Question

y = x4 + ax3 + x2 + 1 is concave along the entire number scale then what can be inferred about a?

Answer 1

The second derivative has no real roots, i.e. no point of inflection.
To find a, equate the discriminant of d^2y/dx^2=0 to less than zero.

Answer 2

\[y = x^4 + ax^3 + x^2 + cx + d\]Concave up means that y'' must be positive...\[y'' = 12x^2+6ax+2 > 0\]\[6x^2 + 3ax + 1 > 0\]\[(\sqrt{6}x \pm \frac{\sqrt{6|a|}}{4})^2 + 1 - \frac{3a^2}{8} > 0\]\[1 > \frac{3a^2}{8} \]\[|a| < \frac{2\sqrt{6}}{3} \]The endpoint values of a = 2*(2/3)^(1/2) have an awkward point of non-concavity at x = 1/sqrt(6).

In the middle step, it is easier to just identify the smallest value of a parabola as its vertex and set x = -(3a)/(2*6) directly, then solve.