−∞lim 3 − e^x=3 ≠ 0

x→−∞lim 2e^x+5

3 − e^x =

x→−∞lim 3 − e^x = 5 3

Short solution:

3 − e^x = 0 + 5 3 − 0 =

5 3

Grading:

●Correct answer (3%)

●Showing e^x→0 (2%), this can be done informally (e.g. → 0)

●Clearly showing that they read the problem wrong, but work and answer is correct (3%)

●Minor simplification mistakes (-1%)

●L’Hospital’s Rule (-5%) (b) Formally:

lim

x→0⁺ln x = −∞

lim

x→0⁺1 + ln x = −∞

Let y = 1 + ln x,

x→0lim⁺tan⁻¹(1 + ln x) = lim

y→−∞tan⁻¹y = −π 2

Short solution:

x→0lim⁺tan⁻¹(1 + ln x) = −π

2 because ln x → −∞

Grading:

●Correct answer (3%)

●Showing ln x → −∞ (2%), this can be done informally (e.g. → −∞ or sketch graph)

●If they sketch graph of tan⁻¹x (and labelling horizontal asymptotes) but didn’t finish (1%)

(2)

x→∞limx ln (1 e+

e x) = −∞

Let y = x ln (¹_e+^e_x),

x→∞lim ( 1 e+

e x)

x

= lim

x→∞e^{x ln(}¹^e⁺^x^e⁾= lim

y→−∞e^y=0

Short solution:

x→∞lim( 1 e+

e x)

x

=0 because 1

e<1 , a^∞ is not an indeterminate form if a < 1.

Grading:

●The main mistake students make is to take the limit twice. Without any further work that will be (0%)

●Correct steps (2%) they do not need to take the natural logarithm, writing (1/e)^∞is also acceptable

●Correct reasoning to get the answer (3%)

● L’Hospital’s Rule (-4%) they get (1%) if they did the natural logarithm correctly but clearly thought ln(1/e) = 0, no points if they didn’t try to evaluate ln(1/e)

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2. (15%) f (x) =

⎧⎪

⎪⎪

⎪

⎨

⎪⎪

⎩

x^x+1 , for x > 0 0 , for x = 0

2(1−cos x)

x , for x < 0 .

(a) (5%) Compute lim

x→0f (x). Is f (x) continuous at x = 0?

(b) (5%) Compute lim

x→0

f (x)−f (0)

x−0 . Is f (x) differentiable at x = 0?

(c) (5%) Compute f^′(x) for x ≠ 0.

Solution:

(a) lim

x→0⁺f (x) = lim

x→0⁺x^x+1. ∵ lim

x→0⁺x = 0, lim

x→0⁺x + 1 = 1 ∴ lim

x→0⁺x^x+1=0¹=0(2 pts) Another solution:

For x > 0, ln f (x) = (x + 1) ln x.

lim

x→0⁺ln f (x) = lim

x→0⁺(x + 1) ln x = −∞.Hence lim

x→0⁺x^x+1=e^−∞=0.

x→0lim⁻f (x) = lim

x→0⁻

2(1 − cos x)

x = lim

x→0⁻

2(1 − cos x)(1 + cos x) x(1 + cos x) = lim

x→0⁻

2 sin²x x(1 + cos x)

= lim

x→0⁻( 2 1 + cos x⋅

sin x

x ⋅sin x) = 1 × 1 × 0 = 0 (2pts) Another solution:

x→0lim⁻f (x) = lim

x→0⁻

2(1 − cos x) x

0 0

ÔÔÔÔÔÔÔÔ

l’Hospital’s Rule lim

x→0⁻

2 sin x 1 =0.

∵ lim

x→0⁺f (x) = 0, lim

x→0⁻f (x) = 0 ∴ lim

x→0f (x) = 0.

lim

x→0f (x) = 0 = f (0) ⇒ f (x) is continuous at x = 0.

⎫⎪

⎪⎪

⎬

⎪⎪

⎪⎭

(1pt)

(b) lim

x→0⁺

f (x)−f (0) x−0 = lim

x→0⁺ f (x)

x = lim

x→0⁺x^x. ln(x^x) =x ln x, lim

x→0⁺x ln x = lim

x→0⁺ ln x

1 x

∞

ÔÔÔÔÔÔÔÔ

l’Hospital’s Rule lim

x→0⁺

1 x

−¹

x2

=0 Hence lim

x→0⁺x^x=e⁰=1.

lim

x→0⁺

f (x)−f (0) x−0 = lim

x→0⁺x^x=1. (2 pts) lim

x→0⁻

f (x)−f (0) x−0 = lim

x→0⁻

2(1−cos x) x² = lim

x→0⁻

2(1−cos x)(1+cos x) x²(1+cos x) = lim

x→0⁻ 2 1+cos x

sin²x

x² =1(2 pts) Another solution:

lim

x→0⁻

f (x)−f (0) x−0 = lim

x→0⁻

2(1−cos x) x²

0 0

ÔÔ lim

x→0⁻ 2 sin x

2x =1.

Hence lim

x→0⁺

f (x)−f (0) x−0 = lim

x→0⁻

f (x)−f (0) x−0 =1.

Therefore lim

x→0

f (x)−f (0)

x−0 =1 and f is differentiable at x = 0. (1 pt) (c) For x > 0, f (x) = x^x+1, ln(f (x)) = (x + 1) ln x...(*)(1 pt)

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3. (15%) Compute the following derivatives.

(a) (5%) Let f (x) = e^{tan x}. Compute f^′(x) and f^′′(x).

(b) (5%) _dx^d (2^x²+x^e^x).

(c) (5%) _dx^d (x sin⁻¹x +√ 1 − x²).

Solution:

(a) Let y = tan x. f^′(x) = (e^y)^′y^′=e^ysec²x = e^{tan x}sec²x.

f^′′(x) = (f^′(x))^′= (e^{tan x})^′sec²x + e^{tan x}(sec²x)^′

=e^{tan x}sec²x ⋅ sec²x + e^{tan x}⋅2 sec x ⋅ sec x tan x

=e^{tan x}sec²x(sec²x + 2 tan x).

Scoring rules:

f^′(x) = e^{tan x}(tan x)^′ (2%), (tan x)^′=sec²x (1%),

f^′′(x) = (e^{tan x})^′sec²x + e^{tan x}(sec²x)^′ (1%), (sec²x)^′=2 sec²x tan x (1%).

(b)

d

dx(2^x²+x^e^x) = d

dx(e^x²^{ln 2}+e^e^x^{ln x})

=e^x²^{ln 2}2 ln 2 ⋅ x + e^e^x^{ln x}(e^xln x + e^x1 x)

=2 ln 2 ⋅ 2^x²x + x^e^xe^x(ln x + 1 x). Scoring rules:

(2^x²+xê^x)^′= (2^x²)^′+ (xê^x)^′ (1%), (2^x²)^′=2^x²(ln 2x²)^′(1%) = 2 ln 2 ⋅ 2^x²x(1%), (xê^x)^′=xê^x(e^xln x)^′(1%) = xê^xe^x(ln x +1

x) (1%).

(c)

d

dx(x sin⁻¹x +

√

1 − x²) = (sin⁻¹x + x 1

√

1 − x²) + 1 2√

1 − x²(−2x)

=sin⁻¹x Scoring rules:

(sin⁻¹x)^′= 1

√ 1 − x²

(2%), (

√

1 − x²)^′= 1 2

√ 1 − x²

(−2x) (1%), (x sin⁻¹x +√

1 − x²)^′= (x sin⁻¹x)^′+ (

√

1 − x²)^′ (1%), (x sin⁻¹x)^′= (sin⁻¹x + x 1

√

1 − x²) (1%).

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4. (12%) Suppose that near the point (3, 8), 3y^2/3+xy = 36 defines a function y = f (x).

(a) (4%) Compute ^dy_dx at (3, 8) which is f^′(3).

(b) (3%) Use the linear approximation to estimate f (3.01).

(c) (5%) Compute ^d_dx²^y₂ at (3, 8). Is the estimation from (b) larger than or smaller than f (3.01)?

Solution:

(a) Consider y as a function of x. Differentiate the equation 3(y(x))^2/3+x ⋅ y(x) = 36 with respect to x. We obtain 3 ⋅²₃y⁻¹³ ⋅y^′+y + x ⋅ y^′=0....(*)(3 pts)

(1 pt for trying to differentiate the equation with respect to x. 2 pts for correct result.) At (3, 8), (*)⇒ y^′+8 + 3y^′=0 ⇒ y^′=2. (1 pt)

(b) The linear approximation of f at x = 3 is

L(x) = f (3) + f^′(3) ⋅ (x − 3) (1 pt)

=8 − 2 ⋅ (x − 3) (1 pt)

f (3.01) ≈ L(3.01) = 8 − 0.02 = 7.98 (1 pt) (c) _dx^d (*) ⇒ −²₃y^−4/3(y^′)²+2y^−1/3y^′′+y^′+y^′+xy^′′=0(3 pts)

(1 pt for trying to differentiate the equation with respect to x. 2 pts for correct result.) At (3, 8), y^′= −2, −²₃×₁₆¹ ×4 + y^′′+ (−4) + 3y^′′=0

⇒ 4y^′′= ²⁵₆ ⇒y^′′=²⁵₂₄ >0 (1 pt)

Because y^′′(x) is continuous near x = 3 and y^′′(3) = ²⁵₂₄>0, we know that y^′′(x) > 0 for x near 3. Hence the curve is concave upward near (3, 8) and the linear approximation is smaller than f (3.01). (1 pt)

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5. (15%) For each limit, state its indeterminate form and use l’Hospital’s rule to compute it.

(a) (5%) lim

x→0

tan⁻¹(x²)

1−cos(3x). (b) (5%) lim

x→0(cos x)^csc(x²⁾. (c) (5%) lim

x→∞x³(_x¹−sin(¹_x)).

Solution:

(a) Formally:

x→0limtan⁻¹(x²) =tan⁻¹(0) = 0 lim

x→01 − cos(3x) = 1 − cos(0) = 0 Indeterminate form: 0

0 L’Hospital’s rule

lim

x→0

tan⁻¹(x²) 1 − cos(3x)=lim

x→0

1 1 + (x²)²

⋅ (2x) 0 − (− sin(3x)) ⋅ 3 =lim

x→0

2x 3(1 + x⁴)sin(3x) From here either

lim

x→0

2x

3(1 + x⁴)sin(3x)=lim

x→0

2 9(1 + x⁴)

⋅ 3x sin(3x)=

2 9 or l’Hospital’s rule again

x→0lim

2x

3(1 + x⁴)sin(3x)=lim

x→0

2

12x³sin(3x) + 9(1 + x⁴)cos(3x)= 2 9

Short solution:

x→0lim

tan⁻¹(x²) 1 − cos(3x)

0 0

L’H

=lim

x→0

2x

3(1 + x⁴)sin(3x) =lim

x→0

2 9(1 + x⁴)

⋅ 3x sin(3x)=

2 9

Grading:

● Correct indeterminate form (1%), they can manipulate the function first to get a different answer (but still correct)

●Show knowledge of l’Hospital’s rule (1%) they get this point as long as they put the function into a fraction, consider whether it is ⁰₀ or ^∞

∞, and start taking derivatives

●(3%) if they identify the indeterminate form and use l’Hospital’s rule correctly once. Finishing the problem from there is worth (2%)

●Clearly showing that they read the problem wrong, but work and answer is correct (at most 3%)

●Using l’Hospital’s rule when it doesn’t apply (-4%) (b) Formally:

lim

x→0cos x = 1

x→0limx²=0⁺ limx→0csc(x²) = ∞ Indeterminate form: 1^∞

Use natural logarithm to obtain indeterminate forms for l’Hospital’s rule

(7)

L’Hospital’s rule

x→0lim

ln(cos x) sin(x²)

=lim

x→0

1

cos x⋅ (−sin x) cos(x²) ⋅ (2x) =lim

x→0

−sin x 2x cos x cos(x²) Either l’Hospital’s rule again or use lim

x→0

sin x

x =1 to get

x→0lim

ln(cos x) sin(x²)

= − 1 2 and

x→0lim(cos x)^csc(x²⁾=e^−1/2

Short solution:

lim

x→0(cos x)^csc(x²⁾

1^∞

= e limx→0

ln(cos x) sin(x²)

0 0

L’H=e limx→0

−sin x

2x cos x cos(x²) =e^−1/2

Grading:

●Using l’Hospital’s rule when it doesn’t apply (-4%) (c) Formally:

x→∞limx³= ∞

x→∞lim 1

x−sin (1 x) =0 Indeterminate form: ∞ ⋅ 0

Direct method:

x→∞lim x³( 1

x−sin (1

x)) = lim

x→∞

1

x−sin (¹_x) x⁻³ L’Hospital’s rule

x→∞lim

1

x−sin (¹_x) x⁻³ = lim

x→∞

−x⁻²−cos (¹_x) ⋅ (−x⁻²)

−3x⁻⁴ = lim

x→∞

1 − cos (¹_x) 3x⁻² L’Hospital’s rule

x→∞lim

1 − cos (¹_x) 3x⁻² = lim

x→∞

sin (_x¹) ⋅ (−x⁻²)

−6x⁻³ = lim

x→∞

sin (_x¹) 6x⁻¹ L’Hospital’s rule

sin (¹) cos (¹) ⋅ (−x⁻²) cos (¹) 1

(8)

lim

y→0⁺

y − sin y y³

0 0

L’H

= lim

y→0⁺

1 − cos y 3y²

0 0

L’H

= lim

y→0⁺

sin y 6y =

1 6

Grading:

●Using l’Hospital’s rule when it doesn’t apply (-4%)

(9)

6. (12%) A manufacturer determines that in order to sell x units of product, the price per unit is given by the law p(x) = 18000 − 0.5x² (in dollars).

On other hand, the cost of manufacturing x units is determined by

C(x) = 1000 + 3000x (in dollars), where C(0) = 1000 is the fixed cost.

(a) (2%) Find the interval of x, denoted by I, on which the price p(x) is non-negative. Note that we require x ≥ 0.

(b) (2%) Let T (x) be the profit function (Revenue = Units × Price, Profit = Revenue − Cost). Please find T (x).

(c) (6%) Find the price per unit such that the profit T is maximized on the interval I.

(d) (2%) Let xmax be the unit where the profit T (xmax) is the maximum value. Will xmax change if we use a different fixed cost C(0)? Please explain your reasoning.

Solution:

(a) (2%)Requires x ≥ 0 and p(x) ≥ 0, that is,

0 ≤ x ≤

√ 18000

0.5 (≈189.73).

Hence

I = [0,√

36000](full credit for the correct answer. No partial credit).

(b) (2%)

T (x) =xp(x) − C(x) = x(18000 − 0.5x²) − (1000 + 3000x)

= −0.5x³+15000x − 1000.

Full credit for the correct answer. No partial credit. It is OK if students do not simplify T (x).

(c) We first look for the critical points inside of I. That is, find x such that 0 = T^′(x) = −1.5x²+15000, i.e., xmax = 100. (2%)

Check that T attains the maximum at xmax. Note that T (0) = −1000, T (√

36000) = −1000 − 3000 ×√

36000 < 0,

T (xmax) = 100(18000 − 0.5 × 100²) − (1000 + 3000 × 100) = 999000 > 0.

Hence T attains the maximum at xmax. (2%). Students do not need to compute the exact values of T at particular points. Determining the signs of T at those points is sufficient.

At xmax = 100, p(xmax) = p(100) = 18000 − 0.5 × 100²=13000. (2%)

(d) (2%)xmax will not change if we use a different fixed value C(0) sinceT^′(x) does not depend on C(0).

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7. (16%) Let

f (x) = ln ∣2x + 1 x − 1∣. (a) (1%) Write down the domain of f (x).

(b) (4%) Compute f^′(x). Write down the interval(s) of increase and interval(s) of decrease of f (x).

(c) (4%) Compute f^′′(x). Write down the interval(s) on which f (x) is concave upward and the interval(s) on which f (x) is concave downward.

(d) (4%) Find all vertical asymptotes and horizontal asymptotes of y = f (x).

(e) (3%) Sketch the graph of y = f (x).

Solution:

(a) The domain of f (x) is (−∞, −¹₂) ∪ (−¹₂, 1) ∪ (1, ∞).

(b)

f^′(x) = x − 1 2x + 1

2(x − 1) − (2x + 1) (x − 1)² =

−3

(2x + 1)(x − 1) = −3 1

2x²−x − 1. (2%) f (x) is increasing on (−¹₂, 1) (1%) and decreasing on (−∞, −¹₂) ∪ (1, ∞) (1%).

(c)

f^′′(x) = −3 ( 1 2x²−x − 1)

′

=3 4x − 1

(2x²−x − 1)². (2%) f (x) is concave upward on (¹₄, ∞) (1%) and concave downward on (−∞,¹₄)(1%).

(d)

limx→1f (x) = ∞, lim

x→−¹₂

f (x) = −∞, lim

x→∞f (x) = lim

x→−∞f (x) = ln 2.

Hence the lines x = 1, x = −¹₂ are vertical asymptotes of y = f (x) (2%), and the line y = ln 2 is a horizontal asymptote of y = f (x) (2%).

(e)