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

Question

Charles Anctil · Accepted Answer

#g(x)# has no critical point and is strictly increasing in its domain.

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]}

Isaiah Allen · Answer

undefined To find the critical points of the function $ g(x) = \frac{x + 2}{7 - 4x} $, you need to find where the derivative of the function is equal to zero or undefined.

First, find the derivative of $ g(x) $ using the quotient rule:

$$ g'(x) = \frac{(7 - 4x)(1) - (x + 2)(-4)}{(7 - 4x)^2} $$

Simplify the expression:

$$ g'(x) = \frac{7 - 4x + 4x + 8}{(7 - 4x)^2} = \frac{15}{(7 - 4x)^2} $$

To find where the derivative is equal to zero, set $ g'(x) = 0 $:

$$ \frac{15}{(7 - 4x)^2} = 0 $$

This equation has no solutions because the numerator is a constant (15) and cannot be zero.

Next, find where the derivative is undefined by setting the denominator equal to zero:

$$ 7 - 4x = 0 $$
$$ 4x = 7 $$
$$ x = \frac{7}{4} $$

So, the critical point of the function $ g(x) $ is $ x = \frac{7}{4} $.