题目

Problem

The function $f$ is defined by

f(x)=\frac{5x}{x^2+7x+12}+\frac{5x}{x+4}\qquad x>0

(a) Show that

f(x)=\frac{5x}{x+3}

(3)

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

(3)

(ii) Hence, state whether $f$ is an increasing or a decreasing function, giving a reason for your answer.

(3)

题目中文翻译

函数 $f$ 定义为

f(x)=\frac{5x}{x^2+7x+12}+\frac{5x}{x+4}\qquad x>0

(a) 证明

f(x)=\frac{5x}{x+3}

(b) 求 $f^{-1}$ 。

(ii) 由此说明 $f$ 是增函数还是减函数，并给出理由。

解答

We have

x^2+7x+12=(x+3)(x+4)

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

Therefore

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

as required.

Let

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

Then

y(x+3)=5x

xy+3y=5x

Collect the $x$ terms:

3y=5x-xy

Thus

3y=x(5-y)

and hence

x=\frac{3y}{5-y}

Therefore

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

Since the domain of $f$ is $x>0$ , the range of $f$ is

0<f(x)<5

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

0<x<5

Hence

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

Using

f(x)=\frac{5x}{x+3}

we get

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

Therefore

\boxed{f'(x)=\frac{15}{(x+3)^2}}

For $x>0$ ,

(x+3)^2>0

f'(x)=\frac{15}{(x+3)^2}>0

Therefore $f$ is an increasing function.