Question

complete the table of values for $f(x) = 6x + 8$ and $g(x) = 2^{x + 1}$.

$x$	$f(x)$	$g(x)$
3
4
5

both $f(x)$ and $g(x)$ grow as $x$ gets larger and larger. which function eventually exceeds the other?
$f(x) = 6x + 8$
$g(x) = 2^{x + 1}$

Explanation:

Part 1: Completing the table of values

For \( f(x) = 6x + 8 \)

Step 1: When \( x = 2 \)

Substitute \( x = 2 \) into \( f(x) \):
\( f(2) = 6(2) + 8 = 12 + 8 = 20 \)

Step 2: When \( x = 3 \)

Substitute \( x = 3 \) into \( f(x) \):
\( f(3) = 6(3) + 8 = 18 + 8 = 26 \)

Step 3: When \( x = 4 \)

Substitute \( x = 4 \) into \( f(x) \):
\( f(4) = 6(4) + 8 = 24 + 8 = 32 \)

Step 4: When \( x = 5 \)

Substitute \( x = 5 \) into \( f(x) \):
\( f(5) = 6(5) + 8 = 30 + 8 = 38 \)

For \( g(x) = 2^{x + 1} \)

Step 1: When \( x = 2 \)

Substitute \( x = 2 \) into \( g(x) \):
\( g(2) = 2^{2 + 1} = 2^3 = 8 \)

Step 2: When \( x = 3 \)

Substitute \( x = 3 \) into \( g(x) \):
\( g(3) = 2^{3 + 1} = 2^4 = 16 \)

Step 3: When \( x = 4 \)

Substitute \( x = 4 \) into \( g(x) \):
\( g(4) = 2^{4 + 1} = 2^5 = 32 \)

Step 4: When \( x = 5 \)

Substitute \( x = 5 \) into \( g(x) \):
\( g(5) = 2^{5 + 1} = 2^6 = 64 \)

Completed Table:

\( x \)	\( f(x) = 6x + 8 \)	\( g(x) = 2^{x + 1} \)
3	26	16
4	32	32
5	38	64

Part 2: Determining which function exceeds the other

• \( f(x) = 6x + 8 \) is a linear function (degree 1), so its growth is constant (slope \( 6 \)).
• \( g(x) = 2^{x + 1} \) is an exponential function (base \( 2 > 1 \)), so its growth accelerates (doubles with each increase in \( x \), since \( 2^{x+1} = 2 \cdot 2^x \)).

From the table:

• At \( x = 4 \), \( f(4) = 32 \) and \( g(4) = 32 \) (equal).
• At \( x = 5 \), \( g(5) = 64 \) exceeds \( f(5) = 38 \).

For large \( x \), exponential functions eventually outpace linear functions because their growth rate becomes infinitely larger. Thus, \( g(x) = 2^{x + 1} \) will eventually exceed \( f(x) = 6x + 8 \).

Final Answers:

Table Completion:

• \( f(2) = 20 \), \( g(2) = 8 \)
• \( f(3) = 26 \), \( g(3) = 16 \)
• \( f(4) = 32 \), \( g(4) = 32 \)
• \( f(5) = 38 \), \( g(5) = 64 \)

Which function exceeds the other?

\( \boldsymbol{g(x) = 2^{x + 1}} \)

Answer:

Part 1: Completing the table of values

For \( f(x) = 6x + 8 \)

Step 1: When \( x = 2 \)

Substitute \( x = 2 \) into \( f(x) \):
\( f(2) = 6(2) + 8 = 12 + 8 = 20 \)

Step 2: When \( x = 3 \)

Substitute \( x = 3 \) into \( f(x) \):
\( f(3) = 6(3) + 8 = 18 + 8 = 26 \)

Step 3: When \( x = 4 \)

Substitute \( x = 4 \) into \( f(x) \):
\( f(4) = 6(4) + 8 = 24 + 8 = 32 \)

Step 4: When \( x = 5 \)

Substitute \( x = 5 \) into \( f(x) \):
\( f(5) = 6(5) + 8 = 30 + 8 = 38 \)

For \( g(x) = 2^{x + 1} \)

Step 1: When \( x = 2 \)

Substitute \( x = 2 \) into \( g(x) \):
\( g(2) = 2^{2 + 1} = 2^3 = 8 \)

Step 2: When \( x = 3 \)

Substitute \( x = 3 \) into \( g(x) \):
\( g(3) = 2^{3 + 1} = 2^4 = 16 \)

Step 3: When \( x = 4 \)

Substitute \( x = 4 \) into \( g(x) \):
\( g(4) = 2^{4 + 1} = 2^5 = 32 \)

Step 4: When \( x = 5 \)

Substitute \( x = 5 \) into \( g(x) \):
\( g(5) = 2^{5 + 1} = 2^6 = 64 \)

Completed Table:

\( x \)	\( f(x) = 6x + 8 \)	\( g(x) = 2^{x + 1} \)
3	26	16
4	32	32
5	38	64

Part 2: Determining which function exceeds the other

• \( f(x) = 6x + 8 \) is a linear function (degree 1), so its growth is constant (slope \( 6 \)).
• \( g(x) = 2^{x + 1} \) is an exponential function (base \( 2 > 1 \)), so its growth accelerates (doubles with each increase in \( x \), since \( 2^{x+1} = 2 \cdot 2^x \)).

From the table:

• At \( x = 4 \), \( f(4) = 32 \) and \( g(4) = 32 \) (equal).
• At \( x = 5 \), \( g(5) = 64 \) exceeds \( f(5) = 38 \).

For large \( x \), exponential functions eventually outpace linear functions because their growth rate becomes infinitely larger. Thus, \( g(x) = 2^{x + 1} \) will eventually exceed \( f(x) = 6x + 8 \).

Final Answers:

Table Completion:

• \( f(2) = 20 \), \( g(2) = 8 \)
• \( f(3) = 26 \), \( g(3) = 16 \)
• \( f(4) = 32 \), \( g(4) = 32 \)
• \( f(5) = 38 \), \( g(5) = 64 \)

Which function exceeds the other?

\( \boldsymbol{g(x) = 2^{x + 1}} \)