The diagram below shows the graph of $y=f(x)$ for $x<\frac{3}{2}$. The graph has stationary points at $(-\frac{\pi}{4}, -2)$ and $(\frac{\pi}{4}, -\frac{2}{5})$, axial intercept at $(0, -\frac{1}{2})$ and asymptotes $y=0$ and $x=\frac{3}{2}$. Sketch, on separate diagrams, the graphs of the following, showing clearly the coordinates of stationary points, axial intercepts and equations of asymptotes where applicable.

1. Determine the coordinates and nature of the stationary points for $y = \frac{1}{f(x)}$.

The stationary points of $y = \frac{1}{f(x)}$ occur at the same $x$-values as $f(x)$.
At $x = -\frac{\pi}{4}$, $f(-\frac{\pi}{4}) = -2$. Thus, $y = \frac{1}{-2} = -\frac{1}{2}$. Since $f(x)$ has a local minimum of a negative value, its reciprocal will have a local maximum. Hence, there is a local maximum at $(-\frac{\pi}{4}, -\frac{1}{2})$.
At $x = \frac{\pi}{4}$, $f(\frac{\pi}{4}) = -\frac{2}{5}$. Thus, $y = \frac{1}{-2/5} = -\frac{5}{2}$. Since $f(x)$ has a local maximum of a negative value, its reciprocal will have a local minimum. Hence, there is a local minimum at $(\frac{\pi}{4}, -\frac{5}{2})$.

2. Determine the new $y$-intercept.

The $y$-intercept of $f(x)$ is $(0, -\frac{1}{2})$.
Taking the reciprocal, the $y$-intercept of $y = \frac{1}{f(x)}$ is $(0, \frac{1}{-1/2}) = (0, -2)$.

3. Determine the end behaviours of the graph at the boundaries of its domain.

As $x \to -\infty$, $f(x) \to 0^-$, so $\frac{1}{f(x)} \to -\infty$.
As $x \to \frac{3}{2}^-$, $f(x) \to -\infty$, so $\frac{1}{f(x)} \to 0^-$.
This means the graph terminates at an open (hollow) circle at $(\frac{3}{2}, 0)$ instead of having a vertical asymptote.

4. Describe the final sketch incorporating all features.

Draw a smooth continuous curve for $x < \frac{3}{2}$ that comes from $-\infty$ in the third quadrant, rises to a local maximum at $(-\frac{\pi}{4}, -\frac{1}{2})$, falls through the $y$-intercept at $(0, -2)$ down to a local minimum at $(\frac{\pi}{4}, -\frac{5}{2})$, and then rises to terminate at an open circle at $(\frac{3}{2}, 0)$. There are no asymptotes to draw.

1. Determine the x-intercepts of the derived function.

The stationary points of $f(x)$ occur at $x = -\frac{\pi}{4}$ and $x = \frac{\pi}{4}$. At these points, the gradient is zero.
Thus, the $x$-intercepts of $y = f'(x)$ are $(-\frac{\pi}{4}, 0)$ and $(\frac{\pi}{4}, 0)$.

2. Determine the intervals where the derivative is positive or negative.

By observing the gradient of $f(x)$:

• For $x < -\frac{\pi}{4}$, $f(x)$ is decreasing, so the gradient is negative ($f'(x) < 0$).
• For $-\frac{\pi}{4} < x < \frac{\pi}{4}$, $f(x)$ is increasing, so the gradient is positive ($f'(x) > 0$).
• For $\frac{\pi}{4} < x < \frac{3}{2}$, $f(x)$ is decreasing, so the gradient is negative ($f'(x) < 0$).

3. Determine the asymptotes of the derived function.

As $x \to -\infty$, $f(x)$ approaches the horizontal asymptote $y=0$ flattening out, meaning its gradient approaches 0. Thus, $y = 0$ is a horizontal asymptote for $f'(x)$.
As $x \to \left(\frac{3}{2}\right)^-$, the curve of $f(x) \to -\infty$ and gets increasingly steep downwards, meaning the gradient $f'(x) \to -\infty$. Thus, $x = \frac{3}{2}$ is a vertical asymptote.