Question

Investigate the family of curves given by \(f\left( x \right) = x{e^{ - cx}}\), where c is a real number. Start by computing the limits as \(x \to \pm \infty \). Identify any transitional values of c where the basic shape change. What happens to the maximum and minimum points and inflection points as c changes? Illustrate by graphing several members of the family.

Short answer

It is observed that \(c = 0\) is a transitional value. With \(\left| c \right|\) increases, the point of inflection becomes closer to the origin.

Step-by-step solution

Identify transitional values of c

The family of the curve given by\(f\left( x \right) = x{e^{ - cx}}\).

When\(c < 0,\)then

\(\begin{aligned}{c}\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } x{e^{ - cx}}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{x}{{{e^{cx}}}}\end{aligned}\)

It is observed that the limit is in indeterminate form\(\frac{\infty }{\infty }\).

Apply l’Hospital rule as shown below:

\(\begin{aligned}{c}\mathop {\lim }\limits_{x \to - \infty } \frac{x}{{{e^{cx}}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{d}{{dx}}\left( x \right)}}{{\frac{d}{{dx}}\left( {{e^{cx}}} \right)}}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{1}{{c{e^{cx}}}}\\ = 0\end{aligned}\)

And\(\mathop {\lim }\limits_{x \to \infty } \frac{x}{{{e^{cx}}}} = \infty \).

When\(c > 0,\)then\(\mathop {\lim }\limits_{x \to - \infty } \frac{x}{{{e^{cx}}}} = - \infty \),

\(\begin{aligned}{c}\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } x{e^{ - cx}}\\ = \mathop {\lim }\limits_{x \to \infty } \frac{x}{{{e^{cx}}}}\end{aligned}\)

It is observed that the limit is in indeterminate form\(\frac{\infty }{\infty }\).

Apply l’Hospital rule as shown below:

\(\begin{aligned}{c}\mathop {\lim }\limits_{x \to \infty } \frac{x}{{{e^{cx}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left( x \right)}}{{\frac{d}{{dx}}\left( {{e^{cx}}} \right)}}\\ = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{c{e^{cx}}}}\\ = 0\end{aligned}\)

When\(c = 0,\)then\(f\left( x \right) = x\). Therefore,\(f\left( x \right) = \pm \infty \)respectively.

It is observed that \(c = 0\) is a transitional value. We can now eliminate the case \(c = 0\) because it is known that how the function works in the case.

What happens to the maximum and minimum point and inflection point as c changes  

Obtain the derivative of the function to find the maxima and minima as shown below:

\(\begin{aligned}{c}f'\left( x \right) = \frac{d}{{dx}}\left( {x{e^{ - x}}} \right)\\ = x\left( { - c{e^{ - cx}}} \right) + {e^{ - cx}}\\ = \left( {1 - cx} \right){e^{ - cx}}\end{aligned}\)

Take\(f'\left( x \right) = 0\)to obtain as shown below:

\(\begin{aligned}{c}\left( {1 - cx} \right){e^{ - cx}} = 0\\1 - cx = 0\\cx = 1\\x = \frac{1}{c}\end{aligned}\)

When\(c < 0\)then, it denotes a minimum value of\(f\left( {\frac{1}{c}} \right) = \frac{1}{{ce}}\)because\(f'\left( x \right)\)varies from negative to positive at\(x = \frac{1}{c}\); When\(c > 0\)then this denotes a minimum value. When\(\left| c \right|\)increases, the maximum or minimum point becomes closer to the origin.

Obtain the second derivative of the function as shown below:

\(\begin{aligned}{c}f''\left( x \right) = \frac{d}{{dx}}\left( {\left( {1 - cx} \right){e^{ - cx}}} \right)\\ = {e^{ - cx}}\left( { - c} \right) + \left( {1 - cx} \right)\left( { - c{e^{ - cx}}} \right)\\ = \left( {cx - 2} \right)c{e^{ - cx}}\end{aligned}\)

When\(f''\left( x \right) = 0\)then

\(\begin{aligned}{c}\left( {cx - 2} \right)c{e^{ - cx}} = 0\\cx - 2 = 0\\x = \frac{2}{c}\end{aligned}\)

Therefore, with \(\left| c \right|\) increases, the point of inflection becomes closer to the origin.

Illustrate by graphing several members of the family

The procedure to draw the graph of the function by using the graphing calculator is as follows:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(f\left( x \right) = x{e^{ - cx}}\)in the tab.
2. Set the values as\(c = - 2, - 1,0,1,2\). Observe the graph of the function.
3. Enter the “GRAPH” button in the graphing calculator.

Visualization of the graph of the functions as shown below: