Points of inflection are points on the function where the second derivative changes sign, that is the graph goes from concave upward to concave downwards or vice-versa.
Let y = f(x)
We must find the second derivative of the function:
(d^2y)/(dx^2) = y''(x)
y'(x) = -6x^5 - 20x^4 + 20x^3, using the power rule
Then,
y''(x) = -30x^4 - 80x^3 + 60x^2
Let us find where the second derivative is zero, to find the critical x-values.
-30x^4 - 80x^3 + 60x^2 = 0
-10x^2(3x^2 + 8x - 6) = 0
x=0 is a solution here, and to find the other two solutions, we solve the following equation:
3x^2 + 8x - 6 = 0
Using the quadratic formula,
x = (-8 ± sqrt(64 - (-72)))/(6) = (-8 ± sqrt(136))/(6)
x approx -3.27698396495, or x approx 0.610317298282
Let us find the concavity of f(x) on these intervals:
on (-infty, -3.27698396495), f''(x)<0
on (-3.27698396495, 0), f''(x)>0
on (0, 0.610317298282), f''(x)>0
on (0.610317298282, infty), f''(x)<0
So in succession on (-infty, infty), the sign line looks like this:
-, +, +, -
Therefore, f''(x) only changes signs at the x-values of -3.27698396495 and 0.610317298282
Since f''(x) changes signs at these values, f(x) has inflection points here as well. The coordinates of the inflection points are:
(-3.27698396495, f(-3.27698396495))
and
(0.610317298282, f(0.610317298282))
Therefore, the coordinates are, rounding to 4 decimal places:
(-3.2770, 849.8146) and (0.6103, 0.3033)