A function g(x) increases by a factor of 7 over every unit interval in x and g(0) = 1. Which could be a function rule for g(x)? Choices: (A)g(x) = 7^x (B)g(x) = 1.07^x (C)g(x) = 1 + 1/7^x (D)g(x) = 1 + 7x

Check g(0): Check g(0) for each function to see if it equals 1. For (A) g(0) = 7^0 = 1. For (B) g(0) = 1.07^0 = 1. For (C) g(0) = 1 + 1/7^0 = 1 + 1/1 = 2. For (D) g(0) = 1 + 7*0 = 1. Options (A), (B), and (D) pass this check, but (C) does not.Check increase by factor: Check the increase by a factor of 7 over one unit interval for the remaining options. For (A) g(1) = 7^1 = 7, which is 7 times g(0). For (B) g(1) = 1.07^1 = 1.07, which is not 7 times g(0). For (D) g(1) = 1 + 7*1 = 8, which is not 7 times g(0). Only option (A) shows the correct increase by a factor of 7.Final function rule: Since option (A) satisfies both conditions, g(x) = 7^x is the correct function rule.

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A function $g(x)$ increases by a factor of $7$ over every unit interval in $x$ and $g(0) = 1$ . $\newline$ Which could be a function rule for $g(x)$ ? $\newline$ Choices: $\newline$ (A) $g(x) = 7^x$ $\newline$ (B) $g(x) = 1.07^x$ $\newline$ (C) $g(x) = 1 + \frac{1}{7^x}$ $\newline$ (D) $g(x)$ = $1$ + $7x$

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Q. A function

g(x)

increases by a factor of

7

over every unit interval in

x

and

g(0) = 1

\newline

Which could be a function rule for

g(x)

\newline

Choices:

\newline

(A)

g(x) = 7^x

\newline

(B)

g(x) = 1.07^x

\newline

(C)

g(x) = 1 + \frac{1}{7^x}

\newline

(D)

g(x)

1

7x

Check $g(0)$ : Check $g(0)$ for each function to see if it equals $1$ . $\newline$ For (A) $g(0) = 7^0 = 1$ . $\newline$ For (B) $g(0) = 1.07^0 = 1$ . $\newline$ For (C) $g(0) = 1 + \frac{1}{7^0} = 1 + \frac{1}{1} = 2$ . $\newline$ For (D) $g(0) = 1 + 7\cdot0 = 1$ . $\newline$ Options (A), (B), and (D) pass this check, but (C) does not.
Check increase by factor: Check the increase by a factor of $7$ over one unit interval for the remaining options. $\newline$ For (A) $g(1) = 7^1 = 7$ , which is $7$ times $g(0)$ . $\newline$ For (B) $g(1) = 1.07^1 = 1.07$ , which is not $7$ times $g(0)$ . $\newline$ For (D) $g(1) = 1 + 7 \times 1 = 8$ , which is not $7$ times $g(0)$ . $\newline$ Only option (A) shows the correct increase by a factor of $7$ .
Final function rule: Since option (A) satisfies both conditions, $g(x) = 7^x$ is the correct function rule.