Question

fill in the table using this function rule.
$f(x)=\sqrt{9 - x}$
simplify your answers as much as possible.
click
ot a real number\ if applicable.

$x$	$f(x)$
$-40$	$\square$
$9$	$\square$
$18$	$\square$

Explanation:

Step1: For \( x = -72 \)

Substitute \( x = -72 \) into \( f(x)=\sqrt{9 - x} \). So \( 9 - (-72)=9 + 72 = 81 \). Then \( f(-72)=\sqrt{81}=9 \).

Step2: For \( x = -40 \)

Substitute \( x = -40 \) into \( f(x)=\sqrt{9 - x} \). So \( 9 - (-40)=9 + 40 = 49 \). Then \( f(-40)=\sqrt{49}=7 \).

Step3: For \( x = 9 \)

Substitute \( x = 9 \) into \( f(x)=\sqrt{9 - x} \). So \( 9 - 9 = 0 \). Then \( f(9)=\sqrt{0}=0 \).

Step4: For \( x = 18 \)

Substitute \( x = 18 \) into \( f(x)=\sqrt{9 - x} \). So \( 9 - 18=-9 \). Since the square root of a negative number is not a real number, \( f(18) \) is not a real number.

Answer:

For \( x = -72 \), \( f(x)=9 \); for \( x = -40 \), \( f(x)=7 \); for \( x = 9 \), \( f(x)=0 \); for \( x = 18 \), \( f(x) \) is not a real number. (Filling the table: when \( x=-72 \), \( f(x)=9 \); \( x = -40 \), \( f(x)=7 \); \( x = 9 \), \( f(x)=0 \); \( x = 18 \), \( f(x) \) is not a real number)