下图显示了 $x<\frac{3}{2}$ 时 $y=f(x)$ 的图像。该图像具有驻点 $(-\frac{\pi}{4}, -2)$ 和 $(\frac{\pi}{4}, -\frac{2}{5})$，与坐标轴的交点为 $(0, -\frac{1}{2})$，渐近线为 $y=0$ 和 $x=\frac{3}{2}$。 在不同的图上分别画出下列函数的图像，并清晰地标出驻点坐标、与坐标轴的交点坐标以及渐近线方程（如果适用）。

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.