函数 ff 和 gg 的定义如下： f:x↦1−1x2−1f : x \mapsto 1-\dfrac{1}{x^{2}-1}，其中 x∈R,  x<−1x \in \mathbb{R},\; x<-1， g:x↦x2+4x+5g : x \mapsto x^{2}+4x+5，其中 x∈R,  x≤1x \in \mathbb{R},\; x\le 1。

  1. (a)

    求 f−1(x)f^{-1}(x) 并写出 f−1f^{-1} 的定义域。

  2. (b)

    指出复合函数 f ⁣gf\!g 和 g ⁣fg\!f 是否存在，并说明理由。由此求出存在的复合函数的值域。

反函数与复合函数存在性条件反函数定义域限制函数概念English
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完整解法

5(a)

1. Set $y = f(x)$ and rearrange to solve for $x^2$.

Let y=1−1x2−1y = 1-\frac{1}{x^{2}-1}

1x2−1=1−y\frac{1}{x^{2}-1} = 1 - y

x2−1=11−yx^{2}-1 = \frac{1}{1-y}

x2=1+11−y=2−y1−yx^{2} = 1 + \frac{1}{1-y} = \frac{2-y}{1-y}

2. Take the square root and choose the correct sign based on the domain of $f$.

x=±2−y1−yx = \pm \sqrt{\frac{2-y}{1-y}}

Since the domain of ff is x<−1x < -1, xx must be negative, so we reject the positive root.
x=−2−y1−yx = -\sqrt{\frac{2-y}{1-y}}

∴f−1(x)=−2−x1−x\therefore f^{-1}(x) = -\sqrt{\frac{2-x}{1-x}}

3. Determine the domain of $f^{-1}$ by finding the range of $f$.

For x<−1x < -1, we have x2>1  ⟹  x2−1>0x^2 > 1 \implies x^2 - 1 > 0.
Since 1x2−1>0\frac{1}{x^2-1} > 0, it follows that 1−1x2−1<11 - \frac{1}{x^2-1} < 1.

As x→−1−x \to -1^-, 1x2−1→∞  ⟹  f(x)→−∞\frac{1}{x^2-1} \to \infty \implies f(x) \to -\infty.
Thus, the range of ff is y<1y < 1.

The domain of f−1f^{-1} is the range of ff, which is x<1x < 1.

答案

f−1(x)=−2−x1−xf^{-1}(x) = -\sqrt{\frac{2-x}{1-x}}, Domain of f−1f^{-1}: x<1x < 1

5(b)

1. Determine the ranges of functions $f$ and $g$.

\text{For } f(x) = 1 - \frac{1}{x^2-1} \text{ with } x < -1: \
x^2 > 1 \implies x^2 - 1 > 0 \implies \frac{1}{x^2-1} > 0 \
\therefore f(x) < 1 \
\text{As } x \to -1^-, f(x) \to -\infty \implies R_f = (-\infty, 1) \
\text{For } g(x) = x^2 + 4x + 5 = (x+2)^2 + 1 \text{ with } x \le 1: \
\text{The minimum is at } x = -2, \text{ where } g(-2) = 1. \
\text{As } x \to -\infty, g(x) \to \infty \implies R_g = [1, \infty)

2. Assess the existence of the composite function $fg$.

\text{For } fg \text{ to exist, we require } R_g \subseteq D_f. \
D_f = (-\infty, -1) \
\text{Since } R_g = [1, \infty) \not\subseteq (-\infty, -1), \
fg \text{ does not exist.}

3. Assess the existence of the composite function $gf$.

\text{For } gf \text{ to exist, we require } R_f \subseteq D_g. \
D_g = (-\infty, 1] \
\text{Since } R_f = (-\infty, 1) \subseteq (-\infty, 1], \
gf \text{ exists.}

4. Calculate the range of the existing composite function $gf$.

\text{The range of } gf \text{ is the range of } g \text{ evaluated over the domain } R_f = (-\infty, 1). \
\text{We consider } g(x) = (x+2)^2 + 1 \text{ for } x \in (-\infty, 1). \
\text{The turning point } x = -2 \text{ lies within this interval.} \
\text{Minimum value} = g(-2) = 1 \
\text{As } x \to -\infty, g(x) \to \infty \
\therefore R_{gf} = [1, \infty)

答案

fgfg does not exist as Rg⊈DfR_g \not\subseteq D_f.
gfgf exists as Rf⊆DgR_f \subseteq D_g.
The range of gfgf is [1,∞)[1, \infty).

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