Chapter 1.3, Problem 37E

Solution

a.

To Determine: The interval in which the function is increasing or decreasing.

The function is increasing on the interval of $\left(-\infty ,\infty \right)$ .

Given information:

  $f\left(x\right)=\frac{3}{1+e^{-x}}$

Graph:

  

From the graph, it can be seen that the function $f\left(x\right)=\frac{3}{1+e^{-x}}$ is always increasing on the interval $\left(-\infty ,\infty \right)$ .

b.

To Determine: The function is odd, even or neither.

The function is neither even nor odd.

Given information:

  $f\left(x\right)=\frac{3}{1+e^{-x}}$

Calculation:

A function is even if $f\left(-x\right)=f\left(x\right)$

Check if $f\left(-x\right)=f\left(x\right)$ :

  $\frac{3}{1+e^x}\ne \frac{3}{1+e^{-x}}$

Since $\frac{3}{1+e^x}\ne \frac{3}{1+e^{-x}}$ , the function is not even.

A function is odd if $f\left(-x\right)=-f\left(x\right)$

Multiply $-1$ by $\frac{3}{1+e^{-x}}$ :

  $-f\left(x\right)=-\frac{3}{1+e^{-x}}$

Since $\frac{3}{1+e^x}\ne \frac{3}{1+e^{-x}}$ , the function is not odd.

So, the function $f\left(x\right)=\frac{3}{1+e^{-x}}$ is neither even nor odd.

c.

To Determine: The extrema of the function, if any.

There are no extrema points.

Given information:

  $f\left(x\right)=\frac{3}{1+e^{-x}}$

Calculation:

There is no local or absolute extreme in the function since it increases continuously between two horizontal asymptotes.

Extremums should be defined over open intervals. Consider the interval $\left(0,1\right)$ .

Hence, $f\left(1\right)$ cannot be a local maximum since it is not within the interval.

Assume $f\left(0,9\right)$ is a local maximum.

When the function is increasing, then $f\left(0.99\right)>f\left(0.9\right)$ .

There are infinitely many local maxima to choose from:

  $0.9,0.99,0.999,0.9999,\dots$

And it goes on without ever reaching $0.\overline{9}=1$ . The same logic can be applied to local minima.

Thus, there is no extrema for the function $f\left(x\right)=\frac{3}{1+e^{-x}}$ .

d.

To Determine: The graph of the function related to a graph of one of the twelve basic functions.

The graph is related to logistic function.

Given information:

  $f\left(x\right)=\frac{3}{1+e^{-x}}$

Calculation:

  

From the above graph, it can be seen that the function $f\left(x\right)=\frac{3}{1+e^{-x}}$ is a logistic function. Because, it vertically scaled by a factor of $3.$