Growth Rates

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constant percentage

$\blacktriangleright$ Linear Functions

The defining characteristic of linear functions is that the have a constant growth rate.

$\frac {L(b)-L(a)}{b-a} = m$

This led to the formula for linear functions: $L(x) = m(x-a) + L(a)$

This tells us that no matter where you are in the domain, of $L$ , if you move the same distance inside the domain, then the value of $L$ changes by the same amount.

Exponential functions have a similar growth, but with percentages.

$\blacktriangleright$ Exponential Functions

Suppose we have a function with this property: No mater where you are in the domain, if you move the same amount in the domain, then the function value changes by the same percent. Then what does the formula look like?

Let’s call our function $E$ .
Let’s call our original position in the domain, $d$ .
Let’s call $D$ the distance moved from $d$ , inside the domain.
Let’s say the percentage change is $p \cdot 100\%$ . (Think of $p$ as a decimal number.)

Inside the domain we are moving from $D$ to $d + D$ , a change of $D$ . How does $E$ change?

In this story, the starting function value is $E(d)$ and the finishing function value is $E(d+D)$ . Therefore, the change in the function value is given by $E(d+D) - E(d)$ .

The supposed property tells us that this change is equal to percentage $p$ of the original value $E(d)$ .

$E(d+D) - E(d) = p \, E(d)$

$E(d+D) = E(d) + p \, E(d)$

$E(d+D) = E(d) (1+p)$

To get the value of $E(d+D)$ , we multiply $E(d)$ by $1+p$ .

If we travel another $D$ , then we move from $d+D$ to $d+2D$ , inside the domain.

We have travelled another distance of $D$ . Our supposed property says that we should get another $p$ percent.

This time our starting value is $E(d+D) = E(d) (1+p)$ and we would like to calculate the value of $E(d+2D)$ .

$E(d + 2D) = E(d+D)(1+p) = E(d) (1+p) \cdot (1+p)$

$E(d + 2D) = E(d) (1+p)^2$

If we travel another $D$ , then we get another $p$ percent of our original starting value, $E(d)(1+p)^2$ .

$E(d + 3D) = E(d)(1+p)^2 \cdot (1+p) = E(d) (1+p)^3$

If we travel ” $x$ ” number of these $D$ distances from $d$ , then we get

$E(d + x \cdot D) = E(d)(1+p)^x$

An exponential function.

Exponential Functions

Exponential functions are those functions that experience a constant percentage growth rate.

Their formulas look like

$exp(x) = a \cdot b^x$

where $a \ne 0$ and $b > 0$ .

$a$ is called the coefficient and $b$ is called the base.

Not Constant Growth

Exponential functions do not have a constant growth rate.

Let $G(t) = 5 \cdot \left (\frac {3}{2}\right )^t$ with domain $R$ .

$G(0) = 5$
$G(1) = \frac {15}{2}$
$G(2) = \frac {45}{4}$

Each time we moved $1$ in the domain, $G(t)$ DID NOT increase by the same amount.

$G(1) - G(0) = \answer {\frac {5}{2}}$

$G(2) - G(1) = \answer {\frac {15}{4}}$

$\frac {5}{2} \ne \frac {15}{4}$ , therefore $G(t)$ does not have a constant rate-of-change.

This example function, $G$ , has a constant percentage rate-of-change.

No matter where you are in the domain, when you move $1$ , the value of $G$ increases by the same percentage (is multiplied by the same factor).

Let $a$ be a real number. The percentage rate-of-change of $G$ over the interval $[a,a+1]$ is

$\frac {5 \left (\frac {3}{2}\right )^{a+1} - 5 \left (\frac {3}{2}\right )^a}{5 \left (\frac {3}{2}\right )^a}$

$\frac {5 \left (\frac {3}{2}\right )^a (\left (\frac {3}{2}\right ) - 1) }{5 \left (\frac {3}{2}\right )^a}$

$\left (\frac {3}{2}\right ) - 1 = \frac {1}{2}$

Every time we move $1$ in the domain, $G$ increases by a factor of $\frac {1}{2}$ . $G$ increases by $50\%$ . Therefore, we multiply by $1 + \frac {1}{2} = \frac {3}{2}$ , which is the base.

The general template for an exponential function is

$exp(t) = a \cdot b^t \, \text { where } \, a \ne 0 \, \text { and } \, b > 0$

Percentage Change

Let $H(t) = 3 \, \cdot \, 2^t$

$H$ is an exponential function, which means it has a constant percentage change.

Values of $H$ :

$H(0) = \answer {3}$
$H(1) = \answer {6}$
$H(2) = \answer {12}$

Change in $H$ :

$H(1) - H(0) = \answer {3}$
$H(2) - H(1) = \answer {6}$

Percentage change in $H$ :

$\frac {H(1) - H(0)}{H(0)} = \answer {1}$ , which is $\answer {100}\%$
$\frac {H(2) - H(1)}{H(1)} = \answer {1}$ , which is $\answer {100}\%$

Every time $t$ increases by $1$ , the value of $H$ gets multiplied by another $2$ , which means its value is doubled, which means it changes by its current value, which means it grows by $100\%$ .

Percentage Change

Let $k(x) = 5 \, \left (\frac {3}{4}\right )^x$

$k$ is an exponential function, which means it has a constant percentage change.

Values of $k$ :

$k(0) = \answer {5}$
$k(1) = \answer {\frac {15}{4}}$
$k(2) = \answer {\frac {45}{16}}$

Change in $k$ :

$k(1) - k(0) = \answer {-\frac {5}{4}}$
$k(2) - k(1) = \answer {-\frac {15}{16}}$

Percentage change in $k$ :

$\frac {k(1) - k(0)}{k(0)} = \answer {-\frac {1}{4}}$ , which is $\answer {-25}\%$
$\frac {k(2) - k(1)}{k(1)} = \answer {-\frac {1}{4}}$ , which is $\answer {-25}\%$

Every time $x$ increases by $1$ , the value of $k$ gets multiplied by another $\frac {3}{4}$ , which means its value decreases. It loses $\frac {1}{4}$ of its value. It has a negative growth rate. It grows by $-25\%$ .

A New Number: e

Consider this function

$e(x) = \left (1 + \frac {1}{x}\right )^x$

Here is a graph

This browser does not support embedded elements.

The graph of this function has a horizontal asymptote. The horizontal asymptote intersects the $y$ -axis at a number. This number is called $e$ . It is a special number and gets its own symbol, much like $\pi$ gets its own symbol.

$e \approx 2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427$

$e$ is an irrational number that has many connections to natural growth. The importance of $e$ will show up in Calculus.

Percentage Change

Let $p(x) = 3 \, e^x$

$p$ is an exponential function, which means it has a constant percentage change.

Values of $p$ :

$p(0) = 3$
$p(1) = 3 e$
$p(2) = 3 e^2$

Change in $p$ :

$p(1) - p(0) = 3 (e-1)$
$p(2) - p(1) = 3 e (e-1)$

Percentage change in $p$ :

$\frac {p(1) - p(0)}{p(0)} = e - 1$ , which is about $171.8\%$
$\frac {p(2) - p(1)}{p(1)} = e-1$ , which is about $171.8\%$

Percentage Change

Let $R(t) = a \cdot r^t$ , with $a$ and $r$ real numbers and $r>0$

$R$ is an exponential function, which means it has a constant percentage change.

Values of $R$ :

$R(0) = a$
$R(1) = a \, r$
$R(2) = a \, r^2$

Change in $R$ :

$R(1) - R(0) = \answer {a (r-1)}$
$R(2) - R(1) = \answer {a r (r-1)}$

Percentage change in $R$ :

$\frac {R(1) - R(0)}{R(0)} = \answer {r-1}$
$\frac {R(2) - R(1)}{R(1)} = \answer {r-1}$

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More Examples of Percent Change