题目

Problem

The function $f$ and the function $g$ are defined by

f(x)=\frac{12}{x+1}\qquad x>0,\ x\in\mathbb{R}

g(x)=\frac{5}{2}\ln x\qquad x>0,\ x\in\mathbb{R}

(a) Find, in simplest form, the value of $fg(e^2)$ .

(2)

(b) Find $f^{-1}$ .

(3)

f^{-1}(x)=f(x)

(3)

题目中文翻译

函数 $f$ 与函数 $g$ 定义为

f(x)=\frac{12}{x+1}\qquad x>0,\ x\in\mathbb{R}

g(x)=\frac{5}{2}\ln x\qquad x>0,\ x\in\mathbb{R}

(a) 求 $fg(e^2)$ 的值，并化为最简形式。

(b) 求 $f^{-1}$ 。

f^{-1}(x)=f(x)

的所有实数解。

解答

We need to find

fg(e^2)

This means

f(g(e^2))

First find $g(e^2)$ :

g(e^2)=\frac52\ln(e^2)

Since

\ln(e^2)=2

we get

g(e^2)=\frac52\cdot 2=5

Now apply $f$ :

f(5)=\frac{12}{5+1}=2

Therefore

\boxed{fg(e^2)=2}

Let

y=f(x)

Then

y=\frac{12}{x+1}

Rearrange:

y(x+1)=12

yx+y=12

Hence

yx=12-y

and therefore

x=\frac{12-y}{y}

x=\frac{12}{y}-1

Replace $y$ by $x$ to write the inverse function:

\boxed{f^{-1}(x)=\frac{12}{x}-1}

Since $x>0$ for the original function, its range is

0<f(x)<12

So the domain of the inverse is

0<x<12

We need to solve

f^{-1}(x)=f(x)

Using part (b),

\frac{12}{x}-1=\frac{12}{x+1}

Multiply by $x(x+1)$ :

12(x+1)-x(x+1)=12x

Expand:

12x+12-x^2-x=12x

-x^2-x+12=0

Therefore

x^2+x-12=0

Factorise:

(x+4)(x-3)=0

x=-4\quad \text{or}\quad x=3

But the equation involves $f(x)$ and $f^{-1}(x)$ , so $x$ must lie in the domain of both functions. In particular, $x>0$ .

Therefore $x=-4$ is rejected.

Hence the only real solution is

\boxed{x=3}