Answer: \(\dfrac{dy}{dx}=\dfrac{25x^2+8x-5}{2\sqrt{5x+2}}\); stationary point \((0.31,-1.7)\), minimum.

(a) Use the product rule on

\(y=(x^2-1)(5x+2)^{\frac12}.\)

Then

\(\frac{dy}{dx} =2x(5x+2)^{\frac12} +(x^2-1)\cdot\frac52(5x+2)^{-\frac12}.\)

Put this over the common denominator \(2\sqrt{5x+2}\):

\(\frac{dy}{dx} =\frac{4x(5x+2)+5(x^2-1)}{2\sqrt{5x+2}}.\)

Simplifying the numerator,

\(4x(5x+2)+5(x^2-1)=20x^2+8x+5x^2-5.\)

So

\(\frac{dy}{dx}=\frac{25x^2+8x-5}{2\sqrt{5x+2}}.\)

Hence

\(A=25,\qquad B=8,\qquad C=-5.\)

(b) A stationary point occurs when \(\frac{dy}{dx}=0\). Since \(2\sqrt{5x+2}\neq0\), solve

\(25x^2+8x-5=0.\)

Using the quadratic formula,

\(x=\frac{-8\pm\sqrt{8^2-4(25)(-5)}}{50} =\frac{-8\pm\sqrt{564}}{50}.\)

The positive solution is

\(x\approx0.315.\)

Substitute this into \(y=(x^2-1)\sqrt{5x+2}\):

\(y\approx-1.70.\)

So, correct to 2 significant figures, the stationary point is

\((0.31,-1.7).\)

(c) The denominator \(2\sqrt{5x+2}\) is positive for \(x\gt0\). The numerator \(25x^2+8x-5\) changes from negative to positive at the positive root. Therefore \(\frac{dy}{dx}\) changes from negative to positive.

Hence the stationary point is a minimum.