Answer:
• (a)f(-2)=2.45.
,
• (b) f(-1) = 2.
,
• (c)f(0) = 1.41
,
• (d)f(1) = 0.
,
• (e)f(2)=1.41i. or DNE
Explanation:
Given the function:
[tex]f(x)=\sqrt{-2x+2}[/tex]
To calculate the given values, replace x with the given value in each case.
(a)f(-2)
[tex]\begin{gathered} f(-2)=\sqrt{-2(-2)+2}=\sqrt{4+2}=\sqrt{6} \\ \implies f(-2)\approx2.45 \end{gathered}[/tex]
The value of f(-2) is approximately 2.45.
(b)f(-1)
[tex]\begin{gathered} f(-1)=\sqrt{-2(-1)+2}=\sqrt{2+2}=\sqrt{4}=2 \\ \implies f(-1)=2 \end{gathered}[/tex]
The value of f(-1) is 2.
(c)f(0)
[tex]\begin{gathered} f(0)=\sqrt{-2(0)+2}=\sqrt{0+2}=\sqrt{2}\approx1.41 \\ \implies f(0)\approx1.41 \end{gathered}[/tex]
The value of f(0) is approximately 1.41
(d)f(1)
[tex]\begin{gathered} f(1)=\sqrt{-2(1)+2}=\sqrt{-2+2}=\sqrt{0}=0 \\ \implies f(1)=0 \end{gathered}[/tex]
The value of f(1) is 0.
(e)f(2)
[tex]\begin{gathered} f(2)=\sqrt{-2(2)+2}=\sqrt{-4+2}=\sqrt{-2}=1.41i \\ \implies f(2)=1.41i \end{gathered}[/tex]
The value of f(2) is 1.41i.
Since this value is not real, we write DNE for f(2).
