The diagram shows the graph of a function $f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 12 - 2014 - Paper 1

The graph of $y = \frac{1}{f(x)}$ will involve understanding the behavior of $f(x)$ at its intercepts and asymptotes. Asymptotes will appear in the graph where $f(x)$ crosses the x-axis. If $f(x) < 0$, then $y = \frac{1}{f(x)}$ will yield negative values, which will be reflected correctly in each quadrant.

Using the identity $4\cos^3\theta - 3\cos\theta = \cos 3\theta$, substitute $x = 2\cos\theta$. Now, we have:
$4 \left( \frac{x}{2} \right)^3 - 3 \left( \frac{x}{2} \right) = \cos 3\theta$
After evaluating using $x = 2\cos\theta$, simplify it to show the required expression.

Given that $\cos 3\theta = \frac{\sqrt{3}}{2}$, find angles $3\theta$ using the inverse cosine function. The values will be $3\theta = \frac{\pi}{6} + 2k\pi$ and $3\theta = \frac{11\pi}{6} + 2k\pi$ for $k \in \mathbb{Z}$. Therefore, divide by 3 to obtain the three angles: $\theta = \frac{\pi}{18}$, $\theta = \frac{11\pi}{54}$, and $\theta = \frac{5\pi}{18}$.

To show that the tangents to the curves $x^2 - y^2 = 5$ and $xy = 6$ are perpendicular, find the derivatives using implicit differentiation. The slopes of each curve at the point $P(x_0, y_0)$ can be computed, and then verify that the product of the slopes is -1.

Evaluate the integral $I_0 = \int_0^1 \frac{2}{x^2 + 2} dx$ using a suitable substitution (possibly $u = x^2 + 2$) and find that $I_0 = \frac{\pi}{4}$.

Use integration by parts on the integral $I_n = \int_0^1 \frac{2x^{2n}}{x^2 + 2} dx$. Set one part as $u = x^{2n}$ and apply the integration technique to arrive at the recurrence relation revealing that $I_n + I_{n-1} = \frac{1}{2n - 1}$.

Using the recurrence relation derived, compute $I_1$ from $I_0$. Then, use these results to find $\int_0^1 \frac{x^4}{x^2 + 2} dx = I_2$ and solve for the integral value.