Calculus Midterm #1 October 14, 2009 Time:8:10-10:10
1. (15 points) Find the limit if it exists. Otherwise, explain why it does not.
(a) lim
x→−3
x2+ x − 6 x + 3 . (b) lim
x→−3
x2+ x − 6 (x + 3)2 . (c) lim
x→−3
x2+ x − 6
√x + 3 .
<sol:> (a) lim
x→−3
x2+ x − 6
x + 3 = lim
x→−3
(x + 3)(x − 2)
x + 3 = lim
x→−3(x − 2) = −5.
(b) lim
x→−3
x2+ x − 6
(x + 3)2 = lim
x→−3
(x + 3)(x − 2)
(x + 3)2 = lim
x→−3
(x − 2) (x + 3). Since lim
x→−3−
(x − 2)
(x + 3) = +∞ and lim
x→−3+
(x − 2)
(x + 3) = −∞, the left limit and right limit are both unbounded as x approaches −3, the limit doesn’t exist.
(c) lim
x→−3
x2+ x − 6
√x + 3 = lim
x→−3
(x + 3)(x − 2)
√x + 3 = lim
x→−3
√x + 3(x − 2) = 0 · (−5) = 0.
2. (15 points) Find the derivatives at all points where they exist.
(a) (7 points) y = (x2+ 3x3)(x−3+ x1/2).
(b) (8 points) y = x2+ 3x3 x−3+ x1/2.
<sol:>
(a) y0 = (x2+ 3x3)0(x−3+ x1/2) + (x2+ 3x3)(x−3+ x1/2)0
= (2x + 9x2)(x−3+ x1/2) + (x2+ 3x3)((−3)x−4+ (1/2)x−1/2).
(b) y0 = (x−3+ x1/2)(x2+ 3x3)0− (x2+ 3x3)(x−3+ x1/2)0
(x−3+ x1/2)2 .
= (x−3+ x1/2)(2x + 9x2) − (x2+ 3x3)((−3)x−4+ (1/2)x−1/2) (x−3+ x1/2)2
3. (10 points) Find the tangent to the graph of g(x) = 7x−3+ x2+ 5 at the point (−1, −1).
<sol:>
首先(−1, −1)為函數圖形上的一點。(g(−1) = 7(−1)−3+ (−1)2+ 5 = −7 + 1 + 5 = −1) 因此可以透過微分求斜率g0(x) = −21x−4+ 2x , 所以過(−1, −1)點的斜率為
m = g0(−1) = −21 − 2 = −23 .
最後過(−1, −1)的切線方程式為y + 1 = −23(x + 1).
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4. (10 points) Write down the definition that a function f (x) is continuous at x = c; and the definition that a function f (x) is differentiable at x = c.
<sol:>
(1) limx→c = f (c).
(或是 (1) 在c點極限存在, 且 (2)f (x)在c點有定義, 且 (3) 極限值=函數值).
(2) limx→cf (x)−f (c) x−c 存在.
5. (10 points) Let
f (x) = x sin (x1) , x 6= 0 0 , x = 0 . (a) (5 points) Find the limit of f (x) at x = 0.
(b) (5 points) Use the definition to show that f (x) is not differentiable at x = 0.
<sol:> (a) Since −1 ≤ sin1x ≤ 1. If x ≥ 0, the inequality become −x ≤ x sin1x ≤ x, then lim
x→0+
−x ≤ lim
x→0+
x sin1
x ≤ lim
x→0+
x. By Squeeze theorem, lim
x→0+
x sin1 x = 0.
Similarly, for x < 0, we have lim
x→0−
x sin1
x = 0. Then the limit lim
x→0x sin1 x = 0 by lim
x→0+x sin1
x = lim
x→0−x sin1 x = 0.
(b) lim
x→0
f (x) − f (0)
x − 0 = lim
x→0
x sin1x − 0
x = lim
x→0sin1
x, the limit value does not exist, and hence the function f (x) is not differentiable at x = 0.
6. (15 points) Let f (x) = x2 and c = 2.
(a) (5 points) Let ε0 = 0.1. Find δ0 > 0 so that |f (x)−4| < ε0, whenever 0 < |x−c| < δ0.
<sol:>
Choose any 0 < δ0 < min{2 −√
4 − 0.1,√
4 + 0.1 − 2}.
(or 0 < δ0 < min{0.02515822 . . . , 0.02454567 . . .}).
(b) (5 points) For ε1 = 0.01 and ε2 = 0.001, find the corresponding δ1 > 0 and δ2 > 0 such that |f (x) − 4| < εi, ∀ 0 < |x − c| < δi, i = 1, 2.
<sol(1):>
Choose any 0 < δ1 < min{2 −√
4 − 0.01,√
4 + 0.01 − 2}.
(or 0 < δ1 < min{0.0025016 . . . , 0.00249844 . . .}).
<sol(2):>
Choose any 0 < δ2 < min{2 −√
4 − 0.001,√
4 + 0.001 − 2}.
(or 0 < δ2 < min{0.000250015626 . . . , 0.00024998 . . .}).
(c) (5 points) For any arbitrarily small ε > 0, give a general formula for δ > 0 such that
|f (x) − 4| < ε, ∀ 0 < |x − c| < δ. Argue that the limit of f (x) at x = c is 4.
<sol:>
For any ε > 0, without loss of generality, we can say ε < 4. Choose 0 < δ <
min{2 −√
4 − ε,√
4 + ε − 2}. then |f (x) − 4| < ε as long as 0 < |x − 2| < δ.
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Since for arbitrary ε > 0 we can find a corresponding δ such that |f (x) − 4| < ε as long as 0 < |x − 2| < δ, hence the limit of f (x) at x = 2 is 4.
7. (10 points) Let
g(x) = 4 − x , x < −1 4x − x2 , x > 1 . Notice that the g(x) is defined everywhere except for x ∈ [−1, 1].
(a) (5 points) Insert a linear function y = ax + b for x ∈ [−1, 1] so that g(x) together with this linear function becomes continuous on every point of R. Call this new function ˜g(x).
(b) (5 points) Find the derivative of ˜g(x) at all differentiable points. You also have to identify all non-differentiable points.
<sol:>
(a) 首先要使整個函數為連續函數, 所以左右極限要等於該點的函數值 ( g(−1) = a(−1) + b = 4 − (−1) =˜ lim
x→−1−g(x) = 4 − (−1) = 5
˜
g(1) = a(1) + b = 4(1) − (1)2 = lim
x→1+g(x) = 4(1) − (1)2 = 3 ⇒ a = −1 b = 4 因此化簡上式可得函數為 ˜g(x) =
4 − x , x < 1 4 − x , −1 ≤ x ≤ 1 4x − x2 , x > 1
= 4 − x , x ≤ 1 4x − x2 , x > 1
(b) 將上式的 ˜g(x) = 4 − x , x ≤ 1
4x − x2 , x > 1 微分可得
˜
g0(x) = −1 , x < 1
4 − 2x , x > 1 . 其中當x = 1時, lim
x→1+
˜
g0(x) = lim
x→1+
(4x − x2) − 3
x − 1 = 2 6= −1 = lim
x→11
(4 − x) − 3
x − 1 = lim
x→1−
˜ g0(x) 所以x = 1時不可微
8. (15 points) At 0◦ Celsius, the heat loss H (in kilocalories per square meter per hour) from a person’s body can be modeled by H = 33(10√
v − v + 10.45), where v is the wind speed (in meters per second).
(a) (7 points) Find the rate of change of the heat loss when the wind is blowing at 2 meters per second. Find also the rate of change of the heat loss when the wind is blowing at 5 meters per second.
(b) (8 points) At which speed of the wind, 2 meters per second or 5 meters per second, will a person suffer from more heat loss? Explain why.
<sol:>
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(a) 要算熱流失的速率, 就要先算其熱流失方程的微分 H0(x) = 33(10(2√1v) − 1) 其中的
H0(2) = 33(10( 1
2√
2) − 1) ≈ 83.6726187 H0(5) = 33(10( 1
2√
5) − 1) ≈ 40.7902424
(b) 本題是在問熱流失的量, 所以是算 max{H(2), H(5)} 其中的 H(2) = 33(10√
2 − 2 + 10.45) ≈ 745.540475 H(5) = 33(10√
5 − 2 + 10.45) ≈ 917.752433
而在本題可以看的出來, 熱流失的速率是在風速v = 2的時候比較快; 但是熱流失的量是在風 速v = 5的時候比較大。 因此流失的速率和流失的量並無直接關係。
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