Question

complete the table of values for $f(x) = 6(2)^x$ and $g(x) = 7x + 9$.

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

both $f(x)$ and $g(x)$ grow as $x$ gets larger and larger. which function eventually exceeds the other?
$f(x) = 6(2)^x$
$g(x) = 7x + 9$

Explanation:

Part 1: Completing the table for \( f(x) = 6(2)^x \) and \( g(x) = 7x + 9 \)

For \( f(x) = 6(2)^x \):

Step 1: When \( x = 1 \)

Substitute \( x = 1 \) into \( f(x) \):
\( f(1) = 6(2)^1 = 6 \times 2 = 12 \)

Step 2: When \( x = 2 \)

Substitute \( x = 2 \) into \( f(x) \):
\( f(2) = 6(2)^2 = 6 \times 4 = 24 \)

Step 3: When \( x = 3 \)

Substitute \( x = 3 \) into \( f(x) \):
\( f(3) = 6(2)^3 = 6 \times 8 = 48 \)

Step 4: When \( x = 4 \)

Substitute \( x = 4 \) into \( f(x) \):
\( f(4) = 6(2)^4 = 6 \times 16 = 96 \)

For \( g(x) = 7x + 9 \):

Step 1: When \( x = 1 \)

Substitute \( x = 1 \) into \( g(x) \):
\( g(1) = 7(1) + 9 = 7 + 9 = 16 \)

Step 2: When \( x = 2 \)

Substitute \( x = 2 \) into \( g(x) \):
\( g(2) = 7(2) + 9 = 14 + 9 = 23 \)

Step 3: When \( x = 3 \)

Substitute \( x = 3 \) into \( g(x) \):
\( g(3) = 7(3) + 9 = 21 + 9 = 30 \)

Step 4: When \( x = 4 \)

Substitute \( x = 4 \) into \( g(x) \):
\( g(4) = 7(4) + 9 = 28 + 9 = 37 \)

Filled Table:

\( x \)	\( f(x) \)	\( g(x) \)
2	24	23
3	48	30
4	96	37

Part 2: Which function eventually exceeds the other?

\( f(x) = 6(2)^x \) is an exponential function (base \( 2 > 1 \)), so it grows exponentially (very rapidly as \( x \) increases).
\( g(x) = 7x + 9 \) is a linear function (degree 1), so it grows at a constant rate.

For large \( x \), exponential functions always outpace linear functions. From the table:

• At \( x = 1 \): \( g(x) > f(x) \)
• At \( x = 2 \): \( f(x) > g(x) \)
• For \( x \geq 2 \), \( f(x) \) (exponential) grows much faster than \( g(x) \) (linear).

Answer:

(Table Values):

\( x \)	\( f(x) \)	\( g(x) \)
2	24	23
3	48	30
4	96	37