f(x) = 1/(x^2-x).

To find the domain of x.

Solution:

We will find the domain or those values of x for which f(x) is real and defined.

f(x) = 1/(x^2-x) = 1/[x(x-1)]. when x = 0 the function becomes discontinuous with a jump.f(x) remains undefined.

When x =1, f(x) becomes discontinous with an unbounded jump. f(x) becomes undefined.

So, (-infinty < x <0) U (0 < x <1 ) U ( 1< x < infnity).

Before impose the domain of the function, we notice that the expression of the function is a ratio and we'll establish the domain knowing the fact that the division by 0 is not allowed.

For this reason, we'll find out first the x values for the denominator is cancelling.

x^2-x = 0

We'll factorize:

x(x-1)=0

We'll put each factor from the product 0:

x=0

x-1=0

x=1

From here we conclude that the function is not defined for x=0 or x=1.

So, the domain of definition is:

(-inf,0)U(0,1)U(1,inf).