# Keys - Solutions to Practice Problems True or False 1F 2F 3...

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Solutions to Practice Problems. True or False: 1F 2F 3 T 4 F 5 F 6T 7T 8F 9F 10 T 11F 12 T 13 T 14 F 15 T 16F 17T 18 F 19 F 20F 21T 22F 23F 24F 25F 26 T 27F 28F 29T 30T 31F Multiple-choice: 1C 2D 3C 4A 5B 6C 7C 8 B 9B 10D 11A 12D 13C 14B 15D 16C 17C 18B19A 20C 21A 22C 23C 24A 25A 26C 27A 28B 29B 30C 31B 32D 33A 34D 35B 36A 37E 38C 39B 40B 41B 42C 43B 44A 45B 46B 47D 48C 49D 50B 51B 52C 11. A From the demand equation 50 - p ( x 2 + 1) = 0, we can get p = 50 x 2 +1 . So the revenue function is R ( x ) = p · x = ( 50 x 2 +1 ) x = 50 x x 2 +1 . 12.D. The marginal revenue function is the ﬁrst derivative of the revenue function, that is. R 0 ( x ) = (50)( x 2 +1) - (50 x )(2 x ) ( x 2 +1) 2 = 50 - 50 x 2 ( x 2 +1) 2 (by using quotient rule). 13. C Method 1: The slope function is equal to the ﬁrst derivative function dy dx = 2 x - 3. So, at point (1 , - 2), the slope is m = 2(1) - 3 = - 1. Therefore the equation of the tangent line to the graph of y = x 2 - 3 x at the point (1 , - 2) is y - ( - 2) = ( - 1)( x - 1), that is y = - x - 1. Method 2: Similarly to Method 1, we can ﬁnd the slope is m = - 1. Now, we suppose the equation is y = - x + b . Since when x = 1, y = - 2, we get - 2 = - (1) + b . Then b = - 1. So the equation of the tangent line to the graph of y = x 2 - 3 x at the point (1 , - 2) is y = - x - 1. 14.B The slope function is equal to the ﬁrst derivative function dy dx = ln x + x ( 1 x ) = ln x + 1 (by using product rule and d dx [ln x ] = 1 x ). When x = 1, the slope is m = ln1 + 1 = 1. Suppose the equation is y = x + b . Since when x = 1, y = 0, we get 0 = (1) + b . Then b = - 1. So the equation of the tangent line to the graph of y = x ln x at the point (1 , 0) is y = x - 1. 15. D NOTICE THAT: d dx [ln f ( x )] = ( 1 f ( x ) ) d dx [ f ( x )]. 1

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The slope function is equal to the ﬁrst derivative function dy dx = ( 1 x 2 )(2 x ) = 2 x x 2 = 2 x . When x = 2, the slope is m = 2 2 = 1. Suppose the equation is y = x + b . Since when x = 2, y = ln4, we get ln 4 = (2) + b . Then b = ln4 - 2. So the equation of the tangent line to the graph of y = ln( x 2 ) at the point (2 , ln 4) is y = x + ln 4 - 2. 16. C NOTICE THAT: d dx [ e f ( x ) ] = ( e f ( x ) ) d dx [ f ( x )]. The slope function is equal to the ﬁrst derivative function dy dx = ( e 2 x - 3 )(2) = 2 e 2 x - 3 . When x = 3 2 , the slope is m = 2 e 2( 3 2 ) - 3 = 2 e 0 = 2. Suppose the equation is y = 2 x + b . Since when x = 3 2 , y = 1, we get 1 = (2)( 3 2 ) + b . 1 = 3 + b. Then b = - 2. So the equation of the tangent line to the graph of y = e 2 x - 3 at the point ( 3 2 , 1) is y = 2 x - 2. 17.C NOTICE THAT: d dx [ e f ( x ) ] = ( e f ( x ) ) d dx [ f ( x )]. The slope function is equal to the ﬁrst derivative function dy dx = ( e - x 2 )( - 2 x ) = - 2 xe - x 2 . When x = 1, the slope is m = - 2(1) e - (1) 2 = - 2 e - 1 = - 2 e . Since the tangent line is through the point (1 , 1 /e ), the equation of the tangent line is y - 1 e = - 2 e ( x - 1)(slope-point form), that is y = - 2 e ( x - 1) + 1 e 18. B. d dx [ x 1 3 + y 1 3 ] = d dx [1] .
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Keys - Solutions to Practice Problems True or False 1F 2F 3...

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