Question

radicals
table for a square root function
fill in the table using this function rule.
$f(x)=\sqrt{x - 3}$
simplify your answers as much as possible.
click
ot a real number\ if applicable.

$x$	$f(x)$
$3$	$\square$
$39$	$\square$
$52$	$\square$

Explanation:

Step1: For \( x = -6 \)

Substitute \( x = -6 \) into \( f(x)=\sqrt{x - 3} \). We get \( f(-6)=\sqrt{-6 - 3}=\sqrt{-9} \). Since the square root of a negative number is not a real number, \( f(-6) \) is not a real number.

Step2: For \( x = 3 \)

Substitute \( x = 3 \) into \( f(x)=\sqrt{x - 3} \). We get \( f(3)=\sqrt{3 - 3}=\sqrt{0}=0 \).

Step3: For \( x = 39 \)

Substitute \( x = 39 \) into \( f(x)=\sqrt{x - 3} \). We get \( f(39)=\sqrt{39 - 3}=\sqrt{36}=6 \).

Step4: For \( x = 52 \)

Substitute \( x = 52 \) into \( f(x)=\sqrt{x - 3} \). We get \( f(52)=\sqrt{52 - 3}=\sqrt{49}=7 \).

Answer:

For \( x = -6 \): Not a real number
For \( x = 3 \): \( 0 \)
For \( x = 39 \): \( 6 \)
For \( x = 52 \): \( 7 \)