Given the function \(\frac{-1}{3(x+1)}+1\)

We need to determine the domain and range of the function.

Domain is the set of input values for which the function is real and defined.

Since it is rational function the denominator should not be 0.

\(3x+4=0\)

\(3x=-4\)

\(x=-\frac{4}{3}\)

excluding \(x=-\frac{4}{3}\), all the values of x are supported

so \(x \langle-\frac{4}{3}\ or\ x\rangle-\frac{4}{3}\)

so the domain of the function is

\(x\ \ln(\infty,-\frac{4}{3})\cup(-\frac{4}{3},\infty)\)

Range is the set of values of the dependent variable for which a function is defined.

So first we need to find the inverse of the function.

\(y=-\frac{1}{3y+4}\)

swap x by y

\(x(3y+4)=-1\)

\(3xy+4x=-1\)

\(3xy=-1-4x\)

\(y=\frac{-1-4x}{3x}\)

so the denominator should not be zero

\(3x=0\)

\(x=0\)

sp excluding \(x=0\) all the values are supported.

so the range is \((-\infty,0)\cup(0,\infty)\)