(a) Given \(\frac{dy}{dx} = x^{-\frac{1}{2}} + k\) and the gradient at (4, -1) is \(-\frac{3}{2}\), substitute \(x = 4\):

\(-\frac{3}{2} = 4^{-\frac{1}{2}} + k\)

\(-\frac{3}{2} = \frac{1}{2} + k\)

Solving for \(k\), we get \(k = -2\).

(b) Integrate \(\frac{dy}{dx} = x^{-\frac{1}{2}} - 2\):

\(y = \int (x^{-\frac{1}{2}} - 2) \, dx = 2x^{\frac{1}{2}} - 2x + c\)

Substitute \(x = 4, y = -1\):

\(-1 = 2(4)^{\frac{1}{2}} - 2(4) + c\)

\(-1 = 4 - 8 + c\)

\(c = 3\)

Thus, the equation is \(y = 2x^{\frac{1}{2}} - 2x + 3\).

(c) Set \(\frac{dy}{dx} = 0\):

\(x^{-\frac{1}{2}} - 2 = 0\)

\(x^{-\frac{1}{2}} = 2\)

\(x = \frac{1}{4}\)

Substitute \(x = \frac{1}{4}\) into the equation:

\(y = 2\left(\frac{1}{4}\right)^{\frac{1}{2}} - 2\left(\frac{1}{4}\right) + 3\)

\(y = 1 - \frac{1}{2} + 3 = \frac{3}{2}\)

The stationary point is \(\left(\frac{1}{4}, \frac{3}{2}\right)\).

(d) Find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = -\frac{1}{2}x^{-\frac{3}{2}}\)

At \(x = \frac{1}{4}\), \(\frac{d^2y}{dx^2} = -\frac{1}{2}\left(\frac{1}{4}\right)^{-\frac{3}{2}} = -4\)

Since \(\frac{d^2y}{dx^2} < 0\), the stationary point is a maximum.