函数 f 的定义为 $f : x \mapsto x^{3} - 1$, $x \in \mathbb{R}$。 (i) 求 $f^{-1}(x)$ 的表达式。 (ii) 在同一幅图上草绘 $y = f(x)$ 和 $y = f^{-1}(x)$ 的图像，并清楚表明这两个图像之间的关系。 (iii) 写出 $f'(x)$，并由此确定 $y = f^{-1}(x)$ 在其与 y 轴相交的点处的斜率。

1. Let $y = f(x)$ and make $x$ the subject of the formula.

$y = x^3 - 1$
$x^3 = y + 1$
$x = \sqrt[3]{y + 1}$

2. Express the inverse function in terms of $x$.

$f^{-1}(x) = \sqrt[3]{x + 1}$

1. Sketch the graph of $y = f(x)$ and label its axial intercepts.

The function $f(x) = x^3 - 1$ is a standard cubic curve translated downwards by $1$ unit.

• When $x = 0$, $y = -1$, giving the $y$-intercept $(0, -1)$.
• When $y = 0$, $x^3 - 1 = 0 \implies x = 1$, giving the $x$-intercept $(1, 0)$.

2. Sketch the line $y=x$ and use it to sketch $y = f^{-1}(x)$, displaying their relationship.

To make the relationship clear, draw the line of reflection $y = x$ (typically dashed).
Sketch $y = f^{-1}(x)$ as the direct reflection of $y = x^3 - 1$ across the line $y = x$.

• The intercept $(1, 0)$ reflects to $(0, 1)$.
• The intercept $(0, -1)$ reflects to $(-1, 0)$.
  Ensure that the two curves intersect exactly on the line $y = x$ in the first quadrant.

1. Write down the derivative of $f(x)$.

$f'(x) = 3x^2$

2. Find the $x$-coordinate on the original curve $y = f(x)$ that corresponds to the $y$-intercept of $y = f^{-1}(x)$.

The graph of $y = f^{-1}(x)$ crosses the $y$-axis when $x = 0$.
Let $y = f^{-1}(0)$. By definition of the inverse function, $f(y) = 0$.
$y^3 - 1 = 0 \implies y = 1$
The corresponding point on the curve $y = f(x)$ is $(1, 0)$.

3. Determine the gradient of $y = f^{-1}(x)$ using the implicit relationship $x = f(y)$.

For the inverse function $y = f^{-1}(x)$, we have $x = f(y)$.
Differentiating implicitly with respect to $y$ gives $\frac{dx}{dy} = f'(y)$.
The gradient of $y = f^{-1}(x)$ is thus $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{f'(y)}$.
Substituting $y = 1$ (the $y$-value when $x = 0$):
$\text{Gradient} = \frac{1}{f'(1)} = \frac{1}{3(1)^2} = \frac{1}{3}$