What are the critical values, if any, of #f(x)= (2-x)/(x+2)^3#?

1 Answer

A critical point on a function #f(x)# occurs at #x_o# if and only if either #f '(x_o)# is zero or the derivative doesn't exist.

Hence we need to compute the first derivative

#df(x)/dx=d((x-2)/(x+2)^3)/dx=(x-4)/(x+2)^3#

which nullifies at #x_o=4#

Using the second derivative test for local extrema we need to compute the second derivative for #f(x)# hence we have that

#d^2f(x)/(d^2x)=-6*(x-6)/(x+2)^5#

Because #f''(4)>0# hence the function has a minimum at #x=4#
which is #f(4)=-1/108#

It does not have a maximum as

#lim_(x->-2) f(x)=oo#