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Calculus I - Fall 2007 - Yahdi Math 111 - Solution of Test 1 Problem 1. The graph of y = f (x) is given below. For each of the graphs 1, 2, 3, 4 and 5 explain how it is obtained from the graph of f (x) and give its corresponding formula. • Graph 1 is obtained from the graph the f (x) by a vertical shift of 3 units upward, therefore the corresponding formula is f (x) + 3 . • Graph 3 is obtained from the graph the f (x) by horizontal shift of 4 units to the right, therefore the corresponding formula is f (x − 4) . • Graph 4 is obtained from the graph the f (x) by a vertical compress by a factor 1 3, therefore the 1 3 f (x) corresponding formula is . • Graph 5 is obtained from the graph the f (x) by a vertical reflection about the x-axis −→ −f (x), then horizontal shift of 4 units to the left, therefore the corresponding formula is −f (x + 4) . • Graph 2 is obtained from the graph the f (x) by horizontal shift of 6 units to the left −→ f (x + 6), then a vertical stretch by a factor 2, therefore the corresponding formula is 2f (x + 6) . √ Problem 2. Find the domain of the function f (x) = x2 − 4x + 3 . x−4 √ x2 − 4x + 3 exists if x2 − 4x + 3 > 0 and x − 4 6= 0. x−4 • Solve x2 − 4x + 3 = 0, then test a number in each interval to find the sign of this expression. x2 − 4x + 3 = 0 ⇐⇒ (x − 1)(x − 3) = 0 ⇐⇒ x = 1 or x = 3. Testing a number in each interval, we get: x = 0 =⇒ 02 −4(0)+3 = 3 positive, x = 2 =⇒ 22 −4(2)+3 = −1 negative, and x = 4 =⇒ 42 −4(4)+3 = 3 positive. Therefore, positive 1 negative 3 positive . • Solving x − 4 = 0, we get x = 4. • So f (x) exists for x 6 1, x > 3, x 6= 4. In other terms, the domain is (−∞, 1] ∪ [3, 4) ∪ (4, +∞) . f (x) = Problem 3. A small-appliance manufacturer finds that it costs $9,000 to produce 1,000 toaster ovens a week and $12,000 to produce 1,500 toaster ovens a week. Assume it is a linear relationship. Express the cost as a function of the number x of toaster ovens produced. 1 2 Let x be the number of toaster ovens produced a week, and let y = f (x) be the cost to produce these x toaster. Since it is a linear function then, using the point-slope form we have y = m(x − x1 ) + y1 . We have: ( x1 = 1, 000toasters −→ y1 = $9, 000, 12, 000 − 9, 000 3, 000 y2 − y1 Therefore, m = = = = 6. Thus x2 − x1 1, 500 − 1, 000 500 x2 = 1, 500toasters −→ y2 = $12, 000. y = 6(x − 1, 000) + 9, 000 = 6x + 3, 000 Problem 4. Find the formula of the function f(x) that best fits the data in the following table: x 0 2 4 6 f(x) 1.21 2.7225 6.12562 13.7826 The data fits the model of an exponential function f (x) = bax because we have constant ratios in y-values 2.7225 6.12562 13.7826 for equally spaced x-values. Indeed, = = = 2.25. We need to choose two points: 1.21 2.7225 6.12562 • (0, 1.21) −→ f (0) = 1.21 −→ ba0 = 1.21 −→ b = 1.21 √ 2.7225 • (2, 2.7225) −→ f (2) = 2.7225 −→ ba2 = 2.7225 −→ a2 = 2.7225 a = 2.25 = 1.5 = = 2.25 =⇒ b 1.21 Conclusion: f (x) = 1.21(1.5)x Problem 5. Under ideal conditions a certain bacteria population is known to triple every three hours. Suppose that there are initially 25 bacteria. (1) Give a formula for the size of this population after t hours. (2) How long will it take for the population to reach 2,500? Denotes by P (t) the bacteria population at the time t, where t is measured in hours and the given initial population is b = P (0) = 25. The population triples every 3 hours, therefore: (1) After 3 hours −→ only 1 period of 3 hours −→ P (3) = 3(25) =⇒ P (3) = 3(25) After 6 hours −→ 2nd period of 3 hours −→ P (6) = 3[3(25)] = 32 (25) We have a pattern: after t hours, we have 3t periods of 3 hours. Since the initial population 25 will t triple each 3 hours, then it will triple 3t times after t hours. Therefore, P (t) = 3 3 25 (2) We have to solve for t equation: P(t) = 2, 500 t t 2, 500 P (t) = 25, 000 =⇒ 3 3 25 = 2, 500 =⇒ 3 3 = = 100 25 Apply the logarithm “ln” to both sides of the last equation and use ln AB = B ln A : t ln 3 3 = ln 100 =⇒ t t ln 100 ln 100 ln 3 = ln 100 =⇒ = =⇒ t = 3 ≈ 12.5754 hours 3 3 ln 3 ln 3 Problem 6. Find a formula for the inverse function f −1 for each of the following functions √ (1) f (x) = x + 1. (2) f (x) = 4 + ln 3x. (1) We need to solve for x, f (x) = y. We have: √ x + 1 = y ⇔ x + 1 = y2 ⇔ x = y2 − 1 Now, interchange y with x, we get: f −1 (x) = y 2 − 1 . (2) We need to solve for x, f (x) = y. We have: 4 + ln 3x = y ⇔ ln 3x = y − 4 ⇔ eln 3x = e(y−4) ⇔ 3x = e(y−4) ⇔ x = Now, interchange y with x, we get: f −1 (x) = e(x−4) . 3 e(y−4) 3