1101模模模組組組01-05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. (21 pts) Compute the limits or show that the limit doesn’t exist.
(a) (7 pts) lim
x→1
2x−2
∣x2−x∣ (b) (7 pts) lim
x→0∣sin x∣ ⋅ sin (1
x) (c) (7 pts) lim
h→0
ln(cos(4h)) h2
Solution:
(a) lim
x→1+
2x−2
∣x2−x∣ = lim
x→1+
2x−2
x(x − 1) (1 pt) Moreover, by the definition of d
dx2x∣
x=1 or the L’Hospital’s Rule,
x→1lim+ 2x−2
x − 1 =2 ln 2 (2 pts) Hence lim
x→1+
2x−2
∣x2−x∣= (lim
x→1+
1
x) ⋅ (lim
x→1+
2x−2
x − 1) =2 ln 2.
Similarly, lim
x→1−
2x−2
∣x2−x∣= lim
x→1−
2x−2
x(1 − x)= −2 ln 2 (3 pts)
∵ lim
x→1+
2x−2
∣x2−x∣ ≠ lim
x→1−
2x−2
∣x2−x∣ ∴ lim
x→1
2x−2
∣x2−x∣ doesn’t exist. (1 pt) (b) ∵ −1 ≤ sin (1
x) ≤1 for all x ≠ 0.
∴ −∣sin x∣ ≤ ∣ sin x∣ sin (1
x) ≤ ∣sin x∣ (3 pts) Since lim
x→0sin x = 0, we have lim
x→0∣sin x∣ = 0 and lim
x→0−∣sin x∣ = 0 (2 pts) Hence by the Squeeze theorem,
x→0lim∣sin x∣ ⋅ sin (1
x) =0 (2 pts) NOTE:
(1) 使用 lim
x→0
sinx1
1 x
=1 or product rule震盪等等之類的說明，最多得1分。
(2) 使用 lim
x→0
sinx1
1 x
=0 扣 2-4分 (主要扣的原因是沒使用到夾擠定理)
(c)
h→0lim
ln(cos(4h)) h2
L’H
Ô Ô0 0
limh→0
− sin(4h) cos(4h) ⋅4
2h
sol 1
ÔÔÔlim
h→0
−8 cos(2h)
sin(4h)
4h = −8 ⋅ 1 = −8(3 pts)
sol 2
ÔÔÔlim
h→0
−2 sin(4h) h cos(4h)
L’H
Ô Ô0 0
h→0lim
−8 cos(4h)
cos(4h) − 4h sin(4h) (2 pts)
=
−8
1 = −8 (1 pt)
sol 3
ÔÔÔlim
h→0
−2 tan(4h) h
L’H
Ô Ô0 0
limh→0
−2 sec2(4h) ⋅ 4
1 = −8
4 pts total. 1 pt for L’H. 1 pt for (ln x)′. 1 pt for (cos x)′. 1 pt for the chain rule constant.
2. (14 pts) Suppose that the equation
x2cos(xy) + ey2−2x + y = 0
is satisfied by a differentiable function y(x) defined on an open interval I containing 1 such that y(1) = 0. Besides, we assume that y′′ exists everywhere on I.
(a) (6 pts) Compute y′(1).
(b) (6 pts) Compute y′′(1).
(c) (2 pts) Does y(x) attain a local extremum at x = 1? if your answer is YES, tell the type of local extremum (local maximum or local minimum) and give your reason.
Solution:
(a) Applying the implicit differentiation, we have
2x cos(xy) − x2sin(xy)(y + xy′) +ey22yy′−2 + y′=0.
Setting x = 1 and y(1) = 0, we obtain that
2 cos(0) − 12⋅sin(0)(0 + y′(1)) + e0⋅2 ⋅ 0 ⋅ y′(1) − 2 + y′(1) = 0, that is, y′(1) = 0.
The full points of this part: 6 points:
(1) (4 points) Completely correct process of implicit differentiation deserves 4 points. If there are some minor algebraic mistakes in the process of implicit differentiation, one may gain at most 2 points.
(2) (2 points) Correct evaluation with the (right or wrong) obtained result of implicit differentiation to obtain y′(1) deserves the rest 2 points. No partial credits will be given here.
(b) Applying the implicit differentiation twice, we have 2 cos(xy) − 2x sin(xy)(y + xy′)
−2x sin(xy)(y + xy′) −x2cos(xy)(y + xy′)2−x2sin(xy)(y′+y′+xy′′) +ey2(2yy′)(2yy′) +ey22y′⋅y′+ey22yy′′
+y′′.
Setting x = 1, y(1) = 0, and y′(1) = 0 we obtain that
2 − 0 − 0 − 0 − 0 + 0 + 0 + 0 + y′′(1), that is y′′(1) = −2. The full points of this part: 6 points:
(1) (5 points) Completely correct process of second time implicit differentiation based on the result ob- tained in (1) deserves 5 points. If there are some minor algebraic mistakes in the process of implicit differentiation, one may gain at most 2 points.
(2) (1 points) Correct evaluation with the (right or wrong) obtained result of second time implicit differ- entiation to obtain y′′(1) deserves the rest 1 points. No partial credits will be given here.
(c) Since y′(1) = 0 and y′′(1) = −2 < 0, y(x) attains a local maximum at x = 1.
Based on the results obtained in (a) and (b) there are the following cases: if the obtained y′(1) ≠ 0, the answer to (c) should be “No,” which deserves 2 points. If the obtained y′(1) = 0 and y′′(1) < 0/ > 0, the answer to (c) should be “y(x) attains a local maximum/minimum at x = 1,” which deserves 2 points. In the second case, the following situations will gain 1 point for partial credit:
(i) To conclude the monotonicity of y′without providing sufficient reasoning about the continuity of y′′. (ii) To conclude the concavity of y without providing sufficient reasoning about the continuity of y′′.
3. (15 pts) Let
f (x) =
⎧⎪
⎪
⎨
⎪⎪
⎩
xx1, x > 0 0, x ≤ 0 (a) (6 pts) Use the definition of derivatives to compute f′(0) as a limit.
(b) (9 pts) Write f′(x) as a piecewise defined function. Is f′(x) a continuous function over all real numbers?
Solution:
(a) By definition, f′(0) = lim
x→0
f (x) − f (0)
x − 0 . It is clear that
xlim→0−
f (x) − f (0) x − 0 = lim
x→0+
0 − 0 x − 0=0.
We claim that
xlim→0+
f (x) − f (0) x − 0 = lim
x→0+
x1/x x = lim
x→0+x1x−1=0. (1)
Let g(x) = xx1−1 and h(x) = ln g(x) = (1
x−1) ln x. Since
x→0lim+ 1
x−1 = ∞ and lim
x→0+ln x = −∞, we have
xlim→0+h(x) = −∞, and hence
x→0lim+g(x) = lim
x→0+eh(x)=0.
●Writing down the form of limit lim
x→0+
x1/x
x or lim
x→0+xx1−1deserves 2 points.
●Showing lim
x→0+x1x−1=0 correctly deserves 4 points.
(b) It is clear that f′(x) = 0 if x < 0 and if x = 0 (by (a)), and this part deserves no points. We first compute f′(x) for x > 0. Let u(x) = ln f (x) = ln x
x . Then, for x > 0, we have f′(x)
f (x) =u′(x) =
1
x⋅x − (ln x) ⋅ 1
x2 =
1 − ln x x2 , and hence
f′(x) = xx1−2(1 − ln x) for every x > 0. (2) In summary, we have
f′(x) = { xx1−2(1 − ln x) if x > 0;
0 if x ⩽ 0.
Finally we show that lim
x→0f′(x) = 0(= f′(0)). Since lim
x→0−f′(x) = lim
x→00 = 0, it suffices to show that lim
x→0+f′(x) = 0, i.e.,
x→0lim+xx1−2(1 − ln x) = 0. (3)
There are many different ways of showing this limit. For example, one may write xx1−2(1 − ln x) as x1x−3x(1 − ln x) and show that
x→0lim+x1x−3=0 and lim
x→0+x(1 − ln x) = lim
x→0+x ln x = 0 ∶ for the first limit, one may proceed similarly as in the proof of (1) since
xlim→0+
1
x−3 = ∞ and lim
x→0+ln x = −∞, and hence
1
as for the second limit, we have
xlim→0+x ln x = lim
x→0+
ln x 1/x
L’H
= lim
x→0+
1/x
−1/x2 = lim
x→0+x = 0.
Note that here it is legitimate to apply the L’Hospital rule to compute the limit lim
x→0+
ln x
1/x since lim
x→0+1/x = ∞ and 1/x2≠0 near x = 0. Here is another way of showing (3), adopted by one of the students. Consider the substitution t = 1/x. Then showing (3) is the same as showing lim
t→∞t−t+2(1 + ln t) = 0, i.e.,
t→∞lim 1 + ln t
tt−2 =0. (4)
To see this, note that lim
t→∞tt−2= ∞and that
the derivative of tt−2 is tt−2(ln t +t − 2 t ), which tends to −∞ as t → ∞, and hence tt−2(ln t +t − 2
t ) ≠0 for t sufficiently large. Therefore one may apply the L’Hospital rule to obtain (4):
t→∞lim 1 + ln t
tt−2 = lim
t→∞
1/t tt−2(ln t +t−2t )
=0.
●The full (2) deserves 4 points.
(1) An intention of applying logarithmic differentiation to compute f′(x) for x > 0 will gain 2 points.
(2) Finishing the process of logarithmic differentiation correctly will gain the rest 2 points.
● Showing (3) with correct reasoning deserves 5 points. Partial credits may be given in the following situations:
(1) Trying to apply the L’Hospital rule to compute the limit in (3) with all conditions checked may gain at most 5 points.
(2) Trying to apply the L’Hospital rule to compute the limit in (3) without checking all conditions may gain at most 2 points.
(3) Some other unsuccessful attempts might gain at most 2 points, depending on the situations.
4. (14 pts) Consider the function f (x) = 3x − tan−1(x − 1).
(a) (6 pts) Show that the equation 3x − tan−1(x − 1) = 3.01 has a unique solution.
(b) (4 pts) Let g(x) be the inverse function of f . Find g(3) and g′(3).
(c) (4 pts) Apply a linear approximation to g to get an estimate of the solution of f (x) = 3.01.
Solution:
(a) The function f is continuous and differentiable.
f (0) =π
4, f (1) = 3, f (2) = 6 −π 4
Intermediate value theorem: f (1) < 3.01 < f (2) implies that there exist at least one solution.
If there are at least two solutions, then by Rolle’s theorem, f′(c) = 0 for some c.
However, f′(x) = 3 − 1
1 + (x − 1)2 ≥2 is never equal to zero. Therefore there is exactly one solution.
(b) g(3) = 1, g′(3) = 1 f′(1)=
1 2 (c) g(3.01) ≈ 1 +1
2(3.01 − 3) = 1.005 Grading scheme: (6 + 4 + 4 = 14 points)
(a) 3 points for existence and 3 points for only one solution. Basically all or nothing unless students make minor mistakes (example: quoting the wrong theorem but stating the condition and conclusion correctly. Just -1 in that case).
(b) 2 points each. All or nothing.
(c) 2 points for showing knowledge of linear approximation. 2 points for answer. If they got (b) wrong they can still get all 4 points here. But if they got (b) wrong but didn’t follow through, then it depends on how they did the problem.
5. (24 pts) f (x) = 1
xe1x for x ≠ 0.
(a) (7 pts) Compute lim
x→0+f (x) and lim
x→0−f (x). Find vertical and horizontal asymptotes of y = f (x).
(b) (7 pts) Compute f′(x). Find critical point(s) of f (x). Find interval(s) of increase and interval(s) of decrease of y = f (x).
(c) (7 pts) Compute f′′(x). Discuss concavity of y = f (x). Find inflection point(s), if any, of y = f (x).
(d) (3 pts) Sketch the curve y = f (x).
Solution:
(a) The right-hand limit is
xlim→0+f (x) = lim
x→0+
1
xex1 = ∞, (1 point) from which the curve y = f (x) has a vertical asymptote x = 0 (1 point).
The left-hand limit is
x→0−lim f (x)y=
1
=x lim
y→−∞
y e−y
l’Hospital
= lim
y→−∞
1
−e−y =0. (2 points) Lastly, since
xlim→∞f (x) = lim
x→∞
1
xe1x =0 (1 point) and
x→−∞lim f (x) = lim
x→−∞
1
xe1x =0, (1 point) the curve y = f (x) has one horizontal asymptote y = 0 (1 point).
(b) The first derivative of f is given by
f′(x) = −ex1 x + 1
x3 . (3 points)
So f has one critical point at −1 (1 point). By the increasing/decreasing test f is increasing on (−1, 0) and decreasing on (−∞, −1) and (0, ∞) (3 points).
(c) The second derivative of f is given by
f′′(x) = ex1 2x2+4x + 1
x5 . (3 points) By the concavity test f is concave upward on (−1 − 1
√2, −1 + 1
√2) ∪ (0, ∞) and concave downward on
(−∞, −1 − 1
√
2) ∪ (−1 + 1
√
2, 0) . (3 points) Thus, f has inflection points at −1 − 1
√2 and −1 + 1
√2 (1 point).
(d) Vertical and horizontal asymptotes (1 point).
Intervals of increasing and decreasing (1 point).
Intervals of concave upward and downward (1 point).
6. (12 pts) A fence 2 m tall is parallel to a tall building at a distance of 2 m from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building?
2 m Building
Fence
2 m Ladder
Solution:
Method 1: θ is the angle between the ladder and the ground.
L(θ) = 2 sec θ + 2 csc θ, 0 < θ < π/2 L′(θ) = 2 sec θ tan θ − 2 csc θ cot θ The only critical number in the domain satisfies
tan3θ = 1, θ = π 4 L′(π/6) < 0, L′(π/3) > 0
With this we verified that the critical number corresponds to the absolute minimum and the shortest length of the ladder would be 4
√ 2 m.
Method 2: x is the distance from the base of the ladder to the fence.
L(x) =
√
(x + 2)2+ (2 + 4 x)
2
, x > 0
L′(x) = x + 2 −x42(2 +x4) (x + 2)
√ 1 + 2x
= 1 −x83
√ 1 +2x The only critical number in the domain satisfies
x3=8, x = 2 L′(1) < 0, L′(3) > 0
With this we verified that the critical number corresponds to the absolute minimum and the shortest length of the ladder would be 4√
2 m.
Method 3: y is the height where the ladder touches the building.
L(y) =
¿ ÁÁ
Ày2+ (2 + 4 y − 2)
2
, y > 2 Similar to above.
Grading scheme: (12 points)
4 points for finding a function to optimize and state its domain.
4 points for solving for critical numbers.
3 points for verifying the critical number is a minimum.
1 point for stating the final answer.
Follow through applies to everything after the function. They only lose points later if their answer doesn’t make sense.
