Access over 35 million academic & study documents

Studypool Answer

Content type
User Generated
Subject
Mathematics
School
Arizona State University
Type
Homework
Rating
Showing Page:
1/14

Sign up to view the full document!

lock_open Sign Up
Showing Page:
2/14

Sign up to view the full document!

lock_open Sign Up
Showing Page:
3/14

Sign up to view the full document!

lock_open Sign Up
End of Preview - Want to read all 14 pages?
Access Now
Unformatted Attachment Preview
Problem 1 a. We first find the derivative of the function.  d d f (x) = 3x4 + 8x3 − 18x2 + 5 dx dx = 12x3 + 24x2 − 36x The points where the derivative is equal to zero is where the function changes from increasing to decreasing and vice versa. So if we set the derivative equal to zero and solve we get the following: 12x3 + 24x2 − 36x = 0 x3 + 2x2 − 3x = 0 x(x2 + 2x − 3) = 0 x(x + 3)(x − 1) = 0 x = −3 x=0 x=1 Now we can pick points around these points to determine if the derivative is positive or negative. x -4 -1 0.5 2 f’(x) -240 48 -10.5 120 Direction Decreasing Increasing Decreasing Increasing So our intervals of increase and decrease are: Increase: (−3, 0) and (1, ∞) Decrease: (−∞, −3) and (0, 1) b. Again we find the derivative of the function.   d d (x − 4) dx (x + 3) − (x + 3) dx (x − 4) d x+3 = 2 dx x − 4 (x − 4) (x − 4)(1 + 0) − (x + 3)(1 − 0) = (x − 4)2 7 =− (x − 4)2 Now if we set this equal to zero and solve we get the following: − 7 =0 (x − 4)2 −7 = 0 1 This doesn’t help us very much. Instead if we solve for when the derivative is negative we can determine when the function is decreasing. − 7 ...
Purchase document to see full attachment
User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.

Anonymous
I was stuck on this subject and a friend recommended Studypool. I'm so glad I checked it out!

Studypool
4.7
Indeed
4.5
Sitejabber
4.4