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

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Answer 1 · 2015-11-04T19:17:17

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

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

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