Question

34. Graph several members of the family of functions

\(f\left( x \right) = \frac{1}{{1 + a{e^{bx}}}}\)

where \(a > 0\). How does the graph change when \(b\) changes? How does it change when \(a\) changes?

Short answer

The several graphs are shows below:

It is observed that as \(a\) decreases, for constant \(b\), the graph of function \(f\left( x \right)\) moves in the right direction and vice versa.

It is also observed that as \(b\) changes values from negative to positive, the curves for different values of a is stretched horizontally and vice versa.

Step-by-step solution

Analyze the given function

The given function is \(f\left( x \right) = \frac{1}{{1 + a{e^{bx}}}}\). At \(x = 0\), the function is defined as \(f\left( 0 \right) = \frac{1}{{1 + a}}\). So, the \(y\) intercept of this function is \(\frac{1}{{1 + a}}\).

As the value of \(a\) increases, the value of \(y\) the intercept of this function will decrease.

Thus, it will move in a downward direction when \(a\) increases and vice-versa, for the constant value of \(b\).

As the value of \(b\) increases, keeping \(a\) as constant, then the value of this function will also decrease as it increases the denominator value.

Thus, it will move in a downward direction when \(b\) increases and vice-versa, for the constant value of \(a\).

Find the function \(f\left( x \right)\) for different values of \(a,b\)

Let the values of \(a\) be 0.1, 1, and 5, for \(b = - 1\).

Substitute these values for \(a,b\) into \(f\left( x \right)\) and simplify to get \(f\left( x \right)\) as \(f\left( x \right) = \frac{1}{{1 + 0.1{e^{ - x}}}}\), \(f\left( x \right) = \frac{1}{{1 + {e^{ - x}}}}\) and \(f\left( x \right) = \frac{1}{{1 + 5{e^{ - x}}}}\) respectively.

Let the values of \(b\) be -0.5, -0.1 and -2, for \(a = 1\).

Substitute these values for \(a,b\) into \(f\left( x \right)\) and simplify to get \(f\left( x \right)\) as \(f\left( x \right) = \frac{1}{{1 + {e^{ - 0.5x}}}}\), \(f\left( x \right) = \frac{1}{{1 + {e^{ - 0.1x}}}}\) and \(f\left( x \right) = \frac{1}{{1 + {e^{ - 2x}}}}\) respectively.

Let the values of \(b\) be 2, 0.5 and 0.1, for \(a = 1\).

Substitute these values for \(a,b\) into \(f\left( x \right)\) and simplify to get \(f\left( x \right)\) as \(f\left( x \right) = \frac{1}{{1 + {e^{2x}}}}\), \(f\left( x \right) = \frac{1}{{1 + {e^{0.5x}}}}\) and \(f\left( x \right) = \frac{1}{{1 + {e^{0.1x}}}}\) respectively.

Draw the graph

Plot the functions \(f\left( x \right) = \frac{1}{{1 + 0.1{e^{ - x}}}}\), \(f\left( x \right) = \frac{1}{{1 + {e^{ - x}}}}\) and \(f\left( x \right) = \frac{1}{{1 + 5{e^{ - x}}}}\) on the same window, by using the graphing calculator as shown below:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(\frac{1}{{1 + 0.1{e^{ - x}}}}\)in the\({Y_1}\)tab and\(\frac{1}{{1 + {e^{ - x}}}}\)in the\({Y_2}\)tab,\(\frac{1}{{1 + 5{e^{ - x}}}}\)in the\({Y_3}\)tab.
2. Enter the “GRAPH” button in the graphing calculator.

Visualization of the graph of functions \(f\left( x \right) = \frac{1}{{1 + 0.1{e^{ - x}}}}\), \(f\left( x \right) = \frac{1}{{1 + {e^{ - x}}}}\) and \(f\left( x \right) = \frac{1}{{1 + 5{e^{ - x}}}}\) is shown below:

Plot the functions \(f\left( x \right) = \frac{1}{{1 + {e^{ - 0.5x}}}}\), \(f\left( x \right) = \frac{1}{{1 + {e^{ - 0.1x}}}}\) and \(f\left( x \right) = \frac{1}{{1 + {e^{ - 2x}}}}\) on the same window, by using the graphing calculator as shown below:

• Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(\frac{1}{{1 + {e^{ - 0.5x}}}}\)in the\({Y_1}\)tab and\(\frac{1}{{1 + {e^{ - 0.1x}}}}\)in the\({Y_2}\)tab,\(\frac{1}{{1 + {e^{ - 2x}}}}\)in the\({Y_3}\)tab.
• Enter the “GRAPH” button in the graphing calculator.

Visualization of graph of functions \(f\left( x \right) = \frac{1}{{1 + {e^{ - 0.5x}}}}\), \(f\left( x \right) = \frac{1}{{1 + {e^{ - 0.1x}}}}\) and \(f\left( x \right) = \frac{1}{{1 + {e^{ - 2x}}}}\) is shown below:

Plot the functions \(f\left( x \right) = \frac{1}{{1 + {e^{2x}}}}\), \(f\left( x \right) = \frac{1}{{1 + {e^{0.5x}}}}\) and \(f\left( x \right) = \frac{1}{{1 + {e^{0.1x}}}}\) on the same window, by using the graphing calculator as shown below:

• Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(\frac{1}{{1 + {e^{2x}}}}\)in the\({Y_1}\)tab and\(\frac{1}{{1 + {e^{0.5x}}}}\)in the\({Y_2}\)tab,\(\frac{1}{{1 + {e^{0.1x}}}}\)in the\({Y_3}\)tab.
• Enter the “GRAPH” button in the graphing calculator.

Visualization of the graph of functions \(f\left( x \right) = \frac{1}{{1 + {e^{2x}}}}\), \(f\left( x \right) = \frac{1}{{1 + {e^{0.5x}}}}\) and \(f\left( x \right) = \frac{1}{{1 + {e^{0.1x}}}}\) is shown below:

Draw the conclusion

On observing the graph carefully, it is observed that as \(a\) decreases, for constant \(b\), the graph of function \(f\left( x \right)\) moves in the right direction and vice versa.

It is also observed that as \(b\) changes from −1 to 0, the curves for different values of a is stretched horizontally and vice versa.

Moreover, if \(b\) is 0, the curve is exactly horizontal.