Find All Values of B for Which F X is Continuous on ž ž

Description

MAT 1308 A – Midterm I
October 2th, 2019. V.1
Professor: Hai Yan Liu(Jack)
NAME
NAME
STUDENT #
Instructions:
• You have 80 minutes to complete this exam.
• Only non-programmable and non-graphic calculators are allowed.
• This is a closed-book exam. No notes, book or any electronic device is allowed.
• Check that you have 8 questions. Do not detach the document.
• Questions 1 to 4 are multiple choice questions, each is worth 2 points, and no partial
marks will be given for these questions. Please write your answers to these questions
in the table below.
• Questions 5 to 8 are long-answer questions, so organise your time accordingly. A correct
answer also needs a clearly written justification written.
• You can use the back of the pages as draft paper, or to continue your answer and work,
but you should indicate it clearly.
• Good luck!
Answers to multiple choice questions:
Q1
Q2
Q3
Q4
Do not write in this table
QCM Q5
Q6
Q7
Q8
Total
/8
/6
/4
/7
/6
/31
2
1. (2 points) Solve the following equation for x ∈ R :
|x − 5| > 7
A. (−2, ∞)
B. (−2, 12)
E. None of these choices
C. (−∞, −2) ∪ (12, +∞)
D. (−2, 1) ∪ (1, +∞)
2. (2 points) Find the equation of the tangent line to the graph of f (x) = (x−3)ex at x = 0.
A. y = 2x + 3
B. y = −2x − 3
E. None of these choices
C. y = 2e2 x + 3e2
D. y = 2e2 x − 3e2
3. (2 points) Find all values of x for which the slope of the tangent line to the graph of
f (x) = (x4 − 18×2 )2 is 0.
√
√
A. x = 0, x = 18, x = −18
B. x = 0, x = 3, x = −3,√ x = 18, √
x = − 18
C. x = 18, x = −18, x = 9, x = −9
D. x = 2, x = 18, x = − 18, x = 9
E. None of these choices
3
4. (2 points) The revenue and cost functions for a given product are the following:
R(x) = −2×2 + 600x,
C(x) = 150 + 10000,
where x is the number of the produced units. What is the marginal profit when x = 110
units?
A. 10$
B. 30$
C. 100$
D. 200$
E. None of these choices
4
5. Evaluate the following limits if they exist:
3x − 3
x→1 x2 − 1
(a) (2 points) lim
2|x − 2|
x→2 (2x − 4)(x + 2)
(b) (2 points) lim
5 − 3×2
x→+∞ 5×2 + 8
(c) (2 points) lim
5
6. (4 points) Let f be defined as
(
4×2
f (x) =
(x − d)2
if x < 2, if x ≥ 2. Find all values of d for which the function is continuous everywhere. Justify. 6 7. (a) (1 point) Give the definition for the derivative of a function f at any point x. √ (b) (6 points) Use the definition given in (a) to find the derivative of f (x) = x − 1. 7 8. (6 points) Find the derivative of each of the following function using appropriate rules, you don’t need to simplify your answer. 5x2 + 6x (a) (3 points) f (x) = 3 3x + 2x + 1 dy at point(1,2) for the following fuction dx 3 2 2 x y + 4x = −y + 10 (b) (3 points) Find MAT 1308 A – Midterm I October 2th, 2019. V.1 Professor: Hai Yan Liu(Jack) NAME NAME STUDENT # Instructions: • You have 80 minutes to complete this exam. • Only non-programmable and non-graphic calculators are allowed. • This is a closed-book exam. No notes, book or any electronic device is allowed. • Check that you have 8 questions. Do not detach the document. • Questions 1 to 4 are multiple choice questions, each is worth 2 points, and no partial marks will be given for these questions. Please write your answers to these questions in the table below. • Questions 5 to 8 are long-answer questions, so organise your time accordingly. A correct answer also needs a clearly written justification written. • You can use the back of the pages as draft paper, or to continue your answer and work, but you should indicate it clearly. • Good luck! Answers to multiple choice questions: Q1 Q2 Q3 Q4 Do not write in this table QCM Q5 Q6 Q7 Q8 Total /8 /6 /4 /7 /6 /31 2 1. (2 points) Solve the following equation for x ∈ R : |x − 5| > 7
A. (−2, ∞)
B. (−2, 12)
E. None of these choices
C. (−∞, −2) ∪ (12, +∞)
D. (−2, 1) ∪ (1, +∞)
Solution: Since |x − 5| > 7 ⇔ x − 5 > 7, or x − 5 < −7; ⇔ x > 12, or x < −2, therefore choose C. 2. (2 points) Find the equation of the tangent line to the graph of f (x) = (x−3)ex at x = 0. A. y = 2x + 3 B. y = −2x − 3 E. None of these choices C. y = 2e2 x + 3e2 D. y = 2e2 x − 3e2 Solution: From y −f (x0 ) = f 0 (x0 )(x−x0 ), we have f (0) = −3, and f 0 (x) = ex +(x−3)ex f 0 (0) = 1 − 3 = −2. Therefore y − (−3) = −2(x − 0). Finally y = −2x − 3, so answer is B. 3. (2 points) Find all values of x for which the slope of the tangent line to the graph of f (x) = (x4 − 18x2 )2 is 0. √ √ A. x = 0, x = 18, x = −18 B. x = 0, x = 3, x = −3,√ x = 18, √ x = − 18 C. x = 18, x = −18, x = 9, x = −9 D. x = 2, x = 18, x = − 18, x = 9 E. None of these choices Solution: f 0 (x) =√2(x4 − 18x2√ )(4x3 − 36x) = 8x3 (x2 − 18)(x2 − 9) = 0, we get x = 0, x = 3, x = −3, x = 18, x = − 18, so the answer is B. 3 4. (2 points) The revenue and cost functions for a given product are the following: R(x) = −2x2 + 600x, C(x) = 150x + 10000, where x is the number of the produced units. What is the marginal profit when x = 110 units? A. 10$ B. 30$ C. 100$ D. 200$ E. None of these choices Solution; Profit is P (x) = R(x) − C(x) = −2x2 + 600x − 150x − 10000 = −2x2 + 450x − 10000, then the marginal profit is P 0 (x) = R0 (x) − C 0 (x) = −4x + 450. Therefore, P 0 (110) = 10 , so the answer is A. 4 5. Evaluate the following limits if they exist: (a) (2 points) lim x→1 3x − 3 x2 − 1 3x − 3 3(x − 1) 3 3 = lim = lim = 2 x→1 x − 1 x→1 (x − 1)(x + 1) x→1 x + 1 2 lim 2|x − 2| x→2 (2x − 4)(x + 2) (b) (2 points) lim We need to discuss both sides, the left side: lim− x→2 2|x − 2| −2(x − 2) = lim− = (2x − 4)(x + 2) x→2 2(x − 2)(x + 2) −2 −1 lim− = . x→2 2(x + 2) 4 2|x − 2| 2(x − 2) 2 1 = lim+ = lim+ = x→2 (2x − 4)(x + 2) x→2 (2(x − 2)(x + 2) x→2 2(x + 2) 4 Therefore, the left side limit is not equal to the right side of limit. So the limit does not exist. The right side: lim+ 5 − 3x2 x→+∞ 5x2 + 8 (c) (2 points) lim 5 x2 ( x52 − 3) −3 5 − 3x2 −3 x2 lim = = lim = lim 8 8 x→+∞ 5x2 + 8 x→+∞ x2 (5 + 2 ) x→+∞ 5 + 2 5 x x 5 6. (4 points) Let f be defined as ( 4x2 f (x) = (x − d)2 if x < 2, if x ≥ 2. Find all values of d for which the function is continuous everywhere. Justify. Solution: Continuous should be both sides continuous. lim− f (x) = 4(2)2 = 16. x→2 2 2 lim f (x) = (2 − d) . Therefore, 16 = (2 − d) , 2 − d = ±4, ⇔ d = −2, d = 6 x→2+ 6 7. (a) (1 point) Give the definition for the derivative of a function f at any point x. √ (b) (6 points) Use the definition given in (a) to find the derivative of f (x) = x − 1. f (x + h) − f (x) h→0 h The definition f 0 (x) = lim √ √ x+h−1− x−1 f (x + h) − f (x) = lim f (x) = lim h→0 h→0 h h √ √ √ √ x+h−1− x−1 x+h−1+ x−1 √ √ = lim h→0 h x+h−1+ x−1 x + h − 1 − (x − 1) h 1 √ √ = lim √ = lim √ = √ h→0 h( x + h − 1 + h→0 2 x−1 x − 1) h( x + h − 1 + x − 1) 0 7 8. (6 points) Find the derivative of each of the following function using appropriate rules, you don’t need to simplify your answer. 5x2 + 6x (a) (3 points) f (x) = 3 3x + 2x + 1 g 0 h − gh0 h2 (10x + 6)(3x3 + 2x + 1) − (5x2 + 6x)(9x2 + 2) = (3x3 + 2x + 1)2 f 0 (x) = dy at point(1,2) for the following fuction dx x3 y 2 + 4x2 = −y + 10 Solution: Both sides derivative about x (b) (3 points) Find 0 (x3 y 2 + 4x2 )x = −yx0 3x2 y 2 + x3 2yyx0 + 8x = −yx0 (2x3 y + 1)y 0 = −(8x + 3x2 y 2 ) dy (8x + 3x2 y 2 ) =− dx (2x3 y + 1) at point(1, 2), the dy −20 = = −4 dx 5 1midterm me Left:1:13:40 Xiaoyang Jia: Attempt 1 Domain Find the domain of function Question 2 (2 points) The domain of () Ã'… 1 4-3e-52 The set of all real numbers x such that O a) x + - In (3/4) Jling Ob)x + - ln(5/3) O x 7 - 1 1n (5/4) O d) d) x + -In (3/5) ex+ - In(4/3) Onx + - In (4/5) Limit Zoom AS - Roo... 2 3 Limit Use limit rule to find limit. un 6 Question 3 (2 points) Find lim f(x) for the function f ( x (2 20 Vx+4–2 a) 2 b) 1/4 O c) 4 O d) o e) 1/2 Break even points Break even points 2 3 5 6 Question 4 (2 points) A company has determined that the following functions are good models for their daily cost C(x) and revenue R(x), where x represents the number of units produced, or sold in one day: C(x) = -14+ 8x, - - R(x) = 2x2 + 202 How many units should the company produce and sell each day in order to break even points. O a) 5 b) 6 c) 7 d) 8 e) 1 3 Question 5 (2 points) Let f(x) be a piecewise function defined as follows, where a, b, and care constants: f(x) = ae + x +1 if x < 0 3 if x = 0 bxtc if x > 0
f(x) is differentiable at x=0.
a) a=3, b=2, c=3
b) a=2, b=3, c=3
O c) a=3, b=3, c=3
d) a=2,c=2, 1-3
e) a=-3.b=2,c=3
Of) a=3, b=3,c=2
Limits
2
3
>
Question 6 (4 points)
5
6
Find following limits
a) lim
+4-
|x-41
22-16
+3-t+1
b) lim
t700 3-4t2-573
I
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2.
3
Question 7 (6 points)
a) Use the definition of derivative to find the derivative of
4
5
6
7
7
f(x) = v1 – 2
()
c). Suppose : 4y² + 2x²y2 – 23 = 5, then find me at
+
t (1, 1).
b). Given function gly
4y2-5y
8y-4
find derivative g'(y)
dy
at
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