A function g(x) increases by a factor of 2 over every unit interval in x and g(0) = 1. Which could be a function rule for g(x)? Choices: (A)g(x) = 0.98^x (B)g(x) = 2^x (C)g(x) = 1 - x/2 (D)g(x) = -2x

Check g(0): Check g(0) for each function to see if it equals 1. (A) g(0) = 0.98^0 = 1 (This matches g(0) = 1, but we need to check if it doubles over each unit interval.) (B) g(0) = 2^0 = 1 (This matches g(0) = 1, and we know that powers of 2 double each time the exponent increases by 1.) (C) g(0) = 1 - 0/2 = 1 (This matches g(0) = 1, but does it increase by a factor of 2 over each unit interval?) (D) g(0) = -2*0 = 0 (This does not match g(0) = 1, so it's not the right function.)Check doubling: Check if the functions double over each unit interval. (A) g(1) = 0.98^1 ≠ 2*g(0), so it doesn't double. (B) g(1) = 2^1 = 2, which is 2*g(0), so it doubles. (C) g(1) = 1 - 1/2 = 1/2, which is not 2*g(0), so it doesn't double. (D) We already know (D) is incorrect because g(0) ≠ 1.Choose correct function: Choose the function that both starts at 1 and doubles over each unit interval. The correct function is (B) g(x) = 2^x.

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A function $g(x)$ increases by a factor of $2$ over every unit interval in $x$ and $g(0) = 1$ . $\newline$ Write a function rule for $g(x)$ .

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Q. A function

g(x)

increases by a factor of

2

over every unit interval in

x

and

g(0) = 1

\newline

Write a function rule for

g(x)

Check $g(0)$ : Check $g(0)$ for each function to see if it equals $1$ . $\newline$ (A) $g(0) = 0.98^0 = 1$ (This matches $g(0) = 1$ , but we need to check if it doubles over each unit interval.) $\newline$ (B) $g(0) = 2^0 = 1$ (This matches $g(0) = 1$ , and we know that powers of $2$ double each time the exponent increases by $1$ .) $\newline$ (C) $g(0) = 1 - 0/2 = 1$ (This matches $g(0) = 1$ , but does it increase by a factor of $2$ over each unit interval?) $\newline$ (D) $g(0)$ $2$ (This does not match $g(0) = 1$ , so it's not the right function.)
Check doubling: Check if the functions double over each unit interval. $\newline$ (A) $g(1) = 0.98^1 \neq 2\cdot g(0)$ , so it doesn't double. $\newline$ (B) $g(1) = 2^1 = 2$ , which is $2\cdot g(0)$ , so it doubles. $\newline$ (C) $g(1) = 1 - \frac{1}{2} = \frac{1}{2}$ , which is not $2\cdot g(0)$ , so it doesn't double. $\newline$ (D) We already know (D) is incorrect because $g(0) \neq 1$ .
Choose correct function: Choose the function that both starts at $1$ and doubles over each unit interval. $\newline$ The correct function is (B) $g(x) = 2^x$ .