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MATH 142 July. 21, 2008
Exam I - Form A
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Instructions:
- Except for the multiple choice questions where the answer must be circled, you must show appropriate legible work to receive full credit. If you use a calculator write down all the steps. DO NOT just put an answer.
- There are 100 possible points. Point values for each problem are as indicated.
- Check and make sure there are eight pages including the cover page, when you begin the exam.
• SCHOLASTIC DISHONESTY WILL NOT BE TOLERATED.
Good Luck!
Multiple Choice: Please circle the letter of the correct answer AND write the letter of the correct answer.
- (5 points) The composite function h(x) = f [g(x)] = ln(
x^3 − 5), if (a) f (x) = ln x − 5 , g(x) =
x^3 (b) f (x) =
x^3 , g(x) = ln x − 5 (c) f (x) = ln
x, g(x) = x^3 − 5 (d) f (x) = ln x^3 , g(x) =
x − 5 (e) f (x) = x^3 − 5 , g(x) = ln
x (f) none of the above
- (5 points) The derivative of f (x) = e^4 ln 4 is
(a) e^4 ln 4 + 14 e^4 (b) 1 (c) 14 e^4 (d) 0 (e) 4e^3 ln 4 + 14 e^4 (f) none of the above
- (6 points) Let f (x) be a function whose domain is (−∞, 1) ∪ (1, ∞) and whose deriva- tive is f ′(x) = (x + 1) (x − 1)^5 . Indicate where f (x) is increasing and where it has local min/max.
(a) increasing in (−∞, −1) ∪ (1, ∞), max at - Note: f is not defined at 1, therefore it has no min there. (b) increasing in (−∞, −1), max at - (c) increasing in (−∞, −1) ∪ (− 1 , ∞), no max or min (d) increasing in (−∞, −1) ∪ (1, ∞), max at -1, min at 1 (e) increasing in (− 1 , 1), minimum at -1, max at 1 (f) none of the above
- (8 points) Sketch the graph of a function that has the following properties.
a) Continuous for all real numbers except x = 2 b) Vertical asymptote at x = 2 c) (^) x→−∞lim f (x) = −2, lim x→∞ f (x) = 3 d) f (−4) = 0, f (0) = 3
- (6 points) Find dy dx
given y = e^5 u, u = ln(2x − 1) Solution 1: dy du = 5e^5 u, du dx
2 x − 1
Therefore dy dx
dy du
du dx
10 e^5 ln(2x−1) 2 x − 1
10 eln(2x−1)^5 2 x − 1 = 10(2x − 1)^4
Solution 2: y(x) = e5 ln(2x−1). Using the chain rule we get dy dx
= e5 ln(2x−1)^ ·
2 x − 1
· 2 = 10(2x − 1)^4
Solution 3: y(x) = e5 ln(2x−1)^ = eln(2x−1)^5 = (2x − 1)^5. Then,
dy dx = 5(2x − 1)^4 · 2 = 10(2x − 1)^4
- (3 points) You are given the following information. Compute f ′(2). If it is not possible, then tell what additional information will be needed. f (2 + h) = 7h^5 − h^2 + 8h + 13 f (2) = 13
Solution: f ′(2) = lim h→∞
f (2 + h) − f (2) h
- (6 points) Find the equation of the tangent line to f (x) =
2 x + 5 at x = 2.
Solution: f (2) = 3, therefore the tangent passes through the point (2, 3). To find the slope copmute f ′(2). f ′(x) =
2 x + 5
⇒ m = f ′(2) = 1/ 3.
Thus, the equation of the tangent is y = 1/3(x − 2) + 3.
- (6 points) $600 is deposited in an account that compounds continuously at 7.25% per- cent per year. How long will it take for the account to be worth $1800?
Solution:
1800 = 600e^0.^0725 ·t^ ⇒ 3 = e^0.^0725 t. Take ln from both sides ⇒
ln 3 = 0. 00725 t ⇒ t = ln 3
- 0725
- (12 points) The price-demand equation and the cost function of a product are given respectively by p = 300 − 0. 1 x and C(x) = 300 + 20x, 0 ≤ x ≤ 3000, where x is the quantity produced and C(x) is cost in dollars.
(a) Find the exact profit of the 11th^ item.
P (x) = R − C = x(300 − 0. 1 x) − (300 + 20x) = − 0. 1 x^2 + 280x − 300 The exact profit of the 11th^ item is P (11) − P (10) = 2767. 9 − 2490 = $277. 9
(b) Approximate the profit of the 11th^ item using marginal analysis.
P ′(x) = 280 − 0. 2 x, so the approximate profit of the 11th^ item is P ′(10) = 278.
(c) Find the marginal average profit function.
The average profit is P (x) = P^ ( xx )= 280 − 0. 1 x − (^300) x. The marginal average profit is (P (x))′^ = − 0 .1 + (^300) x 2
- (18 points) Find the derivative (without simplifying) of:
(a) f (x) = (x^3 − 2 x)^4 √ x^2 + 1
f ′(x) =
4(x^3 − 2 x)^3 (3x^2 − 2)
x^2 + 1 − (x^3 − 2 x)^4 2 √^2 xx (^2) + x^2 + 1
(b) f (x) = (4x^ + 5)(x^3 + 4x)^5
f ′(x) = (4x^ ln 4)(x^3 + 4x)^5 + (4x^ + 5)[5(x^3 + 4x)^4 (3x^2 + 4)]
(c) f (x) = 3
x^5 − ln x^8 + e^7 +
x^4
f (x) = x^5 /^3 − 8 ln x + e^7 + 3x−^4 , f ′(x) =
x^2 /^3 −
x
− 12 x−^5
