函数 f\mathrm{f} 和 g\mathrm{g} 定义为：
f:x↦1x−1,x≠0,g:x↦(x+2)2,x>−2.\begin{aligned} \mathrm{f} &: x \mapsto \frac{1}{x} - 1, \quad x \ne 0, \\ \mathrm{g} &: x \mapsto (x+2)^2, \quad x > -2. \end{aligned}

  1. (i)

    证明 fg(x)=−(x+1)(x+3)(x+2)2\mathrm{fg}(x) = -\frac{(x+1)(x+3)}{(x+2)^2}。写出 fg\mathrm{fg} 的定义域。

  2. (ii)

    由此，用代数方法求解不等式 fg(x)≥0\mathrm{fg}(x) \ge 0。

反函数与复合函数分式不等式函数概念English
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完整解法

2(i)

1. Substitute $\mathrm{g}(x)$ into $\mathrm{f}(x)$ to construct the composite function $\mathrm{fg}(x)$.

fg(x)=f(g(x))\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x))
=f((x+2)2)= \mathrm{f}((x+2)^2)
=1(x+2)2−1= \frac{1}{(x+2)^2} - 1

2. Combine the terms into a single fraction and factorise the numerator using the difference of two squares to achieve the final shown form.

=1−(x+2)2(x+2)2= \frac{1 - (x+2)^2}{(x+2)^2}
Using a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b):
=[1−(x+2)][1+(x+2)](x+2)2= \frac{[1 - (x+2)][1 + (x+2)]}{(x+2)^2}
=(−x−1)(x+3)(x+2)2= \frac{(-x-1)(x+3)}{(x+2)^2}
=−(x+1)(x+3)(x+2)2(Shown)= -\frac{(x+1)(x+3)}{(x+2)^2} \quad \text{(Shown)}

3. Determine the domain of the composite function $\mathrm{fg}$.

The domain of a composite function fg\mathrm{fg} is defined by {x∈Dg∣g(x)∈Df}\{x \in \mathrm{D}_\mathrm{g} \mid \mathrm{g}(x) \in \mathrm{D}_\mathrm{f}\}.
Given Dg=(all x>−2)\mathrm{D}_\mathrm{g} = (\text{all } x > -2) and Df=(all x≠0)\mathrm{D}_\mathrm{f} = (\text{all } x \ne 0).
For any x>−2x > -2, g(x)=(x+2)2>0\mathrm{g}(x) = (x+2)^2 > 0, so the condition g(x)≠0\mathrm{g}(x) \ne 0 is naturally satisfied.
Therefore, the domain of fg\mathrm{fg} is the domain of g\mathrm{g}.
Dfg={x∈R:x>−2}\mathrm{D}_{\mathrm{fg}} = \{x \in \mathbb{R} : x > -2\}

答案

Domain of fg\mathrm{fg} is x>−2x > -2

2(ii)

1. Set up the inequality using the expression from part (i) and find the critical values.

fg(x)≥0\mathrm{fg}(x) \ge 0
−(x+1)(x+3)(x+2)2≥0-\frac{(x+1)(x+3)}{(x+2)^2} \ge 0
(x+1)(x+3)(x+2)2≤0\frac{(x+1)(x+3)}{(x+2)^2} \le 0

The critical values are x=−1x = -1 and x=−3x = -3 from the numerator, and the asymptote is x=−2x = -2 from the denominator.

2. Solve the rational inequality without domain restrictions.

Using a sign test or number line for the expression (x+1)(x+3)(x+2)2\frac{(x+1)(x+3)}{(x+2)^2}:

  • For x<−3x < -3: expression is positive
  • For −3<x<−2-3 < x < -2: expression is negative
  • For −2<x<−1-2 < x < -1: expression is negative
  • For x>−1x > -1: expression is positive

Thus, the solution to the algebraic inequality is:
−3≤x≤−1,x≠−2-3 \le x \le -1, \quad x \ne -2

3. Apply the domain restriction of the composite function to state the final answer.

From part (i), the domain of fg\mathrm{fg} is x>−2x > -2.
Taking the intersection of the general solution (−3≤x≤−1,x≠−2-3 \le x \le -1, x \ne -2) and the domain (x>−2x > -2):
−2<x≤−1-2 < x \le -1

答案

−2<x≤−1-2 < x \le -1

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