answer1 - (1(1(3(3(4(4(2(2 1 First we plot both curves on...

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Unformatted text preview: (1) (1) (3) (3) (4) (4) (2) (2) 1. First, we plot both curves on the same set of axes. A little experimentation gives a reasonable range. I'm defining these as functions so that I don't have to type these out again when I want to plug in later. You could write them as expressions as well. x 1 1 2 3 4 5 The roots can't be found with "solve", so we must use "fsolve" to find numerical solutions. I'm using intervals based on the plot. For the area, I integrate , since between the roots, is larger than . (5) (5) (11) (11) (7) (7) (6) (6) (8) (8) (9) (9) (10) (10) 2.283018609 The above number is the desired area. 2. First I'll define the cubic with the coefficients as variables. Note the command "restart" below. That resets Maple so that variables are unassigned. To have it go through the given points we need and . If these are going to be critical points, we need and . This gives us our four equations in the four unknowns and . So, the desired cubic is . Here's a plot. Notice how I'm using "subs" to put the solution into without typing it over again. (12) (12) x 1 2 3 4 5 1 2 3 3. First, let's plot the function on the given interval to see what's going on. (13) (13) x 1 1 It looks like there is a maximum near .6 and a minimum near the right endpoint. We'll find these by using "fsolve" on . But when using fsolve, you want to plot things (in particular, plot x...
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This note was uploaded on 03/08/2010 for the course MATH 442 taught by Professor Howard during the Spring '08 term at Texas A&M.

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answer1 - (1(1(3(3(4(4(2(2 1 First we plot both curves on...

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