Question

find the domain of the rational function.
$f(z) = \frac{z + 7}{z(z - 1)}$
all real values of z such that $z \
eq 0$ and $z \
eq 1$
all real values of z such that $z \
eq 0$
all real values
all real values of z such that $z \
eq 7$
all real values of z such that $z \
eq 1$

Explanation:

Step1: Recall domain of rational functions

For a rational function \( f(z)=\frac{N(z)}{D(z)} \), the domain excludes values where \( D(z) = 0 \) (since division by zero is undefined).

Step2: Find zeros of denominator

The denominator here is \( D(z)=z(z - 1) \). Set \( D(z)=0 \):
\( z(z - 1)=0 \)
By zero - product property, if \( ab = 0 \), then \( a = 0 \) or \( b = 0 \). So \( z=0 \) or \( z - 1=0\Rightarrow z = 1 \).

Step3: Determine the domain

The domain of \( f(z)=\frac{z + 7}{z(z - 1)} \) is all real numbers except \( z = 0 \) and \( z=1 \) (because at these values the denominator is zero, and the function is undefined).

Answer:

All real values of \( z \) such that \( z
eq0 \) and \( z
eq1 \)