Answer: \(A=3\), \(B=1\); stationary point at \(x=-\frac13\), a minimum.

Given

\(y=(2x-1)\sqrt{4x+3},\)

differentiate using the product rule:

\(\frac{dy}{dx} =2\sqrt{4x+3}+(2x-1)\cdot\frac{2}{\sqrt{4x+3}}.\)

Put the terms over a common denominator:

\(\frac{dy}{dx} =\frac{2(4x+3)+2(2x-1)}{\sqrt{4x+3}}.\)

So

\(\frac{dy}{dx} =\frac{8x+6+4x-2}{\sqrt{4x+3}} =\frac{12x+4}{\sqrt{4x+3}} =\frac{4(3x+1)}{\sqrt{4x+3}}.\)

Hence

\(A=3,\qquad B=1.\)

At a stationary point,

\(\frac{dy}{dx}=0.\)

Since \(\sqrt{4x+3}\gt 0\) in the domain,

\(3x+1=0,\)

so

\(x=-\frac13.\)

The denominator is positive, so the sign of \(\frac{dy}{dx}\) is the sign of \(3x+1\). It changes from negative to positive at \(x=-\frac13\), so the stationary point is a minimum.