We want the conditions necessary so that the graph of
y = x4 - 4p3x + 12
is always above the x-axis.
This is an even-degree polynomial with positive
leading coefficient, thus its extreme right and
left behavior is upward. Polynomials are continuous
at all points. Thus its least relative minimum is
its absolute minimum. We find its relative extrema
by setting its derivative = 0:
y' = 4x3-4p3 = 0
4x3 = 4p3
x = p
So y has but one relative minimum at x = p,
we substitute to find the y-coordinate of the
relative (and absolute) minimum point:
y = x4 - 4p3x + 12
y = p4 - 4p3p + 12
y = p4 - 4p4 + 12
y = -3p4+12
Since we want y to always be positive:
-3p4+12 > 0
-3p4 > -12
p4 < 4
Taking positive square roots of both sides:
p2 < 2
Taking positive square roots of both sides again:
p < √2
That's the restriction on p.
[It's interesting to note that p can be 0 or any negative
value].
As a check, √2 is 1.414...
When p=1.42, just a smidgen above √2,
The graph's minimum point is a smidgen below the x-axis.
So y is NOT always positive. But
when p=1.41, just a smidgen below √2,
Its minimum point is a smidgen above the x-axis
Edwin
