# How do you find the local extrema for f(x)=(x-3)(x-1)(x+2)?

We need to find the null points of the first derivative of f(x)

hence

$f ' \left(x\right) = 3 {x}^{2} - 4 x - 5$

this nullifies at points

${x}_{1} = \frac{1}{3} \left(2 - \sqrt{19}\right)$ and ${x}_{2} = \frac{1}{3} \left(2 + \sqrt{19}\right)$

Using the Second-Derivative Test we have that

f(x) has local maximum at ${x}_{1} = \frac{1}{3} \left(2 - \sqrt{19}\right)$ with value

$f \left(\frac{1}{3} \left(2 - \sqrt{19}\right)\right) = \frac{2}{27} \left(28 + 19 \sqrt{19}\right)$

f(x) has local minimum at ${x}_{1} = \frac{1}{3} \left(2 + \sqrt{19}\right)$ with value

$f \left(\frac{1}{3} \left(2 + \sqrt{19}\right)\right) = \frac{2}{27} \left(28 - 19 \sqrt{19}\right)$