Fill in the table using this function rule:
[tex]\[ f(x) = \sqrt{x} + 5 \][/tex]

Simplify your answers as much as possible. Click "Not a real number" if applicable.

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$f(x)$[/tex] \\
\hline
-4 & [tex]$\square$[/tex] \\
\hline
0 & [tex]$\square$[/tex] \\
\hline
25 & [tex]$\square$[/tex] \\
\hline
49 & [tex]$\square$[/tex] \\
\hline
\end{tabular}



Answer :

Let's fill in the table according to the function rule \( f(x) = \sqrt{x} + 5 \).

1. For \( x = -4 \):
- Since taking the square root of a negative number does not yield a real number, \( f(-4) \) is not a real number.

2. For \( x = 0 \):
- Plug \( x = 0 \) into the function:
[tex]\[ f(0) = \sqrt{0} + 5 = 0 + 5 = 5 \][/tex]
Therefore, \( f(0) = 5 \).

3. For \( x = 25 \):
- Plug \( x = 25 \) into the function:
[tex]\[ f(25) = \sqrt{25} + 5 = 5 + 5 = 10 \][/tex]
Therefore, \( f(25) = 10 \).

4. For \( x = 49 \):
- Plug \( x = 49 \) into the function:
[tex]\[ f(49) = \sqrt{49} + 5 = 7 + 5 = 12 \][/tex]
Therefore, \( f(49) = 12 \).

Now we can fill in the table:
[tex]\[ \begin{tabular}{|c|c|} \hline [tex]$x$[/tex] & [tex]$f(x)$[/tex] \\
\hline
-4 & \text{Not a real number} \\
\hline
0 & 5 \\
\hline
25 & 10 \\
\hline
49 & 12 \\
\hline
\end{tabular}
\][/tex]