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

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### 2 Answers

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).**

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).