题目

Problem

f(x)=3-\frac{x-2}{x+1}+\frac{5x+26}{2x^2-3x-5}\qquad x>4

(a) Show that

f(x)=\frac{ax+b}{cx+d}\qquad x>4

where $a$ , $b$ , $c$ and $d$ are integers to be found.

(4)

(b) Hence find $f^{-1}(x)$ .

(2)

题目中文翻译

f(x)=3-\frac{x-2}{x+1}+\frac{5x+26}{2x^2-3x-5}\qquad x>4

(a) 证明

f(x)=\frac{ax+b}{cx+d}\qquad x>4

其中 $a,b,c,d$ 为待求整数。

(b) 由此求 $f^{-1}(x)$ 。

解答

We have

f(x)=3-\frac{x-2}{x+1}+\frac{5x+26}{2x^2-3x-5}

First factorise

2x^2-3x-5

2x^2-3x-5=(2x-5)(x+1)

f(x)=3-\frac{x-2}{x+1}+\frac{5x+26}{(2x-5)(x+1)}

Use the common denominator

(x+1)(2x-5)

Then

\begin{aligned} f(x) &=\frac{3(x+1)(2x-5)-(x-2)(2x-5)+(5x+26)} {(x+1)(2x-5)} \end{aligned}

Now expand the numerator:

3(x+1)(2x-5)=3(2x^2-3x-5)=6x^2-9x-15

Also,

(x-2)(2x-5)=2x^2-9x+10

So the numerator is

\begin{aligned} &6x^2-9x-15-(2x^2-9x+10)+5x+26\\ &=4x^2+5x+1\\ &=(4x+1)(x+1) \end{aligned}

Therefore

\begin{aligned} f(x) &=\frac{(4x+1)(x+1)}{(x+1)(2x-5)}\\ &=\frac{4x+1}{2x-5} \end{aligned}

Hence

\boxed{f(x)=\frac{4x+1}{2x-5}}

so one possible set of constants is

\boxed{a=4,\quad b=1,\quad c=2,\quad d=-5}

Let

y=f(x)

Then

y=\frac{4x+1}{2x-5}

Rearrange:

y(2x-5)=4x+1

2xy-5y=4x+1

Bring the $x$ terms to one side:

2xy-4x=5y+1

Factorise:

x(2y-4)=5y+1

Therefore

x=\frac{5y+1}{2y-4}

Replace $y$ by $x$ :

\boxed{f^{-1}(x)=\frac{5x+1}{2x-4}}

The domain of $f^{-1}$ is the range of $f$ .

For

f(x)=\frac{4x+1}{2x-5},\qquad x>4

the endpoint approached when $x\to 4^+$ is

f(4)=\frac{17}{3}

As $x\to\infty$ ,

f(x)\to \frac{4}{2}=2

Therefore the range of $f$ is

2<f(x)<\frac{17}{3}

Hence the domain of $f^{-1}$ is

\boxed{2<x<\frac{17}{3}}