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

1 Answer
Dec 20, 2016

Answer:

#g(x)# 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'(x) = frac ((7-4x)+4(x+2)) ((7-4x)^2)=15/((7-4x)^2)#

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

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