Chapter 3.1, Problem 49E

Solution

To graph: The given function and to analyze it.

  

The domain is $\left(-\infty ,\infty \right)$ .

The range is $\left(0,5\right)$ .

It is a continuous function for all real numbers.

The function is always increasing.

No symmetry.

Bounded below and above.

No local extrema.

Horizontal asymptotes are $y=0$ and $y=5$ .

No vertical asymptote.

End behavior: $\underset{x\to -\infty }{\lim }f\left(x\right)=0,\underset{x\to \infty }{\lim }f\left(x\right)=5$ .

Given:

  $f\left(x\right)=\frac{5}{1+4\cdot e^{-2x}}$

Calculation:

Comparing $f\left(x\right)=\frac{5}{1+4\cdot e^{-2x}}$ with $f\left(x\right)=\frac{c}{1+a\cdot e^{-kx}}$ , it follows that

  $c=5,a=4,k=2$ .

Since $k>0$ , therefore it is a logistic growth function.

Now, the graph is

  

The given function is well defined for all real numbers, therefore the domain is $\left(-\infty ,\infty \right)$ .

From the graph, it can be seen that the $y$ -values are ranging from 0 to $5$ .

Therefore, the range is $\left(0,5\right)$ .

From the graph, it can be seen that the curve has no breaks or jumps; therefore it is a continuous function for all real numbers.

From the graph, it can be seen that the value of $y$ increases as the value of $x$ increases from $-\infty$ to $\infty$ .

So, the function is always increasing.

The curve is neither symmetric with the $x$ -axis nor with the $y$ -axis.

So the curve has no symmetry.

Bounded below and above.

There are no turning points for the given curve, so there is no local extrema.

Horizontal asymptotes are $y=0$ and $y=5$ .

No vertical asymptote.

End behavior:

From the graph, it can be seen that $y\to 0$ as $x\to -\infty$ and $y\to 5$ as $x\to \infty$ . So,

  $\begin{array}{c}\underset{x\to -\infty }{\lim }f\left(x\right)=0\\ \underset{x\to \infty }{\lim }f\left(x\right)=5\end{array}$