Question

How can if fine the rang for f(x) and the rang for the inverse f(x)

Answer 1

The given function is

$f(x)=\frac{4}{x^3-1}$

The inverse function is

$f^{-1}(x)=\sqrt[3]{\frac{4}{x}+1}$

Recall that the range of the function is the completely possible set of resulting values of the function.

Consider the function

$f(x)=\frac{4}{x^3-1}$

This is given function rational polynomial function so the range is negative infinity to positive infinity except for the value of x where the denominator is zero.

$x^3-1=0$$x^3=1$$x=1$

Hence the given function is not valid at x=1.

The range of the function is

$(-\infty,\infty)-\mleft\lbrace1\mright\rbrace$

Consider the inverse function

$f^{-1}(x)=\sqrt[3]{\frac{4}{x}+1}$

This is inverse function rational polynomial function so the range is negative infinity to positive infinity except for the value of x where the denominator is zero.

$\sqrt[3]{x}=0$$x=0$

Hence the inverse function is not valid at x=0.

The range of the inverse function is

$(-\infty,\infty)-\mleft\lbrace0\mright\rbrace$