# Question c9e4c

Jun 6, 2015

$f \left(x\right) = \frac{x}{x - 2}$

The domain is $D = \mathbb{R} - \left\{2\right\}$

$f ' \left(x\right) = \frac{\left(x\right) ' \left(x - 2\right) - x \left(x - 2\right) '}{x - 2} ^ 2$

$f ' \left(x\right) = \frac{x - 2 - x}{x - 2} ^ 2 = - \frac{2}{x - 2} ^ 2$

To find the local max an min of $f \left(x\right)$, you need to study the sign of the derivative :

You are absolutely right, there aren't any local max or min.

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

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

To find the intervals in which f(x) is concave up or concave down, you need to study the sign of the second derivative :

The function is concave down for x in ]-oo;+2[ and is concave up for x in ]+2;+oo[#.