Determining Whether \({f(x)=\sqrt{x^2(x-1)}}\) Is Continuous on Its Domain

Determine if the function \(f(x)=\sqrt{x^{2}(x-1)}\) is continuous on its domain.

Solution The domain of \(f(x)=\sqrt{x^{2}(x-1)}\) is \(\{x|x=0\}\cup \{ x|x\geq 1\}\). We need to determine whether \(f\) is continuous at the number \(0\) and whether \(f\) is continuous on the interval \([ 1,\infty)\).

At the number \(0\), there is an open interval containing \(0\) that contains no other number in the domain of \(f\). [For example, use the interval \(\left( -\dfrac{1}{2},\dfrac{1}{2}\right)\!.]\) This means \(\lim\limits_{x\rightarrow 0}f(x)\) does not exist. So, \(f\) is discontinuous at \(0\).

For all numbers \(c\) in the open interval \((1,\infty )\) we have \begin{equation*} f(c)=\sqrt{c^{2}(c-1)} \end{equation*}

and \begin{equation*} \lim_{x\rightarrow c}\sqrt{x^{2}(x-1)}=\sqrt{\lim\limits_{x\rightarrow c} [x^{2}(x-1)] }=\sqrt{c^{2}(c-1)}=f(c) \end{equation*}

So, \(f\) is continuous on the open interval \((1,\infty )\).

Now, at the number \(1\), \begin{equation*} f(1)=0\hbox{ and }\lim_{x\rightarrow 1^{+}}\sqrt{x^{2}(x-1)}=0 \end{equation*}

So, \(f\) is continuous from the right at \(1\).

The function \(f(x)=\sqrt{x^{2}(x-1)}\) is continuous on the interval \([1,\infty )\), but it is discontinuous at \(0\). So, \(f\) is not continuous on its domain.