# How do you find the critical points for f(x) = x - 3ln(x) and the local max and min?

Oct 20, 2016

$\left(3 , 3 - 3 \ln 3\right)$

#### Explanation:

$f \left(x\right) = x - 3 \ln x$

Differentiating wrt $x$:
$f ' \left(x\right) = 1 - \frac{3}{x}$

Differentiating again wrt $x$:
$f ' ' \left(x\right) = - \frac{3}{x} ^ 2 \left(- 1\right) = \frac{3}{x} ^ 2 , \left(> 0 \forall x \in \mathbb{R}\right)$

At critical points, $f ' \left(x\right) = 0 \implies 1 - \frac{3}{x} = 0$
$\therefore \frac{3}{x} = 1$
$x = 3$

$f \left(3\right) = 3 - 3 \ln 3$
$f ' ' \left(3\right) > 0$

Hence, there is one critical point $\left(3 , 3 - 3 \ln 3\right)$ which is a minimum.

graph{x-3lnx [-10, 10, -5, 5]}