Critical numbers are the x values for which f'(x) = 0. These critical points may or may not be maximums/minimums. We first need to find the derivative. We do so using the quotient rule.
f(x) = (2-x)/(x+2)^3
f'(x) = ( ((x+2)^3)(-1) - (2-x)(3)(x+2)^2)/(x+2)^6
f'(x) = ( -(x+2)^3 - 3(2-x)(x+2)^2 ) / (x+2)^6
Set this equal to zero and find which values of x satisfy the equation.
(-(x+2)^3 - 3(2-x)(x+2)^2)/(x+2)^6 = 0
-(x+2)^3 - 3(2-x)(x+2)^2 = 0
-(x+2)^3 = 3(2-x)(x+2)^2
-(x+2) = 3(2-x)
-x - 2 = 6 - 3x
2x = 8
x = 4
Note that x = -2 is also a critical number, since the denominator of f'(x) will be 0 when x = -2. Investigating this critical number, we see that as x -> -2 from the right, the denominator of f(x) gets very small and both the numerator and the denominator remain positive. This implies that the limit of f(x) as x -> -2 from the right explodes to positive infinity. Similarly, as x -> -2 from the left, f(x) explodes to negative infinity. Thus, the global maximum and minimum of the function are infty and -infty, respectively.
Now let's investigate the critical number x = 4. We see that:
f'(0) = (-8 - 3(2)(8))/(64) = -56/(64).
Also,
f'(5) = (-7^3 - 3(-3)(7^2))/(7^6) = (-343 + 441)/(7^6) = 98/7^6.
This tells us that before x = 4, the function is decreasing and after x = 4, the function is increasing. Thus, at x = 4, the function must have a local minimum.
