Question

Hello, just want to check my answers at Part B. Thanks!

Answer 1

The given functions are:

`f(x)=(x-2)(x-1)(x-1),g(x)=\sqrt[3]{x}-2`

Notice that when x=0

`\begin{gathered} f(0)=(0-2)(0-1)(0-1) \\ f(0)=-2 \end{gathered}`

Also

`\begin{gathered} g(0)=\sqrt[3]{0}-2 \\ g(0)=-2 \end{gathered}`

This shows that f(0)=g(0)

Hence there are no breakes in the domain of h(x)