Posted by Z32 on Monday, October 26, 2009 at 12:51pm.
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
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!
Related Questions
math - how do you find the inflection points of a polynomial such as f(x)=12x^5+...
math - f(x)=12x^5+45x^4–80x^3+2 the question asks to find the inflection ...
MATH URGENT - f(x)=12x^5+45x^4–80x^3+2 the question asks to find the ...
calculus - Find all relative extrema and points of inflection of the function: f...
Calculus - Find equations of the tangent lines at all inflection points of the ...
Calculus - The number of people who donated to a certain organization between ...
Calculus - Find all extrema, inflection points, and asymptotes where applicable ...
calculus - Find the points of inflection. f(x) = e^(-x^2) I need to know step by...
calculus - Consider the function f(x)=12x^5+30x^4–160x^3+5 . f(x) has ...
Calculus - Determine the x-value for each inflection point on the graph of the ...
For Further Reading
