Question

fill in the table using this function rule.
$f(x)=\sqrt{x} + 8$
simplify your answers as much as possible.
click
ot a real number\ if applicable.

$x$	$f(x)$
$-1$	$square$
$0$	$square$
$100$	$square$

Explanation:

Step1: Analyze \( x = -12 \)

The function is \( f(x)=\sqrt{x}+8 \). For the square root function \( \sqrt{x} \), the radicand \( x \) must be non - negative (i.e., \( x\geq0 \)) to have a real number output. When \( x = - 12 \), since \( -12<0 \), \( \sqrt{-12} \) is not a real number. So \( f(-12) \) is not a real number.

Step2: Analyze \( x=-1 \)

When \( x = - 1 \), again, the radicand \( x=-1<0 \). So \( \sqrt{-1} \) is not a real number, and thus \( f(-1) \) is not a real number.

Step3: Analyze \( x = 0 \)

Substitute \( x = 0 \) into the function \( f(x)=\sqrt{x}+8 \). We know that \( \sqrt{0}=0 \), so \( f(0)=\sqrt{0}+8=0 + 8=8 \).

Step4: Analyze \( x = 100 \)

Substitute \( x = 100 \) into the function \( f(x)=\sqrt{x}+8 \). Since \( \sqrt{100}=10 \), then \( f(100)=\sqrt{100}+8=10 + 8 = 18 \).

Answer:

For \( x=-12 \): Not a real number
For \( x = - 1 \): Not a real number
For \( x=0 \): \( 8 \)
For \( x = 100 \): \( 18 \)

(In table form, the \( f(x) \) values corresponding to \( x=-12,-1,0,100 \) are "Not a real number", "Not a real number", \( 8 \), \( 18 \) respectively)