Question

let $f(x)=\frac{x}{x + 1}$. find the inverse of $f(x)$ and determine the range of $f(x)$.

• $f^{-1}(x)=\frac{1 - x}{x + 1}$, range of $f:(-infty,-1)cup(-1,infty)$
• $f^{-1}(x)=\frac{x}{1 - x}$, range of $f:(-infty,infty)$
• $f^{-1}(x)=\frac{x + 1}{x}$, range of $f:(-infty,infty)$
• $f^{-1}(x)=\frac{x + 1}{x}$, range of $f:(-infty,0)cup(0,infty)$
• $f^{-1}(x)=\frac{x}{1 - x}$, range of $f:(-infty,1)cup(1,infty)$

Explanation:

Step1: Find the inverse function

Let \( y = f(x)=\frac{x}{x + 1} \). Swap \( x \) and \( y \): \( x=\frac{y}{y + 1} \).
Multiply both sides by \( y + 1 \): \( x(y + 1)=y \).
Expand: \( xy+x=y \).
Move terms with \( y \) to one side: \( xy - y=-x \).
Factor \( y \): \( y(x - 1)=-x \).
Solve for \( y \): \( y=\frac{-x}{x - 1}=\frac{x}{1 - x} \), so \( f^{-1}(x)=\frac{x}{1 - x} \).

Step2: Determine the range of \( f(x) \)

Rewrite \( f(x)=\frac{x}{x + 1}=\frac{x + 1-1}{x + 1}=1-\frac{1}{x + 1} \).
The function \( \frac{1}{x + 1} \) has a vertical asymptote at \( x=-1 \) and never equals \( 0 \). So \( 1-\frac{1}{x + 1} \) never equals \( 1 \).
The range of \( f(x) \) is \( (-\infty,1)\cup(1,\infty) \).

Answer:

\( f^{-1}(x)=\frac{x}{1 - x} \), range of \( f:(-\infty,1)\cup(1,\infty) \) (the last option: \( f^{-1}(x)=\frac{x}{1 - x} \),range of \( f:(-\infty,1)\cup(1,\infty) \))