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3 years ago

15

The function f(x) = ex is called the ---Select--- exponential function. The number e is approximately equal to . (Round your ans

wer to five decimal places.)

1 answer:

kobusy [5.1K]3 years ago

4 0

Answer:

The value of <em>e</em> is approximately 2.71828.

Step-by-step explanation:

An exponential function is of the form:

$y=a^{x}$

Here <em>a</em> is any value that is more than 0.

The natural form of an exponential function is:

$y=e^{x}$

Here <em>e</em> is known as the Euler's number.

The value of <em>e</em> is approximately 2.71828.

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See attached graph.

Step-by-step explanation:

We can graph this with relative ease on the calculator or computer (see attachment) but let's make sure we understand it theoretically and algebraically!

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First of all, this is an absolute value function, so it'll look like a v where every input equals its positive output (2, 2) (-3,3) etc. You can't get a negative number with a graph of y=|x| unless you translate it.

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Answer:

Step-by-step explanation:

Assuming that the differential equation is

$\frac{dP}{dt} = 0.04P\left(1-\frac{P}{500}\right)$ .

We need to solve it and obtain an expression for $P(t)$ in order to complete the exercise.

First of all, this is an example of the logistic equation, which has the general form

$\frac{dP}{dt} = kP\left(1-\frac{P}{K}\right)$ .

In order to make the calculation easier we are going to solve the general equation, and later substitute the values of the constants, notice that $k=0.04$ and $K=500$ and the initial condition $P(0)=100$ .

Notice that this equation is separable, then

$\frac{dP}{P(1-P/K)} = kdt$ .

Now, intagrating in both sides of the equation

$\int\frac{dP}{P(1-P/K)} = \int kdt = kt +C$ .

In order to calculate the integral in the left hand side we make a partial fraction decomposition:

$\frac{1}{P(1-P/K)} = \frac{1}{P} - \frac{1}{K-P}$ .

So,

$\int\frac{dP}{P(1-P/K)} = \ln|P| - \ln|K-P| = \ln\left| \frac{P}{K-P} \right| = -\ln\left| \frac{K-P}{P} \right|$ .

We have obtained that:

$-\ln\left| \frac{K-P}{P}\right| = kt +C$

which is equivalent to

$\ln\left| \frac{K-P}{P}\right|= -kt -C$

Taking exponentials in both hands:

$\left| \frac{K-P}{P}\right| = e^{-kt -C}$

Hence,

$\frac{K-P(t)}{P(t)} = Ae^{-kt}$ .

The next step is to substitute the given values in the statement of the problem:

$\frac{500-P(t)}{P(t)} = Ae^{-0.04t}$ .

We calculate the value of $A$ using the initial condition $P(0)=100$ , substituting $t=0$ :

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This is equivalent to $\frac{3}{8} = e^{-0.04t}$ . Taking logarithms we get $\ln\frac{3}{8} = -0.04t$ . Then,

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