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

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Intermediate Value Theorem

Full solution

Q. A function

g(t)

increases by a factor of

7

over every unit interval in

t

and

g(0) = 1

.

\newline

Which could be a function rule for

g(t)

?

\newline

Choices:

\newline

(A)

g(t) = -7t

\newline

(B)

g(t) = 7^t

\newline

(C)

g(t) = 1 - 7t

\newline

(D)

g(t) = 1.07^t

Check $g(0)$ : Check $g(0)$ for each choice to see if it equals $1$ . $\newline$ (A) $g(0) = -7(0) = 0$ , not $1$ . $\newline$ (B) $g(0) = 7^0 = 1$ , this is correct. $\newline$ (C) $g(0) = 1 - 7(0) = 1$ , this is correct. $\newline$ (D) $g(0) = 1.07^0 = 1$ , this is correct.
Check function increase: Now, check if the function increases by a factor of $7$ over every unit interval in $t$ . $\newline$ (A) is already incorrect. $\newline$ (B) $g(1) = 7^1 = 7$ , which is $7$ times $g(0)$ . $\newline$ (C) $g(1) = 1 - 7(1) = -6$ , which is not $7$ times $g(0)$ . $\newline$ (D) $g(1) = 1.07^1 = 1.07$ , which is not $7$ times $g(0)$ .
Eliminate incorrect options: Eliminate the incorrect options based on the previous step. $\newline$ (A) Incorrect. $\newline$ (C) Incorrect. $\newline$ (D) Incorrect. $\newline$ Only $(B)$ remains as the possible correct answer.

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