# How do you find the critical points of f(x) = x - 3ln(x)?

##### 1 Answer
Apr 1, 2015

I use the definition: a critical point for a function, $f$ is a point (number, value), $c$ in the domain of $f$ at which either $f ' \left(c\right) = 0$ or $f ' \left(c\right)$ does not exist.

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

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

For this function, $f ' \left(x\right)$ does not exist for $x = 0$, but the domain of $f$ is $\left(0 , \infty\right)$, so $0$ is not a critical point for $f$.

$f ' \left(x\right) = 0$ at $x = 3$ which is in the domain of $f$, so $3$ is a critical number for $f$.

(I understand that there are some who would make the critical point $\left(9 , 3 - 3 \ln 3\right)$.)