國立成功大學 微積分 (一) 11 月 9 日, 2018

經濟心理企管交管工資 期中考 13:10-15:00

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Page Points Score

1 20
2 15
3 15
4 25
5 25
Total: 100
Calculus I Midterm Exam Page 1 of 6

1. (5 points) Find the inverse function of the following function.

√
f (x) = 16 − x2 , 0 ≤ x ≤ 4.

√
x= 16 − y 2
√
y = 16 − x2
不部分給分。

2. (5 points) Evaluate the limit √

x2 + 5 − 3
lim .
x→2 x−2

√
x2 + 5 − 3
lim
√x−2
x→2
√
( x2 + 5 − 3)( x2 + 5 + 3)
= lim √
x→2 (x − 2)( x2 + 5 + 3)
x+2
= lim √
x→2 ( x2 + 5 + 3)

2+2 2
= √ =
( 22 + 5 + 3) 3

3. (10 points) Find the derivative of the function

√
f (x) = x x3 + 1.

√ 1
f ′ (x) = x3 + 1 + x · · 3x2 (x3 + 1)−1/2
2
√ 3
= x3 + 1 + x3 (x3 + 1)−1/2
2

通分後答案為
5x3 + 2
√
2 x3 + 1
Calculus I Midterm Exam Page 2 of 6

4. Consider the function

3x
f (x) =
x2+1
(a) (4 points) Find f ′ (x).
(b) (4 points) Find the equation of the tangent line to the graph of f (x) at x = 2.
(c) (7 points) Find all the absolute extreme values of f (x) on the interval [0, 4].

3(x2 + 1) − 3x · 2x 3(1 − x2 )
f ′ (x) = =
(x2 + 1)2 (x2 + 1)2

f (2) = 6/5, f ′ (2) = −9/25. tangent line:

6 −9
y− = (x − 2) or 9x + 25y = 48
5 25

f ′ (x) = 0, ⇒ x = 1.
f (0) = 0, f (1) = 23 , f (4) = 12
17
.
Max: x = 1 or f (1) = 32 .
Min: x = 0 or f (0) = 0.
Calculus I Midterm Exam Page 3 of 6

5. (5 points) Find the tangent line of the curve y 2 + y = x3 + 2x + 3 at the point (1, 2).
2yy ′ + y ′ = 3x2 + 2
′ 3x2 + 2
y =
2y + 1
(1, 2) 代入, y ′ = 1.
tangent line: y − 2 = x − 1 or y = x + 1.

6. (10 points) A company is increasing the production of a product at the rate of 10 units
per week. The demand and cost functions are given by p = 50 − 0.01x and C =
4000 + 40x − 0.02x2 , where x is the number of units produced per week. Find the rate
of change of the profit with respect to time when the weekly sales are x = 800 units..
R = px = 50x − 0.01x2 .
P = R − C = −4000 + 10x + 0.01x2 .

dP dx dx
= 10 + 0.02x
dt dt dt
代入, dP
dt
= 10 · 10 + 0.02 · 800 · 10 = 260.
Calculus I Midterm Exam Page 4 of 6

7. (10 points) A company estimate that the cost of a producing x units of a product can
be modeled by
C = 1600 + 0.05x + 0.0001x3 .
Find the production level that minimizes the average cost per unit.

C̄ = C/x = 1600x−1 + 0.05 + 0.0001x2

C̄ ′ = −1600x−2 + 0.0002x

C ′ = 0, 1600x−2 = 0.0002x, x3 = 8000000, x = 200.

8. (8 points) Determine all vertical and horizontal asymptotes of the function

−2x2 + 7
f (x) = .
x2 − 9

lim f (x) = lim f (x) = −2

x→∞ x→−∞

y = −2. x = 3, x = −3.

√
9. (7 points) Using differential, approximate 1.006.
′ 1 −1/2 ′
f (1) = 1. f (x) = 2
x , f (1) = 0.5.
√ .
1.006 = 1 + 0.5 · 0.006 = 1.003
Calculus I Midterm Exam Page 5 of 6

5 2
10. Consider the function y = f (x) = x 3 − 5x 3 .
(a) (3 points) Determine the interval(s) on which f (x) is increasing.
(b) (3 points) Determine the interval(s) on which f (x) is decreasing.
(c) (5 points) Find all relative extreme values of f (x).
(d) (3 points) Determine the intervals(s) on which f (x) is concave up.
(e) (3 points) Determine the intervals(s) on which f (x) is concave down.
(f) (3 points) Find all point(s) of inflection of f (x).
(g) (5 points) Sketch the graph of y = f (x).

5 2 10 −1 5 −1
f ′ (x) = x 3 − x 3 = x 3 (x − 2)
3 3 3
10 −1 10 −4 10(x + 1)
f ′′ (x) = x3 + x3 = √
3
9 9 9 x4
increasing: x < 0, x > 2.(or (−∞, 0] ∪ [2, ∞))
decreasing: 0 < x < 2.
relative maximum: f (0) = 0. √
5 2
relative minimum: f (2) = 2 3 − 5 · 2 3 = −3 3 4
concave up: (−1, 0) ∪ (0, ∞).
concave down: x < −1.
point of inflection: f (−1) = −1 − 5 = −6, (−1, −6).
Calculus I Midterm Exam Page 6 of 6