Exponential Functions: Free Response

The colony starts at $40$ bacteria, so the starting value is $a = 40$. Doubling every hour means each additional hour multiplies the population by $2$, so the growth factor is $b = 2$. The model is

$f(h) = 40\cdot 2^{h}.$

Evaluate the model at $h=6$:

$f(6) = 40\cdot 2^{6} = 40\cdot 64 = 2560.$

The colony has grown to $2560$ bacteria after $6$ hours.

For any exponential model $a\cdot b^{h}$, setting $h=0$ gives

$f(0) = a\cdot b^{0} = a\cdot 1 = a,$

since $b^{0}=1$ for every allowed base $b$. Here $a=40$, so $f(0)=40$ regardless of what the base happens to be, which is exactly the population before any hour of doubling has passed. This is why the multiplier out front of any exponential rule is always its value at $h=0$.

Write $b^{p}$ as $p$ factors of $b$ multiplied together, and $b^{q}$ as $q$ factors of $b$ multiplied together:

$b^{p} = \underbrace{b\cdots b}_{p}, \qquad b^{q} = \underbrace{b\cdots b}_{q}.$

Multiplying $b^{p}$ by $b^{q}$ simply lines up the two runs of factors one after another, giving $p+q$ copies of $b$ in a row:

$b^{p}\cdot b^{q} = \underbrace{b\cdots b}_{p}\cdot\underbrace{b\cdots b}_{q} = \underbrace{b\cdots b}_{p+q} = b^{p+q}.$

This argument used no particular value of $p$ or $q$, only that each is some positive whole number of factors, so it holds for every such pair.

Split the exponent using the identity, $7 = 4+3$:

$5^{7} = 5^{4}\cdot 5^{3} = 625\cdot 125 = 78125.$

Only one multiplication of two known numbers was needed, not seven copies of $5$.

Set $y=1$ in the identity:

$b^{x+1} = b^{x}\cdot b^{1} = b^{x}\cdot b.$

So moving from any input $x$ to the next input $x+1$ multiplies the output by the fixed number $b$, no matter what $x$ was. That is exactly the constant-ratio test: dividing the output at $x+1$ by the output at $x$ always gives $b$, since

$\frac{b^{x+1}}{b^{x}} = b.$

The table test for exponential data is not a separate rule to memorize; it is the identity itself, read one step at a time.

Subtract each output from the next:

$8-2=6,\qquad 32-8=24,\qquad 128-32=96.$

A linear function would give the SAME difference every step, and these three are not equal, so the data cannot be linear.

Divide each output by the one before it:

$\frac{8}{2}=4,\qquad \frac{32}{8}=4,\qquad \frac{128}{32}=4.$

The ratio is the constant $4$, so the data is exponential with base $b=4$. The output at $x=0$ is $2$, which is the multiplier $a$, giving

$f(x) = 2\cdot 4^{x}.$

Check: $f(3) = 2\cdot 4^{3} = 2\cdot 64 = 128$, matching the table.

The ratio test from part B only certifies that a SINGLE step multiplies the output by $b$; it says nothing about a gap of $7$ steps. By the identity, moving from input $3$ to input $10$ adds $7$ to the exponent, so

$f(10) = f(3+7) = f(3)\cdot b^{7}.$

Dividing both sides by $f(3)$,

$\frac{f(10)}{f(3)} = b^{7}.$

With $b=4$, that is $4^{7} = 16384$, not $4$. A gap of $7$ steps multiplies the output by the base raised to the $7$th power, not by the base itself.

Add $3$ to each value of $g(x)=2^{x}$:

$f(0)=2^{0}+3=1+3=4,\qquad f(1)=2^{1}+3=2+3=5,\qquad f(2)=2^{2}+3=4+3=7.$

Each output of $f$ is exactly $3$ more than the matching output of $g$, since the rule only adds a constant $3$ to whatever $g$ already produced.

A vertical shift moves every output up by the same amount; it never restricts which inputs are allowed, so the domain of $f$ is the same as the domain of $2^{x}$: all real numbers.

The range of $g(x)=2^{x}$ is $y>0$, since an exponential output is never $0$ or negative. Adding $3$ to every one of those outputs raises the whole range by $3$:

$y>0 \;\Longrightarrow\; y+3>3,$

so the range of $f$ is $y>3$.

The horizontal asymptote of $g(x)=2^{x}$ is $y=0$: as $x$ becomes very negative, $2^{x}$ shrinks toward $0$, always staying positive, without ever reaching it.

Every output of $f$ is that same value of $g$ with $3$ added on, since $f(x)=2^{x}+3$. So as $x$ becomes very negative,

$f(x) = 2^{x}+3 \;\longrightarrow\; 0+3=3,$

with $f(x)$ always staying strictly above $3$, exactly as $2^{x}$ always stayed strictly above $0$. The curve creeps toward the horizontal line $y=3$ without ever touching it, so the horizontal asymptote of $f$ is $y=3$.

Continue multiplying $f(1)=6$ by $3$ each step:

$f(2)=18,\quad f(3)=54,\quad f(4)=162.$

Continue adding $20$ to $g(1)=22$ each step:

$g(2)=42,\quad g(3)=62,\quad g(4)=82.$

Compare the matching pairs:

$x=1: 6<22,\quad x=2: 18<42,\quad x=3: 54<62,\quad x=4: 162>82.$

So $g$ is ahead through $x=3$, and $f$ takes over at $x=4$. Once $f$ overtakes, it keeps multiplying by $3$ every step while $g$ only ever adds $20$, so $f$ stays ahead for every larger input too.

One comparison, at one input, says nothing about every OTHER input; the table in parts A and B is proof enough on its own, since $g$ stays ahead not just at $x=1$ but all the way through $x=3$ before $f$ ever takes the lead:

$f(3) = 54 < 62 = g(3), \qquad f(4) = 162 > 82 = g(4).$

A single early check cannot rule out a later crossover.

The correct statement carries a qualifier the quick check leaves out: for a base greater than $1$, an exponential function EVENTUALLY exceeds and stays above any linear function, for all sufficiently large inputs, but it may sit BELOW the linear function over an initial stretch, exactly as $f$ does here through $x=3$. Eventually is not decoration: dropping it turns a true statement into the false one the single check at $x=1$ seemed to support.