This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Then we must find the positive areas above the xaxis. Again, we must use the area formulas for a triangle for Area 4 and Area 8 and a trapezoid for Area 5, 6, and 7. A 4 =(1/2)(5)(10) = 25 *Again note that these area values are now positive because above the xaxis A 5 =(1/2)(5)(10+7.5)=43.75 A 6 =(1/2)(5)(7.5+5)= 31.25 A 7 =(1/2)(5)(5+2.5)=18.75 A 8 =(1/2)(5)(2.5)= 6.25 Now, from knowing these areas, we can find the true value of g(x) at the different xvalues in increments of 5. In order to find the exact position, we take our known value, g(0)=50 and add the areas that we have found. We did this, and our g(x) values can be found in the table below. x 0 5 10 15 20 25 30 35 40 g(x) 50 12.5 100 150 125 81.25 50 31.25 25 Therefore, from plotting these values, our graph of g(x), with critical points at x=15 and x=40, as well as inflection points at x=10 and x=20, looks like: (10,20)10 g(x) 40 15 (20,10) x...
View
Full
Document
This note was uploaded on 04/02/2008 for the course MATH 115 taught by Professor Blakelock during the Fall '08 term at University of Michigan.
 Fall '08
 BLAKELOCK
 Math, Calculus, Division

Click to edit the document details