If I have a function with all positive integer for the coefficients, is there a way to have a lower bound? Zero isn't an option, because I've done the rational root theorem and found all possible roots. If you need, I can provide the function and its list of possible roots below:
$70x^{4}+163x^{3}+109x^{2}+37x+6$
$±1, ±1/2, ±1/5, ±1/7, ±1/10, ±1/14, ±1/35, ±1/70,$ $±2, ±2/5, ±2/7, ±2/35,$ $±3, ±3/2, ±3/5, ±3/7, ±3/10, ±3/14, ±3/35, ±3/70,$ $±6, ±6/5, ±6/7, ±6/35$
On
The polynomial has two negative roots, $-\frac{3}{2}$ and $-\frac{2}{5}$. There are no positive roots according to the Descartes rule of signs. The polynomial is of even degree with a positive leading coefficient so it goes to positive infinity as $x$ tends to infinity. Thus, there should be a lower bound for some $x \in (-\frac{3}{2},-\frac{2}{5})$.
A fairly standard method for finding extrema of a function is known as the first derivative test. Applying this to the given quartic equation yields:
$f'(x)=280x^3+489x^2+218x+37$
Since $37$ is prime, it is unlikely that this factors nicely. Using a graphing calculator app I found that it has 1 real solution at $x\approx -1.183$. This corresponds to the global minimum of the original function, as can readily be verified with a graph. Acquiring an explicit solution for the minimum would require Cardano's method to solve $f'(x)=0$ explicitly. Followed by a very messy calculation with the original quartic to obtain the exact $y$-coordinate.