(1) For the normal density function f (x) = √1
2πe−x22 ,
(a) determine where the function f (x) is increasing or decreasing;
(b) determine where the function f (x) is concave up or concave down;
(c) sketch the graph for f (x) and also specify all local/global extrema; inflection points;
and asymptotes on your picture.
Sol:
(a) 算出函數的一階導數為 f0(x) = −√x2πe−x22 , 可以看出 √12πe−x22 > 0. 所以當 x ≤ 0 時, f0(x) ≥ 0; 當 x > 0 時, f0(x) < 0. 因此 f (x) 在 (−∞, 0] 遞增且 f (x) 在 (0, ∞) 遞減.
(b) 算出函數的二階導數為 f00(x) = √1
2πe−x22 (x2−1). 所以當 x ≥ 1 或 x ≤ −1 時, f00(x) ≥ 0;
當 −1 < x < 1 時, f00(x) < 0. 因此 f (x) 在 (−∞, −1]S[1, ∞) 時為凹向上且在 (−1, 1) 時為凹向下.
(c) lim
x→∞
√1
2πe−x22 = lim
x→−∞
√1
2πe−x22 = 0 ⇒ f 有水平漸近線 y = 0.
f0(0) = 0 且 f00(0) < 0 ⇒ f 在 x = 0 有局部極大值 √1
2π. 因為在 x = 0 左邊為嚴格遞增 且在 x = 0 右邊為嚴格遞減, 因此此點為全域極大值.
f00(1) = f00(−1) = 0 ⇒ f 有反曲點 (1,√1
2πe−12) 與 (−1,√1
2πe−12).
(2) For y = f (x) = lnx−1x4 ,
(a) find the differetial dy at x = 6 with dx = −0.5;
(b) compute ∆y = f (x+dx)−f (x) with the same x and dx. Use a graph to demonstrate dy, ∆y and their difference.
Sol:
(a) 首先先求出y0。
y0 = dy
dx = x − 1
x4 ·(x − 1)4x3 − x4
(x − 1)2 = 3x − 4 x(x − 1) 然後可以推得dy = x(x−1)3x−4 ∗ dx = 3·6−46·5 · (−0.5) = −.2333333333 (b) 接著算出∆y = f (x + dx) − f (x) = f (5.5) − (6) = −.242684992
dy, ∆y的差距在下圖
(3) In a research experiment, a population of fruit flies is increasing in accordance with the exponential growth pattern. After 2 days, there are 100 flies, and after 4 days there are 300 flies. How many flies with there be after 5 days?
Sol:
令此果蠅的數量模型為N (t) = C · ekt, 因此可以得到下列方程式
100 = C · e2k C = 100 100
(4) Find the following limits with clear derivations.
(a) lim
h→0
sin h h .
sol : When h > 0,
sinh ≤ h ≤ tanh
⇒ 1 ≤ sinhh ≤ cosh1
⇒ cosh ≤ sinhh ≤ 1 Use pinching theorem to get:
⇒ lim
h→0+cosh ≤ lim
h→0+
sinh
h ≤ lim
h→0+1
⇒ 1 ≤ lim
h→0+
sinh
h ≤ 1
(1)
When h < 0,
sinh ≥ h ≥ tanh
⇒ 1 ≥ sinhh ≥ cosh1
⇒ cosh ≥ sinhh ≥ 1 Use pinching theorem to get:
⇒ lim
h→0−cosh ≥ lim
h→0−
sinh
h ≥ lim
h→0−1
⇒ 1 ≥ lim
h→0−
sinh
h ≥ 1
(2)
By 1 and 2, we get
h→0lim sinh
h = 1
(b) lim
h→∞
sinh h .
⇒ −1h ≤ sinhh ≤ 1h
Use pinching theorem to get:
h→∞lim
−1
h ≤ lim
h→∞
sinh
h ≤ lim
h→∞
1 h
⇒ 0 ≤ lim
h→∞
sinh
h ≤ 0
Hence lim
h→∞
sinh h = 0
(5) Let y = tan2(3x − 2). Find y00. Sol:
y0 = 2tan(3x − 2)(tan(3x − 2))0
= 2tan(3x − 2)sec2(3x − 2) · (3x − 2)0
= 2tan(3x − 2)sec2(3x − 2) · 3
= 6tan(3x − 2)sec2(3x − 2).
所以
y00 = 6(tan(3x − 2))0· sec2(3x − 2) + 6tan(3x − 2) · (sec2(3x − 2))0
= 6sec2(3x − 2) · (3x − 2)0· sec2(3x − 2) + 6tan(3x − 2) · 2sec(3x − 2) · (sec(3x − 2))0
= 18sec4(3x − 2) + 12tan(3x − 2)sec(3x − 2) · sec(3x − 2)tan(3x − 2) · (3x − 2)0
= 18sec4(3x − 2) + 36tan2(3x − 2)sec2(3x − 2).
(6) Radioactive carbon isotopes have a half-life 5715 years. If 1.5 grams of the isopotopes is present in an object now, at what rate is the amount of the isotopes changing after 15,000 years?
sol : The amount to time model of the isotopes is:
1
(7) The number W (in thousands) of construction workers employed in the united States can be modeled by W = 7594 + 455.2sin(0.41t − 1.713), where t is the time in months, with t = 1 corresponding to January 1,2006 and t = 2 corresponding to February 1,2006 and so on.
(a) Did the number of construction workers exceed 8 million in 2006? If so, during which month(s)?
(b) When is the second time since 2006 that the number of the construction workers exceed 8 million?
sol :
(a) Let K(t) = 7594 + 455.3sin(0.41t − 1.713) − 8000. Want t such that K(t) > 0
, The time t1 such that K(t1) = 0 is
t1 = sin−1(406/455.3) + 1.1713 0.41
= 6.86.
, K(7) > 0 K(8) > 0 K(9) > 0 K(10) < 0
Hence the months on which the number of the workers exceed 8 million are July, August, and September.
K0(t) = (455.3)(0.41)cos(0.41t − 1.713) = 0, t = 0.41kπ + 1.713 =
(b) By the property of sin, we know that the next time t2 when K(t) begin to exceed 0, satisfies that t2− t1 = 2π/0.41 .
= 15.3248422, t2 .
= 22.1848422. Hence the second time since 2006, the number of workers exceed 8 million is t2− 12 = 10.1848, which turns out to be Nov.
