How do you find the critical points if #f'(x)=2-x/(x+2)^3#?

Question

Charles Arocha · Accepted Answer

You need to solve:  #2-x/(x+2)^3 = 0#  Which has no rational and only one real, solution. If you intended to type:  #f'(x) = (2-x)/(x+2)^3#, the we're in better luck. A critical number is a value in the domain of #f# at which the derivative is either #0# or fails to exist. It looks as if the domain for the original #f# was #RR - {-2}#. #f'(-2) does not exist, but #-2# is not in the domain of #f#, so it is not a critical point. #(2-x)/(x+2)^3=0# when #2-x=0# which happens at #x=2#. Assuming that #2# is in the domain of #f#, it is a critical point.

Luke Alvarez · Answer

undefined To find the critical points of $ f'(x) = \frac{2 - x}{(x + 2)^3} $, we first set the derivative equal to zero and solve for $ x $. Then, we check for points where the derivative is undefined.

1. Set $ f'(x) = 0 $:

$$ \frac{2 - x}{(x + 2)^3} = 0 $$

2. Solve for $ x $:

$$ 2 - x = 0 $$
$$ x = 2 $$

3. Check for points where the derivative is undefined. The derivative is undefined when the denominator is equal to zero:

$$ (x + 2)^3 = 0 $$
$$ x = -2 $$

Thus, the critical points of $ f(x) $ occur at $ x = 2 $ and $ x = -2 $.