Question

compare the functions ( f(x) = 50x^2 ) and ( g(x) = 3^x ) by completing parts (a) and (b).
(a) fill in the table below. note that the table is already filled in for ( x = 4 ).
(the aleks calculator can be used to make computations easier.)

( x )	( f(x) = 50x^2 )	( g(x) = 3^x )
5
7
8
9

(b) for ( x geq 8 ), the table suggests that ( f(x) ) is select greater than ( g(x) ).

Explanation:

Step1: Calculate \( f(5) \) and \( g(5) \)

For \( f(x) = 50x^2 \), substitute \( x = 5 \): \( f(5)=50\times5^2 = 50\times25 = 1250 \).
For \( g(x) = 3^x \), substitute \( x = 5 \): \( g(5)=3^5 = 243 \).

Step2: Calculate \( f(7) \) and \( g(7) \)

For \( f(7) \): \( f(7)=50\times7^2 = 50\times49 = 2450 \).
For \( g(7) \): \( g(7)=3^7 = 2187 \).

Step3: Calculate \( f(8) \) and \( g(8) \)

For \( f(8) \): \( f(8)=50\times8^2 = 50\times64 = 3200 \).
For \( g(8) \): \( g(8)=3^8 = 6561 \). Wait, correction: \( 3^8 = 6561 \)? Wait, \( 3^5=243, 3^6=729, 3^7=2187, 3^8=6561 \). Wait, but \( f(8)=50\times64 = 3200 \), so \( g(8)=6561 \).

Step4: Calculate \( f(9) \) and \( g(9) \)

For \( f(9) \): \( f(9)=50\times9^2 = 50\times81 = 4050 \).
For \( g(9) \): \( g(9)=3^9 = 19683 \). Wait, no: \( 3^9 = 19683 \)? Wait, \( 3^8=6561 \), \( 3^9=3\times6561=19683 \).

Wait, let's recheck:

- \( x = 5 \):
\( f(5) = 50 \times 25 = 1250 \)
\( g(5) = 3^5 = 243 \)

- \( x = 7 \):
\( f(7) = 50 \times 49 = 2450 \)
\( g(7) = 3^7 = 2187 \)

- \( x = 8 \):
\( f(8) = 50 \times 64 = 3200 \)
\( g(8) = 3^8 = 6561 \)

- \( x = 9 \):
\( f(9) = 50 \times 81 = 4050 \)
\( g(9) = 3^9 = 19683 \)

For part (b), when \( x \geq 8 \), let's check \( x=8 \): \( f(8)=3200 \), \( g(8)=6561 \) → \( g(x) > f(x) \). Wait, maybe I miscalculated? Wait, \( 3^8 = 6561 \), \( 50 \times 8^2 = 50 \times 64 = 3200 \). So \( g(8) = 6561 > 3200 = f(8) \). Then \( x=9 \): \( g(9)=19683 > 4050 = f(9) \). Wait, the original problem's part (b) says "the table suggests that \( f(x) \) is [Select] greater than \( g(x) \)". Wait, maybe I made a mistake in \( g(x) \). Wait, is \( g(x) = 3^x \) or \( 9^x \)? Wait, the image shows \( g(x) = 3^x \)? Wait, no, maybe it's \( 9^x \)? Wait, the user's image: "g(x) = 3^x" or maybe "g(x) = 9^x"? Wait, let's re-express. Wait, if \( g(x) = 9^x \), then \( 9^5 = 59049 \), which is too big. Wait, the original table for \( x=4 \): \( f(4)=50\times16=800 \), \( g(4)=81 \). So \( g(4)=81=3^4 \), so \( g(x)=3^x \). So for \( x=4 \), \( g(4)=81 \), correct. Then \( x=5 \): \( 3^5=243 \), \( f(5)=1250 \). \( x=7 \): \( 3^7=2187 \), \( f(7)=2450 \) (so \( f(7)=2450 > g(7)=2187 \)). \( x=8 \): \( f(8)=3200 \), \( g(8)=6561 \) (so \( g(8)=6561 > f(8)=3200 \)). \( x=9 \): \( g(9)=19683 > f(9)=4050 \). Wait, but the problem's part (b) says "For \( x \geq 8 \), the table suggests that \( f(x) \) is [Select] greater than \( g(x) \)". Wait, maybe the original \( g(x) \) is \( 3^x \) or \( 9^x \)? Wait, no, \( 3^4=81 \), which matches. So let's redo the calculations:

- \( x = 5 \):
\( f(5) = 50 \times 5^2 = 50 \times 25 = 1250 \)
\( g(5) = 3^5 = 243 \)

- \( x = 7 \):
\( f(7) = 50 \times 7^2 = 50 \times 49 = 2450 \)
\( g(7) = 3^7 = 2187 \)

- \( x = 8 \):
\( f(8) = 50 \times 8^2 = 50 \times 64 = 3200 \)
\( g(8) = 3^8 = 6561 \)

- \( x = 9 \):
\( f(9) = 50 \times 9^2 = 50 \times 81 = 4050 \)
\( g(9) = 3^9 = 19683 \)

Now, for \( x=7 \): \( f(7)=2450 > g(7)=2187 \)
\( x=8 \): \( f(8)=3200 < g(8)=6561 \)
\( x=9 \): \( f(9)=4050 < g(9)=19683 \)

Wait, but the problem's part (b) is asking "For \( x \geq 8 \), the table suggests that \( f(x) \) is [Select] greater than \( g(x) \)". The "Select" dropdown: options are "always", "sometimes", "never"? Wait, the user's image shows a dropdown with "Select" and then "greater than g(x)". Wait, maybe the original problem has a typo, or I misread \( g(x) \). Wait, maybe \( g(x) = 9^x \)? No, \( 9^4=6561 \), which is not 81. So \( g(x)=3^x \) is correct.

Wait, let's re-express the table:

| \( x \) | \( f(x)=50x^2 \) | \( g(x)=3^x \) |
|--------|-----------------|---------------|
| 4 | 800 | 81 |
| 5 | 1250 | 243 |
| 7 | 2450 | 2187 |
| 8 | 3200 | 6561 |
| 9 | 4050 | 19683 |

Now, for \( x \geq 8 \):
- \( x=8 \): \( f(8)=3200 \), \( g(8)=6561 \) → \( g(x) > f(x) \)
- \( x=9 \): \( g(x) > f(x) \)

So the table suggests that for \( x \geq 8 \), \( f(x) \) is **not** greater than \( g(x) \); in fact, \( g(x) \) is greater. But the problem's dropdown is "Select" with "greater than g(x)". Wait, maybe I made a mistake in \( f(x) \). Wait, \( f(x)=50x^2 \): \( x=8 \), \( 50\times64=3200 \), correct. \( x=9 \), \( 50\times81=4050 \), correct. \( g(x)=3^x \): \( 3^8=6561 \), correct. So for \( x \geq 8 \), \( g(x) > f(x) \), so \( f(x) \) is **not** greater than \( g(x) \). But the problem's part (b) is phrased as "the table suggests that \( f(x) \) is [Select] greater than \( g(x) \)". Maybe the original \( g(x) \) is \( 3^x \) but the user made a typo, or maybe \( g(x) = 9^x \) is wrong. Alternatively, maybe \( f(x)=50x^2 \) and \( g(x)=3^x \), but the question is about when \( x \geq 8 \), is \( f(x) \) greater? From the table, at \( x=8 \), \( g(x) \) is greater; at \( x=9 \), \( g(x) \) is greater. So the answer for the dropdown would be "never" (if options are "always", "sometimes", "never"). But the user's image shows a dropdown with "Select" and then "greater than g(x)". Maybe the intended \( g(x) \) is \( 3^x \) but the calculation for \( g(8) \) is wrong? Wait, no, \( 3^8=6561 \) is correct.

Alternatively, maybe the original function is \( g(x) = 9^x \), but \( 9^4=6561 \), which doesn't match \( g(4)=81 \). So \( g(x)=3^x \) is correct.

So for part (a), the table values are:

- \( x=5 \): \( f(5)=1250 \), \( g(5)=243 \)
- \( x=7 \): \( f(7)=2450 \), \( g(7)=2187 \)
- \( x=8 \): \( f(8)=3200 \), \( g(8)=6561 \)
- \( x=9 \): \( f(9)=4050 \), \( g(9)=19683 \)

For part (b), when \( x \geq 8 \), \( g(x) > f(x) \), so \( f(x) \) is not greater than \( g(x) \). But the problem's dropdown is "Select" with "greater than g(x)". Maybe the intended answer is "never" or "not", but the user's image shows a dropdown. Assuming the options are "always", "sometimes", "never", then for \( x \geq 8 \), \( f(x) \) is never greater than \( g(x) \) (since at \( x=8 \) and \( x=9 \), \( g(x) \) is greater).

But let's focus on part (a) first. The table filling:

- \( x=5 \): \( f(5)=50\times25=1250 \), \( g(5)=3^5=243 \)
- \( x=7 \): \( f(7)=50\times49=2450 \), \( g(7)=3^7=2187 \)
- \( x=8 \): \( f(8)=50\times64=3200 \), \( g(8)=3^8=6561 \)
- \( x=9 \): \( f(9)=50\times81=4050 \), \( g(9)=3^9=19683 \)

Answer:

(a) The table values are:

• \( x=5 \): \( f(5)=1250 \), \( g(5)=243 \)
• \( x=7 \): \( f(7)=2450 \), \( g(7)=2187 \)
• \( x=8 \): \( f(8)=3200 \), \( g(8)=6561 \)
• \( x=9 \): \( f(9)=4050 \), \( g(9)=19683 \)

(b) For \( x \geq 8 \), \( f(x) \) is \(\boldsymbol{\text{never}}\) greater than \( g(x) \) (assuming dropdown options include "never").

(Note: If the dropdown options are different, adjust accordingly. But based on calculations, \( g(x) > f(x) \) for \( x \geq 8 \), so \( f(x) \) is not greater.)