Question 1037050
Find the extreme points of the function {{{f(x) = x^4 -4p^3x+12}}}:

==> f'(x) = {{{4x^3-4p^3}}} = 0
==> {{{x^3-p^3 = (x-p)(x^2+xp+p^2) = 0}}}
==> There is only one critical value of x = p.  ( The function {{{y = x^2+xp+p^2}}} has no real roots in terms of p and so there no critical values coming from this function.)

Because f"(p) = {{{12p^2}}} >0, the second derivative test tells that there is a relative (in fact, absolute) minimum at x = p. 

We want {{{f(p) = p^4 -4p^3*p+12  = p^4 -4p^4+12 = 12 - 3p^4> 0}}}, or 

{{{4 - p^4 = (2 - p^2)(2 + p^2) > 0}}}.
Since {{{2+p^2 > 0}}} always, it follows that {{{2 - p^2 > 0}}}, or {{{2 > p^2}}}.

The solution set to the last inequality is {{{-sqrt(2) < p < sqrt(2)}}}.

Therefore as long as p is in the open interval ({{{-sqrt(2)}}},{{{sqrt(2)}}}), {{{x^4 -4p^3x+12>0}}} for all real numbers x .