# What are the critical points of g(x)=(x+2)/(7-4x)?

Dec 20, 2016

$g \left(x\right)$ has no critical point and is strictly increasing in its domain.

#### Explanation:

Critical points are determined as the values of $x$ for which the first derivative vanishes, so:

$g ' \left(x\right) = \frac{\left(7 - 4 x\right) + 4 \left(x + 2\right)}{{\left(7 - 4 x\right)}^{2}} = \frac{15}{{\left(7 - 4 x\right)}^{2}}$

Evidently, the derivative of $g \left(x\right)$is always positive and never equals zero, thus $g \left(x\right)$ has no critical point and is strictly increasing in its domain.

graph{(x+2)/(7-4x) [-10, 10, -5, 5]}