Shape of graph polynomial to the fifth degree

Click For Summary
SUMMARY

The discussion focuses on determining the intervals of increase and decrease for the polynomial function F(x) = -x^(5) + (5/2)x^(4) + (40/3)x^(3) + 5. The derivative F'(x) = -5x^(4) + 10x^(3) + 40x^(2) is factored to F'(x) = -5x^(2)(x - 4)(x + 2). The critical points identified are x = 0, x = 4, and x = -2, with particular emphasis on understanding why x = 0 is a critical point, which is clarified by recognizing that F'(0) = 0.

PREREQUISITES
  • Understanding of polynomial functions and their derivatives
  • Knowledge of critical points and their significance in calculus
  • Ability to factor polynomials and solve for roots
  • Familiarity with the concept of intervals of increase and decrease
NEXT STEPS
  • Study the application of the First Derivative Test for determining intervals of increase and decrease
  • Learn about the implications of critical points on the graph of a polynomial function
  • Explore the concept of concavity and the Second Derivative Test
  • Investigate additional resources on polynomial graphing techniques
USEFUL FOR

Students studying calculus, mathematics educators, and anyone interested in understanding polynomial behavior and graphing techniques.

Iclaudius
Messages
36
Reaction score
0
Hello my friends,

I have this problem and would appreciate someones help:


Determine all intervals where the following function is increasing or decreasing.

F(x) = -x^(5)+(5/2)x^(4)+(40/3)x^(3)+5

Solution
To determine if the function is increasing or decreasing we will need the derivative.

F'(x) = -5x^(4)+10x^(3)+40x^(2)

factored
F'(x) = -5x^(2) (x-4)(x+2)

Ok so here is where i have difficulty, i know x = 0, x = 4, and x = -2 however I do not understand why x = 0.
I understand why x = 4, and x = -2 - from solving simple quadratic at y = 0 to identify where this function is not changing, however where does someone get the x = o from?

I have been looking at some online resources and they have not provided adequate explanation - this particular problem is from http://tutorial.math.lamar.edu/Classes/CalcI/ShapeofGraphPtI.aspx

oh btw i apologize for the messy math notation
Thanks in advance,
Claudius
 
Physics news on Phys.org
Well, you do have two linear factors [ (x-4) and (x+2) ] multiplied by -5x^2.
Don't you think if you set x = 0, then f'(x) = 0 then?
 
Ah yes :D, thank you! :approve:
 

Similar threads

  • · Replies 48 ·
2
Replies
48
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K