# RegionPlot of minimum value

I have a complicated function $f_{a,b}(t)$ of one argument ($t$) whose definition depends on two parameters $a,b$.

For every point in the plane $a-b$ (with $a\in[a_{min}, a_{max}]$ and $b\in[b_{min}, b_{max}]$) I need to find $\min\limits_{t\in\mathbb{R}}{f_{a,b}(t)}$ and verify whether it's positive; all the points $(a,b)$ that satisfy this condition are part of the region that I want to plot.

What's the easiest way to do this?

For example, taking $f_{a,b}(t)=1+at+bt^2$, we can use Minimize[1+at+bt^2, t] and find that the minimum value is (for $b>0$) equal to $\frac{-a^2+4b}{4b}$. Then the plot is given by RegionPlot[(-a^2+4b)/(4b)>0, {a, amin, amax}, {b, bmin, bmax}] for the given values of amin etc.

The issue is that in my case $f_{a,b}(t)$ is more complicated and the minimum must be found numerically (in general).

I have tried with RegionPlot[Minimize[f, x][[1]]> 0, {a, amin, amax}, {b, bmin, bmax}] but the evaluation never completes; replacing Minimize with FindMinimum only gives me a long list of error messages.

Any help is appreciated!

If your function is to complicated to do it analytically, use NMinimize.
RegionPlot[First@NMinimize[f[a, b, t], t] > 0, {a, 0, 3}, {b, 0, 4}]

• This does not work actually, as (for the simple function in my example) I get a long list of error messages such as NMinimize::nnum: The function value 1-0.829053 a+0.687328 b is not a number at {t} = {-0.829053}. Can you provide a minimal working example of the code you used to produce your image? – GioMott Nov 27 '17 at 20:32
• First I defined your simple example function f[a_, b_, t_] = 1 + a t + b t^2, Then the command you see above RegionPlot[First@NMinimize[f[a, b, t], t] > 0, {a, 0, 3}, {b, 0, 4}]. This works fine. – Akku14 Nov 28 '17 at 20:48