# Determine the points of inflection of y=x^3-3x^2-9x+6.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The inflection points could be found by calculating the roots of the second derivative of the function (if there are any).

For the beginning, we'll differentiate the function:

dy/dx=d/dx(x^3-3x^2-9x+6)

dy/dx = d/dx(x^3) - d/dx(3x^2) - d/dx(9x) + d/dx(6)

dy/dx = 3x^2 - 6x - 9

Now, we'll differentiate dy/dx:

d^2y/dx = d/dx(3x^2 - 6x - 9)

d^2y/dx = d/dx(3x^2) - d/dx(6x) - d/dx(9)

d^2y/dx = 6x - 6

or

f"(x) = 6x - 6

After f"(x) calculus, we'll try to determine the roots of f"(x).

f"(x) = 0

6x - 6 = 0

We'll divide by 6:

x - 1 = 0

x = 1

For x = 1, the function has an inflection point.

f(1) = 1^3-3*1^2-9*1+6

f(1) = 1 - 3 - 9 + 6

f(1) = -5

The inflection point is: (1 , -5).

neela | High School Teacher | (Level 3) Valedictorian

Posted on

y(x) = x^3-3x^2-9x+6. To determine the inflection point.

The inflection point is the point at which the curve crosses it tangent, At the point of inflection. d^y/dx^2 = 0.

y' (x)= (x^3-3x^2-9x+6)'

y'(x) = (3x^2-3*2x-9)

y" (x) = (3x^2-6x-9)'

y"(x) = 6x-6. Equating y(x)  to zero, we get:

6x-6 = 0.

x-1 = 0

x =1.

y(1) = {(x^3-3x^2-9x+6) at x= 1 } =  1^3-3*1^2-9*1+6 = -5.

Therefore the point of inflection is at (1 , -5).