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.
