Answer: \(\dfrac{dy}{dx}=\dfrac{3x+4}{2\sqrt{x+2}}\); stationary point \(\left(-\dfrac43,-\dfrac{4\sqrt6}{9}\right)\), minimum.

(a) Write

\(y=x(x+2)^{\frac12}.\)

Using the product rule,

\(\frac{dy}{dx} =x\cdot\frac12(x+2)^{-\frac12}+(x+2)^{\frac12}.\)

Put this over the common denominator \(2\sqrt{x+2}\):

\(\frac{dy}{dx} =\frac{x+2(x+2)}{2\sqrt{x+2}}.\)

So

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

Hence

\(A=3,\qquad B=4.\)

(b) A stationary point occurs when \(\frac{dy}{dx}=0\). Since the denominator is not zero at the stationary point, set the numerator equal to zero:

\(3x+4=0.\)

Thus

\(x=-\frac43.\)

Now substitute into \(y=x\sqrt{x+2}\):

\(y=-\frac43\sqrt{-\frac43+2} =-\frac43\sqrt{\frac23}.\)

Since

\(\sqrt{\frac23}=\frac{\sqrt6}{3},\)

we get

\(y=-\frac{4\sqrt6}{9}.\)

So the stationary point is

\(\left(-\frac43,-\frac{4\sqrt6}{9}\right).\)

(c) For \(x\gt-2\), the denominator \(2\sqrt{x+2}\) is positive. The sign of \(\frac{dy}{dx}\) is therefore controlled by \(3x+4\).

When \(x\lt-\frac43\), \(3x+4\lt0\), and when \(x\gt-\frac43\), \(3x+4\gt0\). So the gradient changes from negative to positive.

Therefore the stationary point is a minimum.