Consider the function f(x)=12x^5+30x^4–300x^3+2. f(x) has inflection points at (reading from left to right) x=D,E, and F. where D is___, E is____, and F is____.

Answer 1

The points of inflection are obtained by setting the second derivative of a function equal to zero
It is relatively easy to show that
f''(x) = 240x^3 + 360x^2 - 600

so
240x^3 + 360x^2 - 600 = 0
2x^3 + 3x^2 - 5 = 0
by a quick trial and error, I notice that x=1 satisfies the equation
so x-1 is a factor
dividing out I got
2x^3 + 3x^2 - 5 = 0
(x-1)(2x^2 + 5x + 5) = 0
so x=1 or x is equal to 2 complex roots.

Your function only has one point of inflection,
at x = 1

check your typing (or check my arithmetic)

Answer 2

Thanks for your help. It helped me figure out the correct answer. The second derivative is 240x^3+360x^2-1800x. From there I factored and got
120x(2x^2+3x-15)=0. then used the quadratic formula to get (-3+sqrt129)/4, (-3-sqrt129)/4, and 0.