Undefined functions

A function is undefined at any input that leads to an impossible operation.
The two you meet most often are dividing by zero and taking the square root of a negative number.

Division by zero is never allowed.
So a fraction $\frac{a}{b}$ is undefined whenever its denominator $b = 0$.
To find the forbidden inputs, set the denominator equal to zero and solve.

An even root of a negative number is also undefined (as a real number).
So in $\sqrt{2x + 3}$, the inside must be zero or positive: $2x + 3 \ge 0$.
And an equation like $\sqrt{2x + 3} = -3$ has no solution, because a square root is never negative.

Watch out when a factor cancels.
$\frac{(x + 1)(x - 3)}{x + 1}$ simplifies to $x - 3$, but it is still undefined at $x = -1$, because that made the original denominator zero.
Cancelling hides the problem but does not remove it.

The domain of a function is all the inputs that are allowed.
So you build it by starting with every number and then removing the values that cause division by zero or a negative under a square root.

For what value of $x$ is $\frac{2}{x - 3}$ undefined?
A fraction is undefined when its denominator is zero.
Setting $x - 3 = 0$ gives $x = 3$, so the expression is undefined at $x = 3$.

Which values must be excluded from the domain of $\frac{2}{x(2x - 3)}$?
The denominator is zero when $x = 0$ or when $2x - 3 = 0$.
Solving the second gives $x = \frac{3}{2}$.
So $x \ne 0$ and $x \ne \frac{3}{2}$.

Does $\sqrt{3x - 6} = -4$ have any real solution?
A square root is always zero or positive, so it can never equal $-4$.
Therefore no real value of $x$ works.