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Homework3 - MA 115 Homework Solutions for Week 3 2.7 #4 (a)...

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Unformatted text preview: MA 115 Homework Solutions for Week 3 2.7 #4 (a) Since g (5) = − 3, the point (5 , − 3) is on the graph of g . Since g ′ (5) = 4, the slope of the tangent line at x = 5 is 4. Using the point-slope form of a line gives us y − ( − 3) = 4( x − 5), or y = 4 x − 23. (b) Since (4 , 3) is on y = f ( x ), f (4) = 3. The slope of the tangent line between (0 , 2) and (4 , 3) is 1 4 , so f ′ (4) = 1 4 . 2.7 #26 v (5) = f ′ (5) = lim h → f (5 + h ) − f (5) h = lim h → [(5 + h ) − 1 − (5 + h )] − (5 − 1 − 5) h = lim h → 1 5+ h − 5 − h − 1 5 + 5 h = lim h → 1 5+ h − h − 1 5 h = lim h → 5 − 5 h (5+ h ) − (5+ h ) 5(5+ h ) h = lim h → 5 − 25 h − 5 h 2 − 5 − h 5 h (5 + h ) = lim h → − 5 h 2 − 26 h 5 h (5 + h ) = lim h → h ( − 5 h − 26) 5 h (5 + h ) = lim h → − 5 h − 26 5(5 + h ) = − 26 25 m/s The speed when t = 5 is vextendsingle vextendsingle − 26 25 vextendsingle vextendsingle = 26 25 = 1 . 04 m/s. 2.7 #27 (a) f ′ ( x ) is the rate of change of the production cost with respect to the number of ounces of gold produced. Its units are dollars per ounce. (b) After 800 ounces of gold have been produced, the rate at which the production cost is increasing is $17 per ounce. So the cost of producing the 800th (or 801st) ounce is about $17. (c) In the short term, the values of f ′ ( x ) will decrease because more efficient use is made of start-up costs as x increases. But eventually f ′ ( x ) might increase due to large-scale operations. 2.8 #3 (a) II. Since from left to right, the slopes of the tangents to graph (a) start out negative, become 0, then positive, then 0, then negative again. The actual function values in graph II follow the same pattern.graph II follow the same pattern....
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This note was uploaded on 04/07/2008 for the course MA 115 taught by Professor Mahalanobis during the Fall '08 term at Stevens.

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Homework3 - MA 115 Homework Solutions for Week 3 2.7 #4 (a)...

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