(i) Differentiate \(y = (x+1)^2 + (x+1)^{-1}\) to find \(\frac{dy}{dx} = 2(x+1) - (x+1)^{-2}\).

Set \(\frac{dy}{dx} = 0\) to find the minimum point: \(2(x+1)^3 = 1\).

Differentiate again to find \(\frac{d^2y}{dx^2} = 2 + 2(x+1)^{-3}\).

Substitute \((x+1)^{-3} = 2\) to find \(\frac{d^2y}{dx^2} = 6\).

(ii) The volume of the solid of revolution is given by \(\pi \int y^2 \, dx\).

Calculate \(y^2 = (x+1)^4 + (x+1)^{-2} + 2(x+1)^2\).

Integrate \(y^2\) from 0 to 1: \(\pi \left[ \frac{(x+1)^5}{5} + (x+1)^{-1} + \frac{2(x+1)^2}{2} \right]_0^1\).

Evaluate the definite integral: \(\pi \left[ \frac{32}{5} - \frac{1}{2} + 4 - \left( \frac{1}{5} + 1 \right) \right]\).

The volume is \(9.7\pi\) or \(30.5\).

(i) To find the points \(P\) and \(Q\), set \(y = \frac{x}{2} + \frac{6}{x} = 4\). Solving gives \(x = 2\) and \(x = 6\). The points are \(P(2, 4)\) and \(Q(6, 4)\).

The derivative of \(y\) is \(\frac{dy}{dx} = \frac{1}{2} - \frac{6}{x^2}\).

At \(x = 2\), the slope \(m = \frac{1}{2} - \frac{6}{4} = -1\). The tangent line is \(y - 4 = -1(x - 2)\) or \(y = -x + 6\).

At \(x = 6\), the slope \(m = \frac{1}{2} - \frac{6}{36} = \frac{1}{3}\). The tangent line is \(y - 4 = \frac{1}{3}(x - 6)\) or \(y = \frac{1}{3}x + 2\).

Solving \(y = -x + 6\) and \(y = \frac{1}{3}x + 2\) gives \(x = 3\), \(y = 3\). The tangents meet at \((3, 3)\), which lies on \(y = x\).

(ii) The volume of revolution is given by \(V = \pi \int_2^6 \left(4^2 - \left(\frac{x}{2} + \frac{6}{x}\right)^2\right) \, dx\).

Integrate \(\pi \int_2^6 \left(16 - \left(\frac{x}{2} + \frac{6}{x}\right)^2\right) \, dx\).

\(= \pi \int_2^6 \left(16 - \left(\frac{x^2}{4} + 6 + \frac{36}{x^2}\right)\right) \, dx\)

\(= \pi \int_2^6 \left(\frac{-x^2}{4} + 10 - \frac{36}{x^2}\right) \, dx\)

Integrate to get \(\left[-\frac{x^3}{12} + 10x + \frac{36}{x}\right]_2^6\).

Calculate: \(\left(-\frac{6^3}{12} + 10 \times 6 + \frac{36}{6}\right) - \left(-\frac{2^3}{12} + 10 \times 2 + \frac{36}{2}\right)\)

\(= \left(-18 + 60 + 6\right) - \left(-\frac{8}{12} + 20 + 18\right)\)

\(= 48 - 37 = 11\)

Volume \(= \pi \times 11 = \frac{10}{3} \pi\) or approximately \(33.5 \pi\).

(i) The area of the shaded region is given by the integral:

\(\frac{1}{2} \int_0^1 (x^4 - 1) \, dx\)

Integrating, we have:

\(\frac{1}{2} \left[ \frac{x^5}{5} - x \right]_0^1\)

\(= \frac{1}{2} \left( \frac{1}{5} - 1 \right) = -\frac{2}{5}\)

The area is \(\frac{2}{5}\).

(ii) The volume when rotated about the x-axis is:

\(\pi \int_0^1 y^2 \, dx = \frac{\pi}{4} \int_0^1 (x^8 - 2x^4 + 1) \, dx\)

Integrating, we have:

\(\frac{\pi}{4} \left[ \frac{x^9}{9} - \frac{2x^5}{5} + x \right]_0^1\)

\(= \frac{\pi}{4} \left( \frac{1}{9} - \frac{2}{5} + 1 \right) = \frac{8\pi}{45}\)

The volume is \(\frac{8\pi}{45}\) or 0.559.

(iii) The volume when rotated about the y-axis is:

\(\pi \int_{-\frac{1}{2}}^0 x^2 \, dy = (\pi) \int_{-\frac{1}{2}}^0 (2y+1)^{1/2} \, dy\)

Integrating, we have:

\((\pi) \left[ \frac{(2y+1)^{3/2}}{3/2} \right]_{-\frac{1}{2}}^0\)

\(= (\pi) \left[ \frac{1}{3} - 0 \right] = \frac{\pi}{3}\)

The volume is \(\frac{\pi}{3}\) or 1.05.

(i) To find the volume of revolution, we use the formula \(V = \pi \int (y + 1) \, dy\) from \(y = 0\) to \(y = h\).

Integrating, we have:

\(V = \pi \left[ \frac{y^2}{2} + y \right]_0^h\)

\(= \pi \left( \frac{h^2}{2} + h \right)\)

Thus, \(V = \pi \left( \frac{1}{2}h^2 + h \right)\).

(ii) To find the area of the shaded region when \(h = 3\), we calculate:

\(\int (y + 1)^{1/2} \, dy\) from \(y = 0\) to \(y = 3\).

Alternatively, \(6 - \int (x^2 - 1) \, dx\) from \(x = 1\) to \(x = 2\).

Integrating, we have:

\(\left[ \frac{2}{3}(y + 1)^{3/2} \right]_0^3\)

\(= \frac{2}{3} \left[ 8 - 1 \right]\)

\(= \frac{14}{3}\)

The volume of the solid of revolution is given by the formula:

\(V = \pi \int (f(x)^2 - g(x)^2) \, dx\)

where \(f(x) = 5 - x\) and \(g(x) = \frac{4}{x}\).

Calculate the volume by integrating from \(x = 1\) to \(x = 4\):

\(V = \pi \left( \int_1^4 (5-x)^2 \, dx - \int_1^4 \frac{16}{x^2} \, dx \right)\)

First, integrate \((5-x)^2\):

\(\int (5-x)^2 \, dx = \frac{(5-x)^3}{3}\)

Evaluate from 1 to 4:

\(\left[ \frac{(5-x)^3}{3} \right]_1^4 = \frac{(5-4)^3}{3} - \frac{(5-1)^3}{3} = \frac{1}{3} - \frac{64}{3} = -\frac{63}{3} = -21\)

Next, integrate \(\frac{16}{x^2}\):

\(\int \frac{16}{x^2} \, dx = -\frac{16}{x}\)

Evaluate from 1 to 4:

\(\left[ -\frac{16}{x} \right]_1^4 = -\frac{16}{4} + \frac{16}{1} = -4 + 16 = 12\)

Subtract the two results:

\(V = \pi (-21 - 12) = \pi (-33) = 33\pi\)

Thus, the volume is \(9\pi\) or approximately 28.3.