What is the domain of f(x)=((2z-6))/(9z+1) ?

The domain of f(x)=((2z-6))/(9z+1) is z -1/9

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domain of f(x)=((2z-6))/(9z+1)

domain

f (x) = \frac{( 2 z - 6 )}{9 z + 1}

z < - \frac{1}{9} or z > - \frac{1}{9}

Interval Notation

(- \infty, - \frac{1}{9}) \cup (- \frac{1}{9}, \infty)

Solution steps

Find undefined (singularity) points:

z = - \frac{1}{9}

The function domain

z < - \frac{1}{9} or z > - \frac{1}{9}

u (x) = x + 1

f (x) = 6 x + 6

6 x + 3 y = - 9

3 - 2 x

f (x) = \frac{( x - 4 ) ( x - 8 )}{x - 5}

What is the domain of f(x)=((2z-6))/(9z+1) ?
The domain of f(x)=((2z-6))/(9z+1) is z<-1/9 \lor z>-1/9