Answer: \(\frac{\mathrm dy}{\mathrm dx}=\frac{1+3x}{\sqrt{1+2x}}\); approximate increase in \(x=\frac{9}{650}\); stationary point has \(x=-\frac13\).

Use differentiation to find the gradient information needed. Stationary points occur when the derivative is zero, and tangents or normals are found from the gradient at the given point.

For \(y=x\sqrt{1+2x}\), use the product rule:

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

Putting over a common denominator gives

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

At \(x=4\),

\(\frac{\mathrm dy}{\mathrm dx}=\frac{13}{3}\).

For small changes, \(\delta y\approx\frac{\mathrm dy}{\mathrm dx}\delta x\). Hence

\(0.06\approx\frac{13}{3}\delta x\), so \(\delta x=\frac{0.06\times3}{13}=\frac{9}{650}=0.0138\ldots\).

For a stationary point, set \(\frac{\mathrm dy}{\mathrm dx}=0\). Since the denominator is non-zero in the interior of the domain,

\(1+3x=0\), so \(x=-\frac13\).

This gives the required answer: \(\frac{\mathrm dy}{\mathrm dx}=\frac{1+3x}{\sqrt{1+2x}}\); approximate increase in \(x=\frac{9}{650}\); stationary point has \(x=-\frac13\).