How do you find all critical point and determine the min, max and inflection given f(x)=2x^3-x^2+1f(x)=2x3x2+1?

1 Answer
Jan 11, 2017

f(x)f(x) has a local maximum for x=0x=0, a local minimum for x=1/3x=13 and an inflection point at x=1/6x=16

Explanation:

We can find the critical points by equating the first derivative to zero:

f(x) = 2x^3-x^2+1f(x)=2x3x2+1

f'(x) = 6x^2-2x =2x(3x-1)

so the critical points are:

x_1 = 0
x_2 = 1/3

To determine if they are local extrema and to find inflection points we calculate the second derivative:

f''(x) = 12x-2 =2(6x-1)

So we have a single inflection point at x=1/6 where f(x) changes from concave down to concave up.

Based on the sign of the second derivative we can also conclude that x_1 is a local maximum and x_2 a local minimum.

graph{2x^3-x^2+1 [-0.3227, 0.9273, 0.7125, 1.3375]}