函数 $\mathrm{f}$ 定义如下：

(a) 求 $\mathrm{f}^{-1}(x)$ 的表达式。

(b) 证明 $1-\frac8{x^2+4}$ 可以表示为 $\frac{x^2-4}{x^2+4}$，并由此写出 $\mathrm{f}$ 的值域。

(c) 解释为什么不能构成复合函数 $\mathrm{ff}$。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

令原函数等于 $y$，交叉相乘并集中含 $x^2$ 的项。原定义域决定取正平方根。

From $y(x^2+4)=x^2-4$, we get $(1-y)x^2=4(1+y)$. The required input is positive, so

先用除法把分式写成一减去一个分式，再通过取倒数来孤立输入的平方。

Rewriting the function gives $y=1-8/(x^2+4)$. Thus

Since the original input is greater than $2$, take the positive square root and interchange the variables:

先通分验证两个表达式相同。再观察分母随 $x>2$ 取遍什么范围。

We have

Since $x^2+4$ takes every value in $(8,\infty)$, the fraction $8/(x^2+4)$ takes every value in $(0,1)$. Therefore the range of $\mathrm f$ is $(0,1)$.

复合时第一次得到的函数值必须能作为下一次的输入。比较值域与定义域。

The first application gives $\mathrm f(x)\in(0,1)$, but the next application would require an input greater than $2$. There is no input for which $\mathrm f(\mathrm f(x))$ is defined.