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A function $f(t)$ increases by a factor of $10$ over every unit interval in $t$ and $f(0) = 1$ . Which could be a function rule for $f(t)$ ? $\newline$ Choices: $\newline$ (A) $f(t) = 1 + \frac{t}{10}$ $\newline$ (B) $f(t) = 1 + 10t$ $\newline$ (C) $f(t) = 10^t$ $\newline$ (D) $f(t) = 1 - 10^t$

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Intermediate Value Theorem

Full solution

Q. A function

f(t)

increases by a factor of

10

over every unit interval in

t

and

f(0) = 1

. Which could be a function rule for

f(t)

?

\newline

Choices:

\newline

(A)

f(t) = 1 + \frac{t}{10}

\newline

(B)

f(t) = 1 + 10t

\newline

(C)

f(t) = 10^t

\newline

(D)

f(t) = 1 - 10^t

Check Option (A): Check option (A) $f(t) = 1 + \frac{t}{10}$ . If $t$ increases by $1$ , $f(t)$ should increase by a factor of $10$ . So, $f(1)$ should be $10 \times f(0) = 10 \times 1 = 10$ . Calculate $f(1)$ for option (A): $f(1) = 1 + \frac{1}{10} = 1.1$ , which is not $10$ times $t$ $0$ .
Check Option (B): Check option (B) $f(t) = 1 + 10t$ . Calculate $f(1)$ for option (B): $f(1) = 1 + 10 \times 1 = 11$ , which is not $10$ times $f(0)$ .
Check Option (C): Check option (C) $f(t) = 10^t$ . Calculate $f(1)$ for option (C): $f(1) = 10^1 = 10$ , which is $10$ times $f(0)$ . This looks promising, but let's check if it holds for another value. Calculate $f(2)$ for option (C): $f(2) = 10^2 = 100$ , which is $10$ times $f(1)$ . This option seems to fit the condition.
Check Option (D): Check option (D) $f(t) = 1 - 10^t$ . Calculate $f(1)$ for option (D): $f(1) = 1 - 10^1 = -9$ , which is not $10$ times $f(0)$ and does not even increase as $t$ increases.

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