I have the integers $k$ and $x$ that are such that $k \geq x \geq 2$.
I would like to find the biggest $x$ which satisfies the minimum of $$\left\lceil \frac{-1+\sqrt{4x^2-12x+8k+1}}{2}\right\rceil.$$
In other words, I'm seeking $\max_x \mathrm{argmin}_x \left\lceil\frac{-1+\sqrt{4x^2-12x+8k+1}}{2}\right\rceil.$
If I compute the derivative in the case of real numbers, I find $x=3/2$. So, I think the solution is $x=2$. But there are possibly other values $x > 2$ that produce the same minimum. What is the correct method to find the maximum of them?
Thank you.
Consider the function $f(x)=\dfrac{-1}{2}+\dfrac{1}{2}\sqrt{4x^2-12x+8k+1}$.
Notice that this function have the same increasing and decreasing intervals as the function $g(x)=4x^2-12x$, since $f(x)$ is the composition of $g(x)$ with monotone functions: adding a constant, taking square root, adding another constant.
The minimum of f(x) is attained when $g(x)=4x^2-12x$ takes its minimum value, and that happens when $x=\frac{3}{2}$ and this minimum value is:
$f\left(\frac{3}{2}\right)=\dfrac{-1}{2}+\dfrac{1}{2}\sqrt{8k-8}$.
Also, the function $f(x)$ is increasing for $x>\frac{3}{2}$, and decreasing for $x<\frac{3}{2}$, so to find the maximum integer $x$ such that $$F(x)=\left\lceil \frac{-1+\sqrt{4x^2-12x+8k+1}}{2}\right\rceil$$ is minimum, we have to solve the equation
$$F(x)=\left\lceil \frac{-1+\sqrt{4x^2-12x+8k+1}}{2}\right\rceil=\left\lceil \dfrac{-1}{2}+\dfrac{1}{2}\sqrt{8k-8}\right\rceil=F\left(\frac{3}{2}\right)$$
This is a hard problem in general (because the ceiling function is so hard to work with) but we can at least give an upper bound for that: notice that such maximum integer $x$ must satisfy $f(x)<f\left(\frac{3}{2}\right)+1$, that means:
\begin{align*} 1-\frac{1}{2}+\frac{1}{2}\sqrt{8k-8}&>-\frac{1}{2}+\frac{1}{2}\sqrt{4x^2-12x+8k+1}\\ 2+\sqrt{8k-8}&>\sqrt{4x^2-12x+8k+1}\\ 4+4\sqrt{8k-8}+8k-8&>4x^2-12x+8k+1\\ 4+4\sqrt{8k-8}&>4x^2-12x+9=(2x-3)^2\\ \sqrt{4+4\sqrt{8k-8}}&>2x-3 \end{align*}
and so, $\displaystyle{x<\dfrac{\sqrt{4+4\sqrt{8k-8}}+3}{2}}$.