10FallHwk08_Soln

10FallHwk08_Soln - Math 2413 Homework Set 8 Solutions 10...

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Math 2413 Homework Set 8 – Solutions 10 Points 1. (2 pts) () () ( )( ) 43 2 32 2 3 1 5 1 2 42 3 0 2 325 fx xx x f x x x ¢ ¢¢ =+- - =+-=+- So, it looks like possible inflection points are : 5 2 3,0, x =- . Here is a number line for the 2 nd derivative. From this we get the following concavity information. 55 22 Concave Up : 3 0 , Concave Down : 3 ,0 x x -< < < < ¥ - << The inflection points are then : 5 2 3,0, x . 4. (2 pts) This is a polynomial and so is continuous and differentiable everywhere, in particular, on the given interval and so the Mean Value Theorem can be used. () ( ) () 2 6 1 4 1 2 3 6 16 g x gg ¢ + - = -= Applying the Mean Value Theorem and solving gives, () ( ) ( ) 2 2 1 6 12 42 6 1 4 1 14 1 23 6 1 4 1 3 0 7 12 7 0.7116 , 3.0449 ff cc fc c -- ¢ - + = = == - - = & = The second solution is not in the interval and the first solution is in the interval and the only value that satisfies the Mean Value Theorem is -0.7116. 8. (2 pts) A quick sketch is to the right. Here are the equations we need. Maximize : Constraint : 10 02 V r h r hr p pp = =+ Solve the constraint for h and plug into the volume equation. ( ) () 2 223 1 11 2 10 0 10 0 50 rr h r V r r r Øø = - = - ºû 2 3 2 5 03 V r r V ¢ = - From the first derivative we get the critical points : 100 3 3.2574 r = =– .
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10FallHwk08_Soln - Math 2413 Homework Set 8 Solutions 10...

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