# 108 商用微積分 第二週小考參考解答

**1.** $f(x)=\sqrt{4-x}\ ,\ g(x)=\frac{1}{x-1}$

**(1)** $f(g(x))=$ $\color{red}{\sqrt{\frac{4x-5}{x-1}}}$
<sol>
$$
\begin{array}{lll}
f(g(x)) & = & f(\frac{1}{x-1}) \\
& = & \sqrt{4-\frac{1}{x-1}} \\
& = & \sqrt{\frac{4x-5}{x-1}} \\
\end{array}
$$

**(2)** $D_{f \circ g}=$ $\color{red}{(-\infty,1)\ \cup\ [\frac{5}4, \infty)}$
<sol>
$$
\begin{array}{lll}
D_{f \circ g} & = & \{x\ |\ x\in D_g\ \wedge\ g(x) \in D_f \} \\
& = & \{x\ |\ x-1\ne\ 0 \wedge\ 4-g(x)\ge 0\} \\
& = & \{x\ |\ x\ne 1\ \wedge\ 4-\frac{1}{x-1}\ge 0\} \\
& = & \{x\ |\ x\ne 1\ \wedge\ \frac{4x-5}{x-1}\ge 0\} \\
& = & \{x\ |\ x\ne 1\ \wedge\ (4x-5)(x-1)\ge 0\} \\
& = & \{x\ |\ x\ne 1\ \wedge\ (x\le 1\ \vee\ x\ge \frac{5}4)\} \\
& = & \{x\ |\ x< 1\ \wedge\ x\ge \frac{5}4\} \\
& = & (-\infty,1)\ \cup\ [\frac{5}4, \infty) \\
\end{array}
$$

---

**2.** $f(x)=5-2x,\ \ f^{-1}(x)=$ $\color{red}{\frac{5-x}2 }$
<sol>
$$
\begin{array}{lll}
& y = f(x) = 5-2x\\
\Longrightarrow & 2x = 5-y\\
\Longrightarrow & x = \frac{5-y}2 \\
\stackrel{x 與 y 對調}{\Longrightarrow} & y = \frac{5-x}2 \\
\Longrightarrow & f^{-1}(x) = \frac{5-x}2 \\
\end{array}
$$

###### tags: `Calculus`