# Exponential and Logistic Graphs

Precalculus - 3.1 Notes: Exponential and Logistic Functions
33:07 — by Scott H.

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

• An exponential function is a function of the form $f \left(x\right) = {a}^{x}$ with $a > 0$ but $a \ne 1$.

For integer and rational $x$, we give definitions of ${a}^{x}$ earlier in algebra classes.
For irrational $x$ we owe you a definition, but one approach to the definition is to describe ${a}^{x}$ for irrational $x$ as the number thar ${a}^{r}$ gets closer to as rational $r$ get close to $x$. (We owe you a proof that there is a unique such number.)

Examples:

$f \left(x\right) = {2}^{x}$

$f \left(x\right) = {5}^{x}$

$f \left(x\right) = {\left(\frac{2}{5}\right)}^{x}$

$f \left(x\right) = {4}^{x} = {\left({2}^{2}\right)}^{x} = {2}^{2 x}$

The last example illustrates why we also consider $f \left(x\right) = {a}^{k x}$ for constant $k \ne 0$ to be exponential functions

We can write $f \left(x\right) = {a}^{k x} = {\left({a}^{k}\right)}^{x}$ and .for $a > 0$ and $k \ne 0$ this $f \left(x\right) = {b}^{x}$ for the "right kind" of $b$. ($b > 0$, and $b \ne 1$)

• Hi there. A logistic graph is like an exponential with an upper limit, so it has two horizontal asymptotes, usually y=0 and y=B, as in the "spread of infection" graph here:

The curve is the solution to the diff eqn $\frac{\mathrm{dy}}{\mathrm{dt}} = r y \left(1 - \frac{y}{B}\right)$ with initial point $\left(t , y\right) = \left(0 , {y}_{0}\right) ,$which can be solved by separation of variables and partial fractions! (Think of the starting point at the lower left.) The solution curve is given by

$y = \frac{B {y}_{0}}{{y}_{0} + \left(B - {y}_{0}\right) {e}^{- r t}}$

If you have the graph, you can read off the $\left(0 , {y}_{0}\right)$, the upper limit $B$, and the inflection point $\left({t}_{\text{.inflect}} , \frac{B}{2}\right)$. (Infection point?)

The next part is to solve for $r$ using the inflection point: that's where the second derivative is 0, so take the derivative of $\frac{\mathrm{dy}}{\mathrm{dt}}$
or the second derivative of the equation for y, and solve!

That part I'll leave for you. You're welcome, from \dansmath/ ;-}

[[Added comment from dansmath: I think the prevailing notation is
y=B/(1 + (B/y_0 - 1)e^(-rt) which is the same equation, just divide top and bottom by ${y}_{0}$.]]

Its carrying capacity for the different elements of the system.

#### Explanation:

A logistic function is a way of measuring the state, at any given time, of a complex co-habitational model.
As the parameters of one denizen or component are modified, the parameters of all the others are also modified to accommodate the change. The limits are reached when the system cannot further accommodate the changes and either the component perishes, or the system grinds to a halt.

Very much like an S or an integral sign.

#### Explanation:

The asymptote is what is known as the Carrying Capacity, that is the limit of a given participant to the viability of the system.

A logistic function is a form of sigmoid function typically found in modelling population growth (see below).

#### Explanation:

Here is the graph of a typical logistic function:

The graph starts at some base population and grows almost exponentially until it begins to approach the population limit imposed by its environment.

Note that logistic models are also used in a variety of other areas (e.g. neural network analysis, etc.) but the growth model application is probably the easiest to visualize.

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