Question 1205395

{{{f(x) = (x^2 - 16)^4}}}


find second derivative


{{{f}}}'{{{(x) =  8x (x^2 - 16)^3}}}

{{{f}}}''{{{(x) = 8 (x^2 - 16)^2 (7x^2 - 16) }}}



So {{{f}}} ′′{{{(x)=0}}} when

{{{8 (x^2 - 16)^2 (7x^2 - 16) =0}}}


critical points of the derivatives are:

{{{x=4}}} => stationary
{{{x=-4}}}=> stationary
{{{x=4/sqrt(7)}}}=> stationary minimum
{{{x=-4/sqrt(7)}}}=> stationary maximum


so, {{{f(x) = (x^2 - 16)^4}}} is

concave {{{up}}}  on {{{x<-4/sqrt(7)}}}
concave {{{down}}}  on {{{-4/sqrt(7)<x<4/sqrt(7)}}}
concave {{{up}}}  on {{{x>4/sqrt(7)}}}


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