Concavity of a function: y=2/3x^3-x^2+3x+5

  • Thread starter Thread starter Hollysmoke
  • Start date Start date
  • Tags Tags
    Function
Click For Summary
SUMMARY

The discussion focuses on determining the concavity of the function y=2/3x^3-x^2+3x+5. The first derivative, y’=2x^2-2x+3, and the second derivative, y’’=4x-2, are calculated to find critical points. The inflection point is identified at x=1/2, where the second derivative changes sign. Substituting x=1/2 back into the original function confirms the inflection point's coordinates.

PREREQUISITES
  • Understanding of calculus, specifically derivatives and concavity.
  • Familiarity with polynomial functions and their properties.
  • Ability to perform algebraic substitutions in equations.
  • Knowledge of critical points and inflection points in functions.
NEXT STEPS
  • Study the application of the second derivative test for concavity.
  • Learn about higher-order derivatives and their significance in function analysis.
  • Explore graphing techniques for visualizing polynomial functions and their concavity.
  • Investigate real-world applications of concavity in optimization problems.
USEFUL FOR

Students and educators in calculus, mathematicians analyzing polynomial functions, and anyone interested in understanding the behavior of functions through derivatives.

Hollysmoke
Messages
185
Reaction score
0
I just wanted to make sure I did this right:

y=2/3x^3-x^2+3x+5
y’=2x^2-2x+3
y’’=4x-2
=2(2x-1)


2x-1=0
2x=1
x=1/2

And then I substitute x into the original equation to find the IP, right?
 
Physics news on Phys.org
Yes, that's correct.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
Replies
5
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K