Determine by inspection whether p(x) has any absolute extrema.?

Question

p(x) = 3x^4 + 4X^3
If so, find them and state where they occur.

Answer 1

Since p(x) has even degree and the leading coefficient is positive, p(x) → ±∞. Thus, there is an absolute minimum but no absolute maximum. Since if f has an absolute extremum on an open interval (a,b) , then it must occur at a critical point of f [applied to the interval (-∞, +∞)], the absolute minimum must occur at a critical point of p. Since p is differentiable everywhere, we can find all critical points by solving the equation p'(x) = 0. This equation is:

12x^3 + 12x^2 = 12x^2(x+1) = 0

from which we conclude that the critical points are x = 0 and x = -1. Evaluating p at these critical points yields:

p(0) = 0 and p(-1) = -1

Therefore, p has an absolute minimum of -1 at x = -1

Answer 2

It should have an absolute minimum because it's a fourth-degree polynomial with a positive leading coefficient.

That means the limit as x goes to negative infinity is positive infinity and the limit as x goes to positive infinity is positive infinity. Somewhere in between there has to be a minimum point.