How find minimum $f(x) = \max_{t\in[-1,1]} \left| t+ \frac{3}{2+t} + x \right|$ How find minimum this function $f(x) = \max_{t\in[-1,1]} \left| t+ \frac{3}{2+t} + x \right|$?
 A: Hint
Name $$g(t)=t+\frac{3}{2+t}$$ You'll verify that $g$ is decreasing on the interval $[-1,-2+\sqrt{3}]$ from $2$ to $2(\sqrt{3}-1)$ and then increasing on the interval $[-2+\sqrt{3},+1]$ from $2(\sqrt{3}-1)$ to $2$.
Hence $$f(x) = \max_{u\in[2(\sqrt{3}-1),2]} \left| u + x \right|$$
Can you move forward from there?
A: First of all we can simplify the expression of the function $f$:
$$f(x)=\max_{t\in[-1,1]}\left|t+2+\frac{3}{t+2}+x-2\right| =\max_{t\in[1,3]}\left|t+\frac{3}{t}+x-2\right|=\max_{t\in[1,3]}\left|h(t)+x\right|$$
where $h(t)= t +\frac{1}{ t}-2$, and we know that $h$ is continue differentiable, hence we can determine its minimum and its maximum in the interval [1,3] and prove that:
$$ h(2\sqrt 3)+x \leq h(t)+x\leq h(3)+x$$
finally $$f(x)=\max\left(\left|h\left(2\sqrt 3\right) +x\right|,\left|x+h(3)\right|\right)$$
The minimum of this function is attainable at the point $x$ such that: $x+h(3)=-x-h(2\sqrt 3)$ which is 
$$x=\frac{h(3)+h(2\sqrt3)}{2} $$
A: This is not a rigorous answer, but too long for a comment.
The output of the Maple code
   f := x-> maximize(abs(t+3/(2+t)+x), t = -1 .. 1, location)[1] :
   plot(f, -3 .. 0); 


DirectSearch:-GlobalOptima(f, {x >= -2, x <= 0});

$$   [.267949192431140,[ -1.73205080756887], 20]$$
suggests that the minimum of $f$ which equals $2-\sqrt{3}$ is attained at $x=-\sqrt{3}.$
