math team hw 6_1#18 - Then we must find the positive areas...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
18. We know: g(0)=50 g’ is graphed below: Using this graph we can form a chart for g(x). We will divide the graph into 8 divisions with a Δx of 5. (These divisions are marked by dashed lines on the graph). Therefore, we will have 8 areas that we will need to compute. In order to compute these areas, we need two formulas: The area of a trapezoid= (1/2)(h)(b 1 +b 2 ) The area of a triangle= (1/2)(b)(h) First we can find the area between the curve and the x-axis from x=0 to x=5, this being area 1. Area 1 is a trapezoid with bases of 15, 10, and a height of 5. Therefore, plugging our values into the formula, we find: A 1 = (1/2)(5)(15+10)= -62.5 *this area becomes negative because it is under the x-axis We use the same steps to find Area 2 and Area 3, which is the trapezoid from x=5 to x=10 and the triangle from x=10 to x=15. Note that since all of these areas are still under the x-axis, they will be negative. A 2 =(1/2)(5)(15+20)= -87.5 A 3 =(1/2)(5)(20)= -50
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Then we must find the positive areas above the x-axis. 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 x-axis 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 x-values 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.

Page1 / 2

math team hw 6_1#18 - Then we must find the positive areas...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online