How does $$ f(t)=4 \cdot 8^{t} $$ change over the interval from $$ t=3 $$ to $$ t=4 $$ ? $$ f(t) $$ decreases by $$ 8 \% $$ $$ f(t) $$ decreases by a factor of 8 $$ f(t) $$ increases by 8 $$ f(t) $$ increases by a factor of 8

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1. First, compute the value of $$ f(t) $$ at $$ t = 3 $$:
   $$
   f(3) = 4 \cdot 8^3 = 4 \cdot 512 = 2048.
   $$

2. Next, compute the value of $$ f(t) $$ at $$ t = 4 $$:
   $$
   f(4) = 4 \cdot 8^4 = 4 \cdot 4096 = 16384.
   $$

3. Calculate the factor by which $$ f(t) $$ increases from $$ t = 3 $$ to $$ t = 4 $$:
   $$
   \text{Factor} = \frac{f(4)}{f(3)} = \frac{16384}{2048} = 8.
   $$

4. Therefore, $$ f(t) $$ increases by a factor of 8.

The correct answer is: $$ f(t) $$ increases by a factor of 8.
$$ f(t) $$ increases by a factor of 8.

How does change over the interval from to ?
decreases by
decreases by a factor of 8
increases by 8
increases by a factor of 8

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How does change over the interval from to ? decreases by decreases by a factor of 8 increases by 8 increases by a factor of 8