SOLUTION: Find the equations on the real number p for which {{{x^4 -4p^3x+12>0}}} for all real numbers x .

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Question 1037050: Find the equations on the real number p for which x%5E4+-4p%5E3x%2B12%3E0 for all real numbers x .
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Find the extreme points of the function f%28x%29+=+x%5E4+-4p%5E3x%2B12:
==> f'(x) = 4x%5E3-4p%5E3 = 0
==> x%5E3-p%5E3+=+%28x-p%29%28x%5E2%2Bxp%2Bp%5E2%29+=+0
==> There is only one critical value of x = p. ( The function y+=+x%5E2%2Bxp%2Bp%5E2 has no real roots in terms of p and so there no critical values coming from this function.)
Because f"(p) = 12p%5E2 >0, the second derivative test tells that there is a relative (in fact, absolute) minimum at x = p.
We want f%28p%29+=+p%5E4+-4p%5E3%2Ap%2B12++=+p%5E4+-4p%5E4%2B12+=+12+-+3p%5E4%3E+0, or
4+-+p%5E4+=+%282+-+p%5E2%29%282+%2B+p%5E2%29+%3E+0.
Since 2%2Bp%5E2+%3E+0 always, it follows that 2+-+p%5E2+%3E+0, or 2+%3E+p%5E2.
The solution set to the last inequality is -sqrt%282%29+%3C+p+%3C+sqrt%282%29.
Therefore as long as p is in the open interval (-sqrt%282%29,sqrt%282%29), x%5E4+-4p%5E3x%2B12%3E0 for all real numbers x .