Question

What is the domain of the function$f\left(x\right)=\sqrt{x^3-3x^2}?$

Short answer

The domain of the function is a set of all real numbers greater than or equal to$3.$

Step-by-step solution

Step 1. Given Information 

The given function is$f\left(x\right)=\sqrt{x^3-3x^2}$

We have to find the domain.

Step 2. Finding the domain

As the given function is a square root function.

To find the domain first, find the real zeros of f(x).

So, $$\begin{gathered} \sqrt{x^3-3x^2}=0 \\ {x}^3-3x^2=0 \\ {x}^2\left(x-3\right)=0 \\ {x}^2=0andx-3=0 \\ x=0x=3 \end{gathered}$$

Step 3. Determining intervals

Now, by using the real zeros the intervals we get are

$\left(-\infty ,0\right),\left(0,3\right),\left(3,\infty \right).$

Step 4. Check the points

Let $x=-2$and substitute in the given function

$$\begin{gathered} f(-2)=\sqrt{{\left(-2\right)}^3-3{\left(-2\right)}^2} \\ =\sqrt{-8-12} \\ =\sqrt{-20} \end{gathered}$$

So, it is not true.

Let $x=2$and substitute in the given function

$$\begin{gathered} f\left(2\right)=\sqrt{{\left(2\right)}^3-3{\left(2\right)}^2} \\ =\sqrt{8-12} \\ =\sqrt{-4} \end{gathered}$$

So, it is not true.

Let $x=4$and substitute in the given function

$$\begin{gathered} f\left(4\right)=\sqrt{{\left(4\right)}^3-3{\left(4\right)}^2} \\ =\sqrt{64-48} \\ =\sqrt{16} \\ =4 \end{gathered}$$

So, it is true.

Therefore, the domain of the function is all real numbers greater than or equal to$3.$