To find \(f(x)\), we need to integrate \(f'(x) = 3x^{\frac{1}{2}} + 3x^{-\frac{1}{2}} - 10\).

Integrating term by term:

\(\int 3x^{\frac{1}{2}} \, dx = 2x^{\frac{3}{2}} + C_1\)

\(\int 3x^{-\frac{1}{2}} \, dx = 6x^{\frac{1}{2}} + C_2\)

\(\int -10 \, dx = -10x + C_3\)

Thus, \(f(x) = 2x^{\frac{3}{2}} + 6x^{\frac{1}{2}} - 10x + c\).

We know the curve passes through the point \((4, -7)\), so substitute \(x = 4\) and \(f(x) = -7\):

\(-7 = 2(4)^{\frac{3}{2}} + 6(4)^{\frac{1}{2}} - 10(4) + c\)

\(-7 = 16 + 12 - 40 + c\)

\(-7 = -12 + c\)

\(c = 5\)

Therefore, \(f(x) = 2x^{\frac{3}{2}} + 6x^{\frac{1}{2}} - 10x + 5\).

To find the equation of the curve, we need to integrate \(\frac{dy}{dx} = 2(3x + 4)^{\frac{3}{2}} - 6x - 8\).

Integrate each term separately:

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

Let \(u = 3x + 4\), then \(du = 3 \, dx\) or \(dx = \frac{du}{3}\).

\(\int 2(3x + 4)^{\frac{3}{2}} \, dx = \frac{2}{3} \int u^{\frac{3}{2}} \, du = \frac{2}{3} \cdot \frac{2}{5} u^{\frac{5}{2}} = \frac{4}{15} (3x + 4)^{\frac{5}{2}}\)

2. \(\int -6x \, dx = -3x^2\)

3. \(\int -8 \, dx = -8x\)

Thus, the integral is:

\(y = \frac{4}{15} (3x + 4)^{\frac{5}{2}} - 3x^2 - 8x + c\)

Given the stationary point \((-1, 5)\), substitute into the equation:

\(5 = \frac{4}{15}(3(-1) + 4)^{\frac{5}{2}} - 3(-1)^2 - 8(-1) + c\)

\(5 = \frac{4}{15}(1)^{\frac{5}{2}} - 3 + 8 + c\)

\(5 = \frac{4}{15} + 5 + c\)

\(c = 5 - 5 - \frac{4}{15} = -\frac{4}{15}\)

Therefore, the equation of the curve is:

\(y = \frac{4}{15}(3x + 4)^{\frac{5}{2}} - 3x^2 - 8x - \frac{4}{15}\)

To find the equation of the curve, we need to integrate the given derivative \(\frac{dy}{dx} = -\frac{8}{x^3} - 1\).

Integrate \(\frac{dy}{dx}\) with respect to \(x\):

\(y = \int \left( -\frac{8}{x^3} - 1 \right) \, dx\)

\(y = \int -8x^{-3} \, dx - \int 1 \, dx\)

\(y = -8 \int x^{-3} \, dx - \int 1 \, dx\)

\(y = -8 \left( \frac{x^{-2}}{-2} \right) - x + c\)

\(y = 4x^{-2} - x + c\)

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

Given that the point (2, 4) lies on the curve, substitute \(x = 2\) and \(y = 4\) to find \(c\):

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

\(4 = 1 - 2 + c\)

\(4 = -1 + c\)

\(c = 5\)

Thus, the equation of the curve is \(y = \frac{4}{x^2} - x + 5\).

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

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

Let \(u = x-1\), then \(du = dx\). The integral becomes \(\int -\frac{2}{u^3} \, du = \frac{1}{u^2} = \frac{1}{(x-1)^2}\).

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

Given \((0, 3)\), substitute to find \(c\):

\(3 = 0 + \frac{1}{1} + c \Rightarrow c = 2\).

Therefore, \(y = x + (x-1)^2 + 2\).

(b) The gradient of the tangent at \(x = 0\) is \(f'(0) = 3\).

The equation of the tangent is \(y - 3 = 3(x - 0) \Rightarrow y = 3x + 3\).

Set \(3x + 3 = x + (x-1)^2 + 2\) to find intersection:

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

Multiply through by \((x-1)^2\):

\((2x + 1)(x-1)^2 = 1 \Rightarrow (2x + 1)(x-1)^2 - 1 = 0\).

(c) Substitute \(x = \frac{3}{2}\) into \((2x + 1)(x-1)^2 - 1 = 0\):

\((2 \times \frac{3}{2} + 1)(\frac{3}{2} - 1)^2 - 1 = 0\).

\((3 + 1)(\frac{1}{2})^2 - 1 = 0 \Rightarrow 4 \times \frac{1}{4} - 1 = 0\).

Thus, \(x = \frac{3}{2}\) satisfies the equation.

Find \(y\)-coordinate: \(y = \frac{3}{2} + (\frac{3}{2} - 1)^2 + 2\).

\(y = \frac{3}{2} + \frac{1}{4} + 2 = \frac{7}{2}\).

Given \(\frac{dy}{dx} = \frac{2}{\sqrt{x}} - 1\), we need to find \(y\) by integrating.

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

\(y = \int \left( \frac{2}{\sqrt{x}} - 1 \right) \, dx\)

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

\(y = 2 \cdot \frac{x^{1/2}}{1/2} - x + c\)

\(y = 4\sqrt{x} - x + c\)

Use the point \(P(9, 5)\) to find \(c\):

\(5 = 4\sqrt{9} - 9 + c\)

\(5 = 12 - 9 + c\)

\(5 = 3 + c\)

\(c = 2\)

Thus, the equation of the curve is \(y = 4\sqrt{x} - x + 2\).