Question

Please help me with these questions (lesson 2.06 and below)

Answer 1

N 2.06

we have the function

`H(x)=-2\sqrt[3]{x}+1`

For each value of x, substitute in the given function to obtain the value of H(x)

so

For x=-8

substitute

`\begin{gathered} H(x)=-2\sqrt[3]{-8}+1 \\ H(x)=-2(-2)+1 \\ H(x)=5 \end{gathered}`

For x=0

`\begin{gathered} H(x)=-2\sqrt[3]{0}+1 \\ H(x)=-2(0)+1 \\ H(x)=1 \end{gathered}`

For x=1

`\begin{gathered} H(x)=-2\sqrt[3]{1}+1 \\ H(x)=-2(1)+1 \\ H(x)=-1 \end{gathered}`