题目

Problem

Given that

f(x)=\frac{4}{3x+5}\qquad x>0

g(x)=\frac{1}{x}\qquad x>0

(a) state the range of $f$ ,

(2)

(b) find $f^{-1}$ ,

(3)

(1)

(d) Show that the equation $fg(x)=gf(x)$ has no real solutions.

(4)

题目中文翻译

已知

f(x)=\frac{4}{3x+5}\qquad x>0

g(x)=\frac{1}{x}\qquad x>0

(a) 写出 $f$ 的值域；

(b) 求 $f^{-1}$ ；

(d) 证明方程 $fg(x)=gf(x)$ 没有实数解。

解答

For

f(x)=\frac{4}{3x+5},\qquad x>0

we have $3x+5>5$ , so $f(x)$ is positive and less than $\frac45$ .

Also, as $x$ becomes very large, $f(x)$ gets closer to $0$ but never reaches $0$ .

Therefore the range is

\boxed{0<f(x)<\frac45}

Let

y=\frac{4}{3x+5}

Rearrange to make $x$ the subject:

\begin{aligned} y(3x+5)&=4 \\ 3xy+5y&=4 \\ 3xy&=4-5y \\ x&=\frac{4-5y}{3y} \end{aligned}

\boxed{f^{-1}(x)=\frac{4-5x}{3x}},\qquad 0<x<\frac45

这里反函数的定义域就是原函数 $f$ 的值域。

Here $fg(x)$ means $f(g(x))$ .

Since

g(x)=\frac1x

we get

\begin{aligned} fg(x) &=f\left(\frac1x\right) \\ &=\frac{4}{3\left(\frac1x\right)+5} \\ &=\frac{4}{\frac3x+5} \end{aligned}

\boxed{fg(x)=\frac{4}{\frac3x+5}}

Equivalently,

fg(x)=\frac{4x}{3+5x}

First find $gf(x)$ :

\begin{aligned} gf(x) &=g(f(x)) \\ &=\frac{1}{f(x)} \\ &=\frac{1}{\frac{4}{3x+5}} \\ &=\frac{3x+5}{4} \end{aligned}

If $fg(x)=gf(x)$ , then

\frac{4}{\frac3x+5}=\frac{3x+5}{4}

Since

\frac3x+5=\frac{3+5x}{x}

we have

\frac{4x}{3+5x}=\frac{3x+5}{4}

Cross-multiply:

\begin{aligned} 16x&=(3+5x)(3x+5) \\ 16x&=15x^2+34x+15 \\ 0&=15x^2+18x+15 \end{aligned}

Divide by $3$ :

5x^2+6x+5=0

Its discriminant is

b^2-4ac=6^2-4(5)(5)=36-100=-64

Since the discriminant is negative, this quadratic has no real roots.

Therefore the equation $fg(x)=gf(x)$ has no real solutions.