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

Jan 11, 2017

$f \left(x\right)$ has a local maximum for $x = 0$, a local minimum for $x = \frac{1}{3}$ and an inflection point at $x = \frac{1}{6}$

#### Explanation:

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

$f \left(x\right) = 2 {x}^{3} - {x}^{2} + 1$

$f ' \left(x\right) = 6 {x}^{2} - 2 x = 2 x \left(3 x - 1\right)$

so the critical points are:

${x}_{1} = 0$
${x}_{2} = \frac{1}{3}$

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

$f ' ' \left(x\right) = 12 x - 2 = 2 \left(6 x - 1\right)$

So we have a single inflection point at $x = \frac{1}{6}$ where $f \left(x\right)$ 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]}