函数 f\mathrm{f} 由下式给出：
f:x↦14−(2x+1)2,x∈R, x≤−12, x≠−32.\mathrm{f} : x \mapsto \frac{1}{4 - (2x + 1)^2}, \quad x \in \mathbb{R}, \ x \le -\frac{1}{2}, \ x \ne -\frac{3}{2}.

  1. (i)

    在同一张图中画出 y=f(x)y = \mathrm{f}(x) 和 y=f−1(x)y = \mathrm{f}^{-1}(x) 的图像，并给出渐近线的方程（若有）。以类似的格式定义 f−1\mathrm{f}^{-1}。

  2. (ii)

    解释为什么第 (i) 问中曲线交点的 xx 坐标满足方程 4x3+4x2−3x+1=04x^3 + 4x^2 - 3x + 1 = 0，并求出该交点的坐标。

  3. (iii)

    函数 g\mathrm{g} 定义为
    g:x↦k−ex,x∈R, x≤0,其中 k 是一个整数。\mathrm{g} : x \mapsto k - \mathrm{e}^x, \quad x \in \mathbb{R}, \ x \le 0, \quad \text{其中 } k \text{ 是一个整数}。

    用 kk 表示 g\mathrm{g} 的值域。确定使复合函数 fg\mathrm{fg} 存在的最大的 kk 值，并求其值域。

反函数与复合函数反函数图像存在性条件函数概念解方程反函数定义域限制渐近线对称性与限制English
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完整解法

6(i)

1. Find the algebraic expression for the inverse function $\mathrm{f}^{-1}(x)$ by making $x$ the subject.

Let y=14−(2x+1)2\text{Let } y = \frac{1}{4 - (2x + 1)^2}
4−(2x+1)2=1y4 - (2x + 1)^2 = \frac{1}{y}
(2x+1)2=4−1y(2x + 1)^2 = 4 - \frac{1}{y}
2x+1=±4−1y2x + 1 = \pm\sqrt{4 - \frac{1}{y}}
Since the domain of f is x≤−12, we have 2x+1≤0.\text{Since the domain of } \mathrm{f} \text{ is } x \le -\frac{1}{2}, \text{ we have } 2x + 1 \le 0.
Thus, we reject the positive root:\text{Thus, we reject the positive root:}
2x+1=−4−1y2x + 1 = -\sqrt{4 - \frac{1}{y}}
x=−12−124−1yx = -\frac{1}{2} - \frac{1}{2}\sqrt{4 - \frac{1}{y}}
∴f−1(x)=−12−124−1x\therefore \mathrm{f}^{-1}(x) = -\frac{1}{2} - \frac{1}{2}\sqrt{4 - \frac{1}{x}}

2. Determine the domain of $\mathrm{f}^{-1}$ by finding the range of $\mathrm{f}$.

Consider the domain of f:x∈(−∞,−32)∪(−32,−12]\text{Consider the domain of } \mathrm{f}: x \in \left(-\infty, -\frac{3}{2}\right) \cup \left(-\frac{3}{2}, -\frac{1}{2}\right]
For x∈(−32,−12],0≤(2x+1)2<4\text{For } x \in \left(-\frac{3}{2}, -\frac{1}{2}\right], \quad 0 \le (2x+1)^2 < 4
  ⟹  0<4−(2x+1)2≤4  ⟹  f(x)≥14\implies 0 < 4 - (2x+1)^2 \le 4 \implies \mathrm{f}(x) \ge \frac{1}{4}
For x∈(−∞,−32),(2x+1)2>4\text{For } x \in \left(-\infty, -\frac{3}{2}\right), \quad (2x+1)^2 > 4
  ⟹  4−(2x+1)2<0  ⟹  f(x)<0\implies 4 - (2x+1)^2 < 0 \implies \mathrm{f}(x) < 0
The range of f is (−∞,0)∪[14,∞).\text{The range of } \mathrm{f} \text{ is } (-\infty, 0) \cup \left[\frac{1}{4}, \infty\right).
Hence, the domain of f−1 is x<0 or x≥14.\text{Hence, the domain of } \mathrm{f}^{-1} \text{ is } x < 0 \text{ or } x \ge \frac{1}{4}.

3. Define the inverse function $\mathrm{f}^{-1}$ formally in the same notation as the given function.

f−1:x↦−12−124−1x,x∈R, x<0 or x≥14\mathrm{f}^{-1} : x \mapsto -\frac{1}{2} - \frac{1}{2}\sqrt{4 - \frac{1}{x}}, \quad x \in \mathbb{R}, \ x < 0 \text{ or } x \ge \frac{1}{4}

4. Identify the equations of the asymptotes for both functions.

For y=f(x):\text{For } y = \mathrm{f}(x):
Vertical asymptote: x=−32\text{Vertical asymptote: } x = -\frac{3}{2}
Horizontal asymptote: y=0 (as x→−∞)\text{Horizontal asymptote: } y = 0 \text{ (as } x \to -\infty)
For y=f−1(x), the asymptotes are reflected across y=x:\text{For } y = \mathrm{f}^{-1}(x), \text{ the asymptotes are reflected across } y=x:
Vertical asymptote: x=0\text{Vertical asymptote: } x = 0
Horizontal asymptote: y=−32\text{Horizontal asymptote: } y = -\frac{3}{2}

5. Sketch the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{f}^{-1}(x)$ on the same diagram.

Sketch y=f(x) as a curve with an endpoint at (−12,14), asymptotic to x=−32 and y=0.\text{Sketch } y = \mathrm{f}(x) \text{ as a curve with an endpoint at } \left(-\frac{1}{2}, \frac{1}{4}\right), \text{ asymptotic to } x = -\frac{3}{2} \text{ and } y = 0.
Sketch y=f−1(x) by reflecting y=f(x) in the line y=x.\text{Sketch } y = \mathrm{f}^{-1}(x) \text{ by reflecting } y = \mathrm{f}(x) \text{ in the line } y = x.
It has an endpoint at (14,−12) and is asymptotic to x=0 and y=−32.\text{It has an endpoint at } \left(\frac{1}{4}, -\frac{1}{2}\right) \text{ and is asymptotic to } x = 0 \text{ and } y = -\frac{3}{2}.
Both curves intersect on the line y=x.\text{Both curves intersect on the line } y = x.

答案

Asymptotes of y=f(x)y = \mathrm{f}(x): x=−32, y=0x = -\frac{3}{2}, \ y = 0
Asymptotes of y=f−1(x)y = \mathrm{f}^{-1}(x): x=0, y=−32x = 0, \ y = -\frac{3}{2}

f−1:x↦−12−124−1x,x∈R, x<0 or x≥14\mathrm{f}^{-1} : x \mapsto -\frac{1}{2} - \frac{1}{2}\sqrt{4 - \frac{1}{x}}, \quad x \in \mathbb{R}, \ x < 0 \text{ or } x \ge \frac{1}{4}

(A sketch should be drawn showing y=f(x)y = \mathrm{f}(x) and y=f−1(x)y = \mathrm{f}^{-1}(x) as reflections of each other across the line y=xy = x. y=f(x)y = \mathrm{f}(x) has an endpoint at (−12,14)\left(-\frac{1}{2}, \frac{1}{4}\right) and asymptotes x=−32x = -\frac{3}{2} and y=0y = 0. y=f−1(x)y = \mathrm{f}^{-1}(x) has an endpoint at (14,−12)\left(\frac{1}{4}, -\frac{1}{2}\right) and asymptotes x=0x = 0 and y=−32y = -\frac{3}{2}.)

6(ii)

1. Explain why the intersection lies on $y=x$ and set up the equation.

The point of intersection between a function y=f(x)y = \mathrm{f}(x) and its inverse y=f−1(x)y = \mathrm{f}^{-1}(x) lies on the line y=xy = x. Therefore, the xx-coordinate of this point satisfies f(x)=x\mathrm{f}(x) = x.

2. Substitute the expression for $\mathrm{f}(x)$ and simplify to obtain the given cubic equation.

14−(2x+1)2=x\frac{1}{4 - (2x+1)^2} = x
1=x[4−(4x2+4x+1)]1 = x[4 - (4x^2 + 4x + 1)]
1=x(3−4x−4x2)1 = x(3 - 4x - 4x^2)
1=3x−4x2−4x31 = 3x - 4x^2 - 4x^3
4x3+4x2−3x+1=0(shown)4x^3 + 4x^2 - 3x + 1 = 0 \quad \text{(shown)}

3. Solve the cubic equation using a graphing calculator and state the coordinates.

Using a graphing calculator to solve 4x3+4x2−3x+1=04x^3 + 4x^2 - 3x + 1 = 0:
x≈−1.5764x \approx -1.5764
Since the intersection point lies on y=xy = x, the yy-coordinate is also approximately −1.5764-1.5764.
The coordinates of the point of intersection are (−1.58,−1.58)(-1.58, -1.58) (to 3 s.f.).

答案

(−1.58,−1.58)(-1.58, -1.58)

6(iii)

1. Find the range of g in terms of k.

Given g(x)=k−ex\mathrm{g}(x) = k - \mathrm{e}^x for x≤0x \le 0.
Since x≤0x \le 0, we have 0<ex≤10 < \mathrm{e}^x \le 1.
Multiplying by −1-1 gives −1≤−ex<0-1 \le -\mathrm{e}^x < 0.
Adding kk, we obtain k−1≤k−ex<kk - 1 \le k - \mathrm{e}^x < k.
Thus, the range of g\mathrm{g} is Rg=[k−1,k)\mathrm{R}_\mathrm{g} = [k - 1, k).

2. Use the condition for the existence of composite function fg to find the largest integer k.

For the composite function fg\mathrm{fg} to exist, the range of g\mathrm{g} must be a subset of the domain of f\mathrm{f}.
Rg⊆Df\mathrm{R}_\mathrm{g} \subseteq \mathrm{D}_\mathrm{f}
[k−1,k)⊆(−∞,−1/2]∖{−3/2}[k - 1, k) \subseteq (-\infty, -1/2] \setminus \{-3/2\}
This requires k<−1/2k < -1/2. Since kk is an integer, k≤−1k \le -1.
If k=−1k = -1, Rg=[−2,−1)\mathrm{R}_\mathrm{g} = [-2, -1). This interval contains −3/2-3/2, which is not in Df\mathrm{D}_\mathrm{f}, so k=−1k = -1 is rejected.
If k=−2k = -2, Rg=[−3,−2)\mathrm{R}_\mathrm{g} = [-3, -2). This interval is strictly less than −1/2-1/2 and does not contain −3/2-3/2, which satisfies the condition.
Thus, the largest integer value is k=−2k = -2.

3. Find the range of fg by applying the function f to the range of g.

With k=−2k = -2, we have Rg=[−3,−2)\mathrm{R}_\mathrm{g} = [-3, -2).
The range of fg\mathrm{fg} is the set of values of f(u)\mathrm{f}(u) for u∈[−3,−2)u \in [-3, -2).
−3≤u<−2-3 \le u < -2
−5≤2u+1<−3-5 \le 2u + 1 < -3
9<(2u+1)2≤259 < (2u + 1)^2 \le 25
−21≤4−(2u+1)2<−5-21 \le 4 - (2u + 1)^2 < -5
Taking the reciprocal reverses the inequalities for values of the same sign:
−15<14−(2u+1)2≤−121-\frac{1}{5} < \frac{1}{4 - (2u + 1)^2} \le -\frac{1}{21}
Therefore, the range of fg\mathrm{fg} is (−15,−121]\left(-\frac{1}{5}, -\frac{1}{21}\right].

答案

Range of g\mathrm{g} is [k−1,k)[k - 1, k). Largest k=−2k = -2. Range of fg\mathrm{fg} is (−15,−121]\left(-\frac{1}{5}, -\frac{1}{21}\right].

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