Hi!!!
When finding the local max, min and inflection points of f(x) = x^2 e^x
When I'm finding f''(x) = 0
I have to use quadratic forumula. But I end up having to square root minus 4. Which you can't do. So does anyoine know a way I can find the inflection points??
Could someone please tell me if I'm doing it right.
f'(x) = 2x e^x + x^2 x e^x
= xe^x (2+x)
x = 0 and -2
f(0) = 0
f(-2) = 0.54134
So local max and min are (0,0) and (-2, 0.54)
Then
f''(x) = 2e^x + 2xe^x + 2xe^x + x^2 e^x
= 2e^x + x^2e^x + 4xe^x
e^x (x^2 + 4x + 2)
Quadratic forumula:
-4 +- [sqrt]-4 / 2
= syntax error....
PLEASE HELp![Smile :) :)](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Dr_Doom
When finding the local max, min and inflection points of f(x) = x^2 e^x
When I'm finding f''(x) = 0
I have to use quadratic forumula. But I end up having to square root minus 4. Which you can't do. So does anyoine know a way I can find the inflection points??
Could someone please tell me if I'm doing it right.
f'(x) = 2x e^x + x^2 x e^x
= xe^x (2+x)
x = 0 and -2
f(0) = 0
f(-2) = 0.54134
So local max and min are (0,0) and (-2, 0.54)
Then
f''(x) = 2e^x + 2xe^x + 2xe^x + x^2 e^x
= 2e^x + x^2e^x + 4xe^x
e^x (x^2 + 4x + 2)
Quadratic forumula:
-4 +- [sqrt]-4 / 2
= syntax error....
PLEASE HELp
Dr_Doom