(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.

(a) We have \(f'(x) = \frac{-3}{(x+2)^4}\) and the gradient of the tangent is \(-\frac{16}{27}\) at \(x = a\). Set \(\frac{-3}{(a+2)^4} = -\frac{16}{27}\).

Equating gives \(16(a+2)^4 = 81\).

Solving for \((a+2)^2\), we get \((a+2)^2 = \frac{9}{4}\).

Thus, \(a+2 = \pm \frac{3}{2}\).

Solving for \(a\), we find \(a = \frac{1}{2}\) or \(a = -\frac{7}{2}\).

(b) To find \(f(x)\), integrate \(f'(x) = \frac{-3}{(x+2)^4}\).

\(f(x) = \int \frac{-3}{(x+2)^4} \, dx = \frac{-1}{(x+2)^3} + c\).

Given \(f(-1) = 5\), substitute to find \(c\):

\(5 = \frac{-1}{1^3} + c \Rightarrow 5 = -1 + c \Rightarrow c = 6\).

Thus, \(f(x) = \frac{-1}{(x+2)^3} + 6\).

(a) To find the equation of the curve, integrate \(\frac{dy}{dx} = 3x^{\frac{1}{2}} - 3x^{-\frac{1}{2}}\).

Integrating, we get:

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

\(y = \left[ \frac{3}{\frac{3}{2}} x^{\frac{3}{2}} \right] - \left[ \frac{3}{\frac{1}{2}} x^{\frac{1}{2}} \right] + c\)

\(y = 2x^{\frac{3}{2}} - 6x^{\frac{1}{2}} + c\)

Using the point \((3, 5)\) to find \(c\):

\(5 = 2(3)^{\frac{3}{2}} - 6(3)^{\frac{1}{2}} + c\)

\(5 = 2 \times 3\sqrt{3} - 6 \times \sqrt{3} + c\)

\(5 = 6\sqrt{3} - 6\sqrt{3} + c\)

\(c = 5\)

Thus, the equation is \(y = 2x^{\frac{3}{2}} - 6x^{\frac{1}{2}} + 5\).

(b) To find the \(x\)-coordinate of the stationary point, set \(\frac{dy}{dx} = 0\):

\(3x^{\frac{1}{2}} - 3x^{-\frac{1}{2}} = 0\)

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

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

\(x = 1\)

(c) \(y\) increases as \(x\) increases when \(\frac{dy}{dx} > 0\):

\(3x^{\frac{1}{2}} - 3x^{-\frac{1}{2}} > 0\)

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

\(x > 1\)