Note that the global minimum of a continuous function $f(x)$ in a closed interval $[a,b]$ is the minimum over the following candidates:
- Evaluating $f$ at the endpoints: $f(a),f(b)$.
- Evaluating $f$ at the critical points: $f(c)$, where $f'(c)=0$ or $f$ is not differentiable at $c$.
Hence, since $f(-5) = 18$ and $f(10) = 108$ are both larger than $f(-1/2) = -9/4$, it follows that the local minimum at $x=-1/2$ is also a global minimum, as desired.
Problem can be attacked by meta-cheating based on the idea that there is only 1 correct answer. Clearly $ax^4$ is the dominant term, so if $a$ is positive, there is no global maxima and if $a$ is negative, there is no global minima.
It remains to
- Discuss the other options, for completeness
- Discuss your work.
Options A and C are easily dismissed, because for the generic cubic represented by $p'(x)$, how in the world could it be guaranteed that the cubic is always less than zero or always greater than zero.
Option B is similarly dismissed. How in the world could one guarantee that the pertinent cubic has no real root?
From what I can see, your overall analysis seems sound: I saw no obvious mistakes.
Exception
for the second part ($a$ is positive):
According to the derivative $p'(x)=4ax^3+9x^2+6$ the function increases for every x.
The issue is (for example) whether it can be guaranteed that $p'(x)$ is always positive, regardless of the value of $a$. If $a$ is positive (non-zero), then for large negative $x$, $p'(x)$ will be negative and for large positive $x$, $p'(x)$ will be positive.
By focusing on whether $p'(x)$ is increasing or decreasing, you are mistakenly focusing on whether $p''(x)$ is positive or negative, rather than focusing on whether $p'(x)$ is positive or negative.
Best Answer
We easily get $f(x,y)\ge 0$ and $$f(x,y)=0\iff x^2=4\;\wedge\;y=0\iff (x,y)=(-2,0)\;\vee\;(x,y)=(2,0).$$ Obviously $f(\pm 2,0)=0$, so local minima are also global.