题目

Problem

The functions $f$ and $g$ are defined by

f(x)=5-\frac{4}{3x+2}\qquad x\ge 0

g(x)=4\sin\left(\frac{x}{3}+\frac{\pi}{6}\right)\qquad x\in\mathbb{R}

(a) Find the range of $f$

(2)

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

(ii) Write down the domain of $f^{-1}$

(3)

(2)

题目中文翻译

函数 $f$ 和 $g$ 定义为

f(x)=5-\frac{4}{3x+2}\qquad x\ge 0

g(x)=4\sin\left(\frac{x}{3}+\frac{\pi}{6}\right)\qquad x\in\mathbb{R}

(a) 求 $f$ 的值域。

(b) (i) 求 $f^{-1}(x)$ 。

(ii) 写出 $f^{-1}$ 的定义域。

解答

We have

f(x)=5-\frac{4}{3x+2} \qquad x\ge0

When $x=0$ ,

f(0)=5-\frac42=3

As $x$ increases,

\frac{4}{3x+2}

decreases towards $0$ , so $f(x)$ increases towards $5$ .

Therefore the range is

\boxed{3\le f(x)<5}

Let

y=5-\frac{4}{3x+2}

Then

\frac{4}{3x+2}=5-y

3x+2=\frac{4}{5-y}

Thus

3x=\frac{4}{5-y}-2

and hence

x=\frac13\left(\frac{4}{5-y}-2\right)

Therefore

\boxed{f^{-1}(x)=\frac13\left(\frac{4}{5-x}-2\right)}

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

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

\boxed{3\le x<5}

First find $g(-\pi)$ :

g(-\pi)=4\sin\left(-\frac{\pi}{3}+\frac{\pi}{6}\right)

g(-\pi)=4\sin\left(-\frac{\pi}{6}\right)

Hence

g(-\pi)=4\left(-\frac12\right)=-2

Therefore

fg(-\pi)=f(-2)

Now

f(-2)=5-\frac{4}{3(-2)+2}

f(-2)=5-\frac{4}{-4}=6

Thus

\boxed{fg(-\pi)=6}