Question

1. $f(x) = 2^x$

$x$	$f(x)$
$-1$
$0$
$1$
$2$

Explanation:

Step1: Calculate for \( x = -2 \)

Substitute \( x = -2 \) into \( f(x) = 2^x \). Using the rule \( a^{-n}=\frac{1}{a^n} \), we have \( f(-2)=2^{-2}=\frac{1}{2^2}=\frac{1}{4} \).

Step2: Calculate for \( x = -1 \)

Substitute \( x = -1 \) into \( f(x) = 2^x \). Using the rule \( a^{-n}=\frac{1}{a^n} \), we have \( f(-1)=2^{-1}=\frac{1}{2^1}=\frac{1}{2} \).

Step3: Calculate for \( x = 0 \)

Substitute \( x = 0 \) into \( f(x) = 2^x \). Using the rule \( a^0 = 1 \) (for \( a
eq0 \)), we have \( f(0)=2^0 = 1 \).

Step4: Calculate for \( x = 1 \)

Substitute \( x = 1 \) into \( f(x) = 2^x \). We have \( f(1)=2^1 = 2 \).

Step5: Calculate for \( x = 2 \)

Substitute \( x = 2 \) into \( f(x) = 2^x \). We have \( f(2)=2^2 = 4 \).

Now we can fill the table:

\( x \)	\( f(x) \)
-1	\( \frac{1}{2} \)
0	1
1	2
2	4

Answer:

The filled table is as above with \( f(-2)=\frac{1}{4} \), \( f(-1)=\frac{1}{2} \), \( f(0) = 1 \), \( f(1)=2 \), \( f(2)=4 \).