f'(x) = 60x^4 + 180x^3 - 1080x^2
f''(x) = 240x^3 + 540x^2 - 2160x
= 0 for points of inflection
240x^3 + 540x^2 - 2160x = 0
divide by 60
4x^3 + 9x^2 - 36x = 0
x(4x^2 + 9x - 36) = 0
Can you take it from there ?
There are 3 solutions for x
Find the inflection points for f(x) = 12x^5 + 45x^4 - 360x^3 + 7
3 answers
I got 0 for one of my inflections since a 0 in the place of the "x"(4x^2 + 9x - 36) would make it zero. But I'm not sure what would need to go into the other x's to make it 0. I tried 4/9 9/36 and 36/9 but I couldn't get it.
Oh, okay. I found out I had to use the quadratic formula. Thanks for the help!
3 answers