Question

How to fine the rang for f(x) and the rang for the inverse functon

Answer 1

$\begin{gathered} \text{range of f(x)= (-}\infty,0)\cup(0,\infty) \\ \text{range of f(x)=x}<0\text{ or x}>0 \\ \text{rangeof f}^{-1}(x)=(-\infty,1)\cup(1,\infty) \\ \text{rangeof f}^{-1}(x)=x<1\text{ or x}>1 \end{gathered}$

Explanation

Step 1

find the inverse of the function

a) let

$y=\frac{4}{x^3-1}$

now, swap the variables and isolate

$\begin{gathered} y=\frac{4}{x^3-1}\leftrightarrow x=\frac{4}{y^3-1} \\ \end{gathered}$$\begin{gathered} x=\frac{4}{y^3-1} \\ x(y^3-1)=4 \\ y^3-1=\frac{4}{x} \\ y^3=\frac{4}{x}+1 \\ y=\sqrt[3]{\frac{4}{x}+1} \\ \end{gathered}$

find the DOMAIN of the inverse, this would be the range for f(x)

so

$\begin{gathered} y=\sqrt[3]{\frac{4}{x}+1} \\ \text{the function is defined in all ream numbers, except when x= 0} \\ so \\ \text{domain of inverse: (-}\infty,0)\cup(0,\infty) \end{gathered}$

therefore, the range of f(x)

$\begin{gathered} \text{range of f(x)= (-}\infty,0)\cup(0,\infty) \\ \text{range of f(x)=x}<0\text{ or x}>0 \end{gathered}$

Step 2

Now, as the range of a functino is the domain of its inverse,

the range of the inverse will be the domain of f(x), hence

a) find the domain of f(x)

$y=\frac{4}{x^3-1}$

the domains is all real numbers except the number/s that make the denominator equals zero,s o

$\begin{gathered} x^3-1=0 \\ x^3=1 \\ x=1 \end{gathered}$

therefore, the range of the inverse function is

$\begin{gathered} \text{domain f(x)=rangeof f}^{-1}(x) \\ \text{rangeof f}^{-1}(x)=(-\infty,1)\cup(1,\infty) \\ \text{rangeof f}^{-1}(x)=x<1\text{ or x}>1 \end{gathered}$

I hope this helps you