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

To sketch the graph of $y = \frac{1}{f(x)}$:

1. Identify Asymptotes: If $f(x)$ has any zeros (where it crosses the x-axis), then these will become vertical asymptotes in the graph of $y = \frac{1}{f(x)}$.

2. Reciprocal Values: For values where $f(x)$ is positive, the graph will approach $0$ as $f(x)$ increases. Conversely, for negative values of $f(x)$, $y = \frac{1}{f(x)}$ will decrease toward negative infinity.

3. Draw Intercepts: If $f(a) = 1$, then $(a, 1)$ will be a point on the graph of $y = \frac{1}{f(x)}$.

4. Final Sketch: Combine the behavior shown to represent the graph accurately.

Substitution: Given $x = 2\cos\theta$, substitute this into $x^3 - 3x = \sqrt{3}$. This leads to:

$(2\cos\theta)^3 - 3(2\cos\theta) = \sqrt{3}.$(1)

Expand and Simplify: From (1), we get:

$8\cos^3\theta - 6\cos\theta = \sqrt{3}.$(2)

Rearrange: Rearranging (2) gives:

$8\cos^3\theta - 6\cos\theta - \sqrt{3} = 0.$ Now we need to relate this to $\cos 3\theta$. The relation states:

$\cos 3\theta = 4\cos^3\theta - 3\cos\theta.$ Substituting the above will give:

$\cos 3\theta = 4\cos^3\theta - 3\cos\theta = \frac{\sqrt{3}}{2}.$

From our result in part (i), where $\cos 3\theta = \frac{\sqrt{3}}{2}$, we find:

1. Identify Angles: The angles where cosine takes the value $\frac{\sqrt{3}}{2}$ are:

   • $3\theta = \frac{\pi}{6}$
   • $3\theta = \frac{11\pi}{6}$
   • $3\theta = \frac{7\pi}{6}$.

2. Solve for $\theta$:

   • $\theta_1 = \frac{\pi}{18}$
   • $\theta_2 = \frac{11\pi}{18}$
   • $\theta_3 = \frac{7\pi}{18}$.

3. Find corresponding $x$ values: Using $x = 2\cos\theta$, compute:

   • $x_1 = 2\cos(\theta_1)$,
   • $x_2 = 2\cos(\theta_2)$,
   • $x_3 = 2\cos(\theta_3)$.