# How do you find the domain and range of f(x)=sqrt(3x-2)?

Nov 7, 2017

Visualize the graph (often easier by graphing the function). Then, determine all possible values of $x$ and $y$.

#### Explanation:

Let's list the transformations of the function before we graph it out. By graphing it out, we get a visual of the function so it's much easier to determine the domain and range.

• Horizontal compression by a factor of $\frac{1}{3}$.
• Horizontal translation requires us to isolate $x$ within the radical. We get $\frac{2}{3}$ to the right.

Now let's graph this.

The easiest way to do this is to sub in values for $x$ and solve for $y$.

You get this:

graph{y=sqrt(3x-2) [-10, 10, -5, 5]}

Now we can visually see the function's domain and range.

The domain is a set of all the possible $x$ values, while range is for $y$.

Because $x$ can only be values that is equal to or greater than $2$, the domain is: $\left\{x \in \mathbb{R} | x \ge \frac{2}{3}\right\}$

On the other hand, the range can only be values equal to or greater than $0$. Thus, the range is: $\left\{y \in \mathbb{R} | y \ge 0\right\}$

Hope this helps :)