Question

the function ( f ) is defined by ( f(x) = 2^x ) and the function ( g ) is defined by ( g(x) = x^2 + 16 ).

1. find the values of ( f ) and ( g ) when ( x ) is 4, 5, and 6.
2. are the values of ( f(x) ) always greater than ( g(x) ) for all ( x )? explain how you know.

Explanation:

Part 1:

Step 1: Find \( f(x) \) for \( x = 4, 5, 6 \)

The function \( f(x)=2^x \). Substitute \( x = 4, 5, 6 \) into \( f(x) \):

• For \( x = 4 \): \( f(4)=2^4 = 16 \)
• For \( x = 5 \): \( f(5)=2^5 = 32 \)
• For \( x = 6 \): \( f(6)=2^6 = 64 \)

Step 2: Find \( g(x) \) for \( x = 4, 5, 6 \)

The function \( g(x)=x^2 + 16 \). Substitute \( x = 4, 5, 6 \) into \( g(x) \):

• For \( x = 4 \): \( g(4)=4^2 + 16 = 16 + 16 = 32 \)
• For \( x = 5 \): \( g(5)=5^2 + 16 = 25 + 16 = 41 \)
• For \( x = 6 \): \( g(6)=6^2 + 16 = 36 + 16 = 52 \)

To determine if \( f(x)=2^x \) is always greater than \( g(x)=x^2 + 16 \) for all \( x \), we can test values of \( x \) less than 4 (since we tested \( x = 4, 5, 6 \) in part 1).

• Let's test \( x = 3 \):
• \( f(3)=2^3 = 8 \)
• \( g(3)=3^2 + 16 = 9 + 16 = 25 \)

Here, \( f(3)=8 < g(3)=25 \).

• Let's test \( x = 0 \):
• \( f(0)=2^0 = 1 \)
• \( g(0)=0^2 + 16 = 16 \)

Here, \( f(0)=1 < g(0)=16 \).

Since we found values of \( x \) (e.g., \( x = 3 \), \( x = 0 \)) where \( f(x) < g(x) \), \( f(x) \) is not always greater than \( g(x) \) for all \( x \).

Answer:

• When \( x = 4 \): \( f(4) = 16 \), \( g(4) = 32 \)
• When \( x = 5 \): \( f(5) = 32 \), \( g(5) = 41 \)
• When \( x = 6 \): \( f(6) = 64 \), \( g(6) = 52 \)

Part 2: