SOLUTION: I am having a hard timfiguaring out what the relative max and min. is. how to caculate it? I have a TI- 83 plus. question might be: f(x)= 2-x^2 or f(x)= x^3 +4x^2

Algebra ->  Linear-equations -> SOLUTION: I am having a hard timfiguaring out what the relative max and min. is. how to caculate it? I have a TI- 83 plus. question might be: f(x)= 2-x^2 or f(x)= x^3 +4x^2      Log On


   

Question 614453: I am having a hard timfiguaring out what the relative max and min. is. how to caculate it? I have a TI- 83 plus. question might be: f(x)= 2-x^2 or f(x)= x^3 +4x^2
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
I am having a hard timfiguaring out what the relative max and min. is. how to caculate it? I have a TI- 83 plus. question might be:
f(x)= 2-x^2
f'(x) = -2x 1st derivative
-2x = 0
x = 0 --> max, since the coeff of the x^2 term is negative.
-----------
or
f(x)= x^3 +4x^2
f'(x) = 3x^2 + 8x = 0
x = 0, x = -8/3
----
f"(x) = 6x + 8 2nd deriviative
----
f"(0) = 8 --> local min
----
f"(-8/3) = -8 --> local max
===============================
dl the FREE graph software to see these at
http://www.padowan.dk