# How do you find the critical points for f(x)=x^3-2x^2+3x?

Jun 28, 2017

$x = \frac{1}{3} \left(2 - i \sqrt{5}\right) \mathmr{and} x = \frac{1}{3} \left(2 + i \sqrt{5}\right)$

#### Explanation:

To find the critical points you first take the derivative using the power:

$\frac{d}{\mathrm{dx}} = {x}^{n} = n {x}^{n - 1}$

The derivative of $f \left(x\right) = {x}^{3} - 2 {x}^{2} + 3 x$ is:

$\frac{d}{\mathrm{dx}} = 3 {x}^{2} - 4 x + 3$

Now you set it equal to zero and solve:

$3 {x}^{2} - 4 x + 3 = 0$

When you solve it, it will yield complex roots:

$x = \frac{1}{3} \left(2 - i \sqrt{5}\right) \mathmr{and} x = \frac{1}{3} \left(2 + i \sqrt{5}\right)$

These two solutions are your critical numbers.