Question

complete the table of values for $f(x) = x + 9$ and $g(x) = 2(3)^x$.

$x$	$f(x)$	$g(x)$
2
3
4

both $f(x)$ and $g(x)$ grow as $x$ gets larger and larger. which function eventually exceeds the other?
$f(x) = x + 9$ $g(x) = 2(3)^x$

Explanation:

Part 1: Completing the Table of Values

For \( f(x) = x + 9 \):

Step 1: When \( x = 1 \)

Substitute \( x = 1 \) into \( f(x) \):
\( f(1) = 1 + 9 = 10 \)

Step 2: When \( x = 2 \)

Substitute \( x = 2 \) into \( f(x) \):
\( f(2) = 2 + 9 = 11 \)

Step 3: When \( x = 3 \)

Substitute \( x = 3 \) into \( f(x) \):
\( f(3) = 3 + 9 = 12 \)

Step 4: When \( x = 4 \)

Substitute \( x = 4 \) into \( f(x) \):
\( f(4) = 4 + 9 = 13 \)

For \( g(x) = 2(3)^x \):

Step 1: When \( x = 1 \)

Substitute \( x = 1 \) into \( g(x) \):
\( g(1) = 2(3)^1 = 2 \times 3 = 6 \)

Step 2: When \( x = 2 \)

Substitute \( x = 2 \) into \( g(x) \):
\( g(2) = 2(3)^2 = 2 \times 9 = 18 \)

Step 3: When \( x = 3 \)

Substitute \( x = 3 \) into \( g(x) \):
\( g(3) = 2(3)^3 = 2 \times 27 = 54 \)

Step 4: When \( x = 4 \)

Substitute \( x = 4 \) into \( g(x) \):
\( g(4) = 2(3)^4 = 2 \times 81 = 162 \)

Completed Table:

\( x \)	\( f(x) \)	\( g(x) \)
2	11	18
3	12	54
4	13	162

Part 2: Which Function Eventually Exceeds the Other?

• \( f(x) = x + 9 \) is a linear function (degree 1), so its growth is constant (slope = 1).
• \( g(x) = 2(3)^x \) is an exponential function (base \( 3 > 1 \)), so its growth accelerates (multiplies by 3 for each increase in \( x \)).

From the table:

• At \( x = 1 \), \( f(x) > g(x) \) (10 > 6).
• At \( x = 2 \), \( g(x) > f(x) \) (18 > 11).
• For \( x \geq 2 \), \( g(x) \) grows much faster than \( f(x) \) (e.g., \( x = 4 \): 162 vs. 13).

Exponential functions eventually outpace linear functions, so \( g(x) = 2(3)^x \) will exceed \( f(x) = x + 9 \) as \( x \) becomes very large.

Final Answers:

Table Completion:

• \( f(1) = 10 \), \( g(1) = 6 \)
• \( f(2) = 11 \), \( g(2) = 18 \)
• \( f(3) = 12 \), \( g(3) = 54 \)
• \( f(4) = 13 \), \( g(4) = 162 \)

Which Function Exceeds the Other?

\( \boldsymbol{g(x) = 2(3)^x} \)

Answer:

Part 1: Completing the Table of Values

For \( f(x) = x + 9 \):

Step 1: When \( x = 1 \)

Substitute \( x = 1 \) into \( f(x) \):
\( f(1) = 1 + 9 = 10 \)

Step 2: When \( x = 2 \)

Substitute \( x = 2 \) into \( f(x) \):
\( f(2) = 2 + 9 = 11 \)

Step 3: When \( x = 3 \)

Substitute \( x = 3 \) into \( f(x) \):
\( f(3) = 3 + 9 = 12 \)

Step 4: When \( x = 4 \)

Substitute \( x = 4 \) into \( f(x) \):
\( f(4) = 4 + 9 = 13 \)

For \( g(x) = 2(3)^x \):

Step 1: When \( x = 1 \)

Substitute \( x = 1 \) into \( g(x) \):
\( g(1) = 2(3)^1 = 2 \times 3 = 6 \)

Step 2: When \( x = 2 \)

Substitute \( x = 2 \) into \( g(x) \):
\( g(2) = 2(3)^2 = 2 \times 9 = 18 \)

Step 3: When \( x = 3 \)

Substitute \( x = 3 \) into \( g(x) \):
\( g(3) = 2(3)^3 = 2 \times 27 = 54 \)

Step 4: When \( x = 4 \)

Substitute \( x = 4 \) into \( g(x) \):
\( g(4) = 2(3)^4 = 2 \times 81 = 162 \)

Completed Table:

\( x \)	\( f(x) \)	\( g(x) \)
2	11	18
3	12	54
4	13	162

Part 2: Which Function Eventually Exceeds the Other?

• \( f(x) = x + 9 \) is a linear function (degree 1), so its growth is constant (slope = 1).
• \( g(x) = 2(3)^x \) is an exponential function (base \( 3 > 1 \)), so its growth accelerates (multiplies by 3 for each increase in \( x \)).

From the table:

• At \( x = 1 \), \( f(x) > g(x) \) (10 > 6).
• At \( x = 2 \), \( g(x) > f(x) \) (18 > 11).
• For \( x \geq 2 \), \( g(x) \) grows much faster than \( f(x) \) (e.g., \( x = 4 \): 162 vs. 13).

Exponential functions eventually outpace linear functions, so \( g(x) = 2(3)^x \) will exceed \( f(x) = x + 9 \) as \( x \) becomes very large.

Final Answers:

Table Completion:

• \( f(1) = 10 \), \( g(1) = 6 \)
• \( f(2) = 11 \), \( g(2) = 18 \)
• \( f(3) = 12 \), \( g(3) = 54 \)
• \( f(4) = 13 \), \( g(4) = 162 \)

Which Function Exceeds the Other?

\( \boldsymbol{g(x) = 2(3)^x} \)