Question

complete the table of values for $f(x) = 2^x + 1$ and $g(x) = x + 5$.

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

both $f(x)$ and $g(x)$ grow as $x$ gets larger and larger. which function eventually exceeds the other?
$f(x) = 2^x + 1$
$g(x) = x + 5$

Explanation:

Part 1: Completing the table for \( f(x) = 2^x + 1 \) and \( g(x) = x + 5 \)

For \( f(x) = 2^x + 1 \):

Step 1: When \( x = 1 \)

\( f(1) = 2^1 + 1 = 2 + 1 = 3 \)

Step 2: When \( x = 2 \)

\( f(2) = 2^2 + 1 = 4 + 1 = 5 \)

Step 3: When \( x = 3 \)

\( f(3) = 2^3 + 1 = 8 + 1 = 9 \)

Step 4: When \( x = 4 \)

\( f(4) = 2^4 + 1 = 16 + 1 = 17 \)

For \( g(x) = x + 5 \):

Step 1: When \( x = 1 \)

\( g(1) = 1 + 5 = 6 \)

Step 2: When \( x = 2 \)

\( g(2) = 2 + 5 = 7 \)

Step 3: When \( x = 3 \)

\( g(3) = 3 + 5 = 8 \)

Step 4: When \( x = 4 \)

\( g(4) = 4 + 5 = 9 \)

Part 2: Which function eventually exceeds the other?

\( f(x) = 2^x + 1 \) is an exponential function, and \( g(x) = x + 5 \) is a linear function. Exponential functions grow faster than linear functions as \( x \) becomes large. From the table:

• At \( x = 1 \): \( f(1) = 3 \), \( g(1) = 6 \) ( \( g \) is larger)
• At \( x = 2 \): \( f(2) = 5 \), \( g(2) = 7 \) ( \( g \) is larger)
• At \( x = 3 \): \( f(3) = 9 \), \( g(3) = 8 \) ( \( f \) is larger)
• At \( x = 4 \): \( f(4) = 17 \), \( g(4) = 9 \) ( \( f \) is larger)

As \( x \) increases further, the exponential growth of \( 2^x \) will make \( f(x) \) grow much faster than \( g(x) \). So \( f(x) = 2^x + 1 \) will eventually exceed \( g(x) = x + 5 \).

Completed Table:

\( x \)	\( f(x) = 2^x + 1 \)	\( g(x) = x + 5 \)
2	5	7
3	9	8
4	17	9

Answer:

for the exceeding function:
\( f(x) = 2^x + 1 \)