# If f(x)=3x^5+15x^4-10x^3-90x^2+mx+n demonstrate that the inflexion points of the f(x) graph belong to a line .

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The concavity or convexity is determiminrd by the sign of the 2nd diffrential coefficient. When the sign of the 2nd diffrential coefficient changes the curve changes its convexity to concavity or otherwise.At changing point of sign of d2y/dx^2 the value of d2y/dx^2= 0. The point is the inflexion point of the curve. So solve for f''(x) = o to get the inflection point.

f'(x) = (3x^5+15x^4-10x^3-90x^2+mx+n )'

=3*5x^4+15*4x^3-10*3x^2-90*2x+m

=15x^4+60x^3-30x^2-180x+mx

f''(x) = (15x^4+60x^3-30x^2-180x+mx)'

=15*4x^3+60*3x^2-30*2x-180.

So f(x) = 0 gives: 60x^3+180x^2-60x-180 = 0.Reducing by 60, we get:

x^3+3x^2-x-3 = 0.

x^2(x+3)-1(x+3) = 0.

(x+3)(x^2-1) = 0.

(x+3)(x+1)(x-1) = 0.

x =-3, x=-1 and x = 1 are the inflection points.

Then f(-3) = 3*(-3)^5+15(-3)^4-10(-3)^3-90(-3)^2+m(-3)+n = -729+1215+270-810-3m+n = -54-3m+n.

f(-1) = 3(-1)^5 +15(-1)^4-10(-1)^3-90(-1)^2 +m(-1)+n

=-3+15+10-90-m+n = -68-m+n

f(1) = 3+15-10-90+m+n = -82+m+n

Thus (-3 , -54-3m+n), (-1, -68,-m+n) and (1, -82+m+n) are on the line.

y - (-54-3m+n) = [(-68-m+n-(-54-3m+n))/(-1--3)]{x- -3)

y+54+3m-n = [(-14+2m)/2]{x+3}

y = (m-7)(x+3)-54-3m+n

y = (m-7)(x) +3m-21-54-3m +n

Tally we check whether (1, -82+m+n) lies in the line:

y = (m-7)x -75+n. Putting x= 1, we get the 3rd point y = (m-7)-75+n= -82+m+n .

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!