For the beginning, we'll calculate the second derivative of the function and, after that, we'll determine the roots of the second derivative, in order to find out the inflexion points of the function.
f'(x) = 3*5*x^4 + 15*4*x^3 - 10*3*x^2 - 90*2*x + m
f"(x) = 3*5*4*x^3 + 15*4*3*x^2 - 10*3*2*x - 90*2
We'll factorize:
3*5*4*(x^3 + 3*x^2 - x - 3) = 0
We'll divide the expression by the product 3*5*4:
x^3 + 3*x^2 - x - 3 = 0
We'll group the first and the second term together, and the last 2 terms together.
x^2*(x + 3) - (x + 3) = 0
(x + 3)*(x^2 - 1) = 0
(x + 3)*(x - 1)*(x + 1) = 0
(x + 3) = 0 for x = - 3
x - 1 = 0 for x = 1
x + 1 = 0 for x = - 1
So, x = - 3, x = - 1, x = 1 are inflexion points for the function given.
To prove that these inflexion points are on the same line, we have to build the determinative, formed by the inflexion points.
-3 f(-3) 1
-1 f(-1) 1
1 f(1) 1
We have to calculate it and if it's cancelling, then the inflexion points belong to the same line.
We'll calculate it using the "triangle" rule.
-3*f(-1)*1 + -1*f(1)*1 + 1*f(-3)*1 - 1*1*f(-1) + f(-3) + 3*f(1)
f(-1) = -3 + 15 + 10 - 90 - m + n = -68 - m + n
-3*f(-1) = -3*(- 68 - m + n) = 204 + 3m - 3n
f(1) = 3 + 15 - 10 - 90 + m + n = - 82 + m + n
3*f(1) = - 246 + 3m + 3n
f(-3) = - 729 + 1215 + 270 - 810 - 3m + n = - 54 - 3m + n
204 + 3m - 3n + 82 - m - n - 54 - 3m + n + 68 + m - n - 54 -3m + n - 246 + 3m + 3n = 0 q.e.d.
The inflexion points belong to the same line!
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