Final-solutions.pdf

007 100points find the value of lim x 2 x ln x 6 x 1

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007 10.0points Find the value of lim x 0 + 2 x - ln x 6 x . 1. limit = -∞ 2. limit = 0 3. limit = 6 4. limit = correct 5. limit = 1 3 6. limit = 2 7. none of the other answers Explanation: Let’s first check if the given limit is an indeterminate form. Now ln x is defined for x > 0 and the graph of ln x has a vertical asymptote at x = 0; in addition, lim x 0 + ln x = -∞ . On the other hand, 2 x - ln x 6 x = 1 3 - ln x 6 x . Thus lim x 0 + 2 x - ln x 6 x = 1 3 + 0 , which is not an indeterminate form. In fact, since the second term is positive, we see that lim x 0 + 2 x - ln x 6 x = . 008 10.0points If f is a linear function whose graph has slope m and y -intercept b , evaluate the inte- gral I = integraldisplay 1 0 f ( x ) dx . 1. I = 1 2 m + 1 2 b

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Version 121 – Final – chen – (53405) 4 2. I = 1 2 m 3. I = m + b 4. I = m 5. I = m + 1 2 b 6. I = 1 2 m + b correct Explanation: Since the graph of f has slope m and y - intercept b , f ( x ) = mx + b . But then by the Fundamental Theorem of Calculus, integraldisplay 1 0 f ( x ) dx = integraldisplay 1 0 ( mx + b ) dx = bracketleftBig 1 2 mx 2 + bx bracketrightBig 1 0 . Consequently, I = 1 2 m + b . 009 10.0points Evaluate the integral I = integraldisplay 1 0 5 x 5 radicalbig 1 - x 2 dx . 1. I = - 5 6 2. I = 5 6 3. I = - 25 12 4. I = 25 12 correct 5. I = 25 6 6. I = - 25 6 Explanation: Set u = 1 - x 2 ; then du = - 2 x dx , while x = 0 = u = 1 x = 1 = u = 0 . Thus I = 5 integraldisplay 1 0 (1 - x 2 ) 1 / 5 · x dx = - 5 2 integraldisplay 0 1 u 1 / 5 du = 5 2 integraldisplay 1 0 u 1 / 5 du . Consequently, I = 5 2 bracketleftBig 5 6 u 6 / 5 bracketrightBig 1 0 = 25 12 . 010 10.0points Determine F ( x ) when F ( x ) = integraldisplay x 5 6 cos t t dt . 1. F ( x ) = 6 cos x x 2. F ( x ) = - 6 sin x x 3. F ( x ) = - 6 sin x x 4. F ( x ) = - 3 cos( x ) x 5. F ( x ) = 3 sin( x ) x 6. F ( x ) = - 6 sin( x ) x 7. F ( x ) = 3 cos x x 8. F ( x ) = 3 cos( x ) x correct
Version 121 – Final – chen – (53405) 5 Explanation: By the Fundamental Theorem of Calculus and the Chain Rule, d dx parenleftBig integraldisplay g ( x ) a f ( t ) dt parenrightBig = f ( g ( x )) g ( x ) . When F ( x ) = integraldisplay x 5 6 cos t t dt , therefore, F ( x ) = 6 cos( x ) x parenleftBig d dx x parenrightBig . Consequently, F ( x ) = 3 cos( x ) x , since d dx x = 1 2 x . 011 10.0points Use linear appproximation to estimate the value of 14 1 / 4 . ( Hint : (16) 1 / 4 = 2.) 1. 14 1 / 4 2 2. 14 1 / 4 61 32 3. 14 1 / 4 15 8 4. 14 1 / 4 63 32 5. 14 1 / 4 31 16 correct Explanation: Set f ( x ) = x 1 / 4 , so that f (16) = 2 as the hint indicates. Then df dx = 1 4 x 3 / 4 . By differentials, therefore, we see that f ( a + Δ x ) - f ( a ) df dx vextendsingle vextendsingle vextendsingle x = a Δ x = Δ x 4 a 3 / 4 . Thus, with a = 16 and Δ x = - 2, 14 1 / 4 - 2 = - 1 16 . Consequently, 14 1 / 4 31 16 . 012 10.0points Determine lim x 0 + x (5 - 7 ln( x )) . 1. limit = -∞ 2. none of the other answers 3. limit = - 7 4. limit = 5. limit = 0 correct 6. limit = - 2 Explanation: Since the limit is of the form 0 · ∞ use of L’Hospital’s Rule is suggested. First write x (5 - 7 ln( x )) = 5 - 7 ln( x ) 1 x and set f ( x ) = 5 - 7 ln( x ) , g ( x ) = 1 x .

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