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)#