# Why RegionPlot does not show the negative part of a function?

I have this function

$$f(x)=\frac{1-25 x^2}{\left| 25 x^2-1\right| }\;\frac{2500 x}{\sqrt{\left(25 x^2-1\right)^2 \sinh (\pi x)+9}}+\cosh ^{-1}\left(\frac{1000 x^2+\left(25 x^2-1\right)^2}{\left(25 x^2+1\right)^2}\right)$$

 f[x_] := ArcCosh[(1000 x^2 + (-1 + 25 x^2)^2)/(1 + 25 x^2)^2] + (
1 - 25 x^2)/Abs[-1 + 25 x^2] (2500 x )/ Sqrt[
9 + (-1 + 25 x^2)^2 Sinh[π  x]];


for $$0. I plot this function and I get As can be seen, for $$0.2, function is decreasing first and then increasing. Therefore $$f'(x)$$ must be negative and positive. Also, $$f'(x)$$ must be zero at one point. But when I use RegionPlot to see where $$f'(x)$$ is negative, it says nowhere. Also, I try to Plot $$f'(x)$$, but I get no answer. Does someone know where I am making mistake?

RegionPlot[a (f'[x]) < 0, {x, 0.2, 1}, {a, 0, 1}]

Plot[f'[x], {x, 0.2, 1}]

ContourPlot[a (f'[x]) == 0, {x, 0, 1}, {a, 0, 1}]


The problem is that Abs has no built in derivative.

One solution is setting

Abs'[x_] := HeavisideTheta[x]


Then you can do

Plot[f'[x], {x, 0, 1}] • Or to use RealAbs. – Chip Hurst Jul 20 at 19:40
• Or the replacement Abs[x_] :> Sqrt[x^2] – Bob Hanlon Jul 20 at 20:24