Answer:
Explanation:
Given the below function;
[tex]f(x)=\sqrt{-2x+2}[/tex]
*To find f(-2), we'll go ahead and substitute x with -2 and evaluate;
[tex]\begin{gathered} f(-2)=\sqrt{-2(-2)+2}=\sqrt{4+2}=\sqrt{6}=2.45 \\ \therefore f(-2)=2.45 \end{gathered}[/tex]
So f(-2) = 2.45
*To find f(-1), we'll go ahead and substitute x with -1 and evaluate;
[tex]\begin{gathered} f(-1)=\sqrt{-2(-1)+2}=\sqrt{2+2}=\sqrt{4}=2 \\ \therefore f(-1)=2 \end{gathered}[/tex]
So f(-1) = 2
*To find f(0), we'll go ahead and substitute x with 0 and evaluate;
[tex]\begin{gathered} f(0)=\sqrt{-2(0)+2}=\sqrt{0+2}=\sqrt{2}=1.41 \\ f(0)=1.41 \end{gathered}[/tex]
So f(0) = 1.41
*To find f(1), we'll go ahead and substitute x with 1 and evaluate;
[tex]\begin{gathered} f(1)=\sqrt{-2(1)+2}=\sqrt{-2+2}=\sqrt{-4} \\ f(1)=NDE \end{gathered}[/tex]
Since we have a negative root, the output doeswe'll use NDE
