Guided Example #1 - Domain and Range

To determine the domain of ${\displaystyle f\left(x\right)=\frac{1}{x-3}}$ we set the denominator equal to zero and solve for ${\displaystyle x.x-3=0\to x=3}$.

To determine the domain of ${\displaystyle f\left(x\right)=\frac{1}{x-3}}$ we set the denominator equal to zero and solve for ${\displaystyle x.x-3=0\to x=3}$.

In other words, if we substitute 3 for x, the denominator would equal 0, and we know that division by 0 is undefined. In other words ${\displaystyle f\left(x\right)=\frac{1}{x-3}}$ is a function for the set of real numbers if we remove 3 from the domain. Therefore, the domain is ${\displaystyle (-\infty ,3)\cup (3,\infty )}$.

One simple way to obtain the range is to solve the equation for x and find the domain for this new equation.
Starting with ${\displaystyle y=\frac{1}{x-3}}$,

Cross multiply and solve for x
${\displaystyle y(x-3)=1\to yx-3y=1\to yx=1+3y\to x=\frac{1+3y}{y}}$

Now set the denominator equal to zero and solve for y. Here we simply get ${\displaystyle y=0}$. So if we substitute 0 for y, the denominator will be zero. Therefore, the range is all real values except 0.

To determine the domain of ${\displaystyle f\left(x\right)=\frac{1}{x-3}}$ we set the denominator equal to zero and solve for ${\displaystyle x.x-3=0\to x=3}$.

To determine the domain of ${\displaystyle f\left(x\right)=\frac{1}{x-3}}$ we set the denominator equal to zero and solve for ${\displaystyle x.x-3=0\to x=3}$.

In other words, if we substitute 3 for x, the denominator would equal 0, and we know that division by 0 is undefined. In other words ${\displaystyle f\left(x\right)=\frac{1}{x-3}}$ is a function for the set of real numbers if we remove 3 from the domain. Therefore, the domain is ${\displaystyle (-\infty ,3)\cup (3,\infty )}$.

One simple way to obtain the range is to solve the equation for x and find the domain for this new equation.
Starting with ${\displaystyle y=\frac{1}{x-3}}$,

Cross multiply and solve for x
${\displaystyle y(x-3)=1\to yx-3y=1\to yx=1+3y\to x=\frac{1+3y}{y}}$

Now set the denominator equal to zero and solve for y. Here we simply get ${\displaystyle y=0}$. So if we substitute 0 for y, the denominator will be zero. Therefore, the range is all real values except 0.