Find the inflection point(s).

Question

f(x)=10—12x+6x²—x³

Answer 1

We need to find the second derivative of f(x) and set it equal to zero. This means we need to find the derivative of the derivative. When we solve for the x values, they will be a location of the points of inflection. A point of infection is where the graph changes shape (i.g.  concave up to concave down).  This method is known as the second derivative test.

 

f(x) = 10 - 12x + 6x² - x³

 

d/dx[f(x)] = 12 + 12x - 3x²

 

d/dx(d/dx)[f(x)] = d²/dx²[f(x)] = 12 - 6x

 

Note that the notation d²/dx² is second derivative.  We can also use the notation f''(x).

 

12 - 6x = 0

6(2 - x) = 0

 

x = 2

 

The location of the point of inflection is at x = 2.

 

Next, we perform a test point.  We will use the point x=1 and x=3 as our test points and evaluate them in the second derivative.

 

f''(1) = 6(2 - 1)

        = 6(1)

        = 6

 

f''(3) = 6(2 - 3)

        = 6(-1)

        = -6

 

Since the f''(1) is positive, the graph will be concave up and (-∞, 2).  Since f"(3) is negative, the graph is concave down at (2,∞).  Therefore, the shape of the graph changes, indicating the existence of a point of inflection.

 

Evaluate x = 2 in the original function f(x) to find the point of inflection.

 

f(2) = 10 - 12(2) + 6(2)² - (2)³

      = 10 - 24 + 24 - 8

      = 2

 

The point of inflection is (2, 2).