Question

Describe how the graph of f varies as c varies. Graph several members of the family to illustrate the trends that you discover. In particular, you should investigate how maximum and minimum points and inflection points move when c changes. You should also identify any transitional values of c at which the basic shape of the curve changes.

31. \(f\left( x \right) = {e^x} + c{e^{ - x}}\)

Short answer

The value 0 is the transitional value of \(c\). With \(c\) increases from \( - \infty \) to 0, the \(x - \)intercept and inflection point is \(\frac{1}{2}\left( { - \ln c} \right)\) and decrease from \(\infty \) to \( - \infty \).

Step-by-step solution

Identify the transitional values of c

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

\(\begin{array}{c}{e^x} + c{e^{ - x}} = 0\\c{e^{ - x}} = - {e^x}\\c = - {e^{2x}}\\2x = \ln \left( { - c} \right)\\x = \frac{1}{2}\ln c\end{array}\)

The derivative of the function is\(f'\left( x \right) = {e^x} - c{e^{ - x}}\).

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

\(\begin{array}{c}{e^x} - c{e^{ - x}} = 0\\c{e^{ - x}} = {e^x}\\c = {e^{2x}}\\2x = \ln c\\x = \frac{1}{2}\ln c\end{array}\)

The second derivative of the function is\(f''\left( x \right) = {e^x} + c{e^{ - x}} = f\left( x \right)\).

It is observed that 0 is the transitional value of\(c\). With\(c\)increases from\( - \infty \)to 0, the\(x - \)intercept and inflection point is\(\frac{1}{2}\left( { - \ln c} \right)\)and decrease from\(\infty \)to\( - \infty \). Moreover,\(f' > 0\), therefore, the function\(f\)is increasing.

If\(c = 0\), then\(f\left( x \right) = f'\left( x \right) = f''\left( x \right) = {e^x}\)with\(f\)is positive, increasing and concave upward. Moreover,\(f = f'' > 0\), therefore the function\(f\)is positive and concave upward.

The\(y - \)intercept is\(f\left( 0 \right) = 1 + c\), and these values increases when\(c\) from\( - \infty \)to\(\infty \).

Graph of 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\(f\left( x \right) = {e^x} + c{e^{ - x}}\)in the\({Y_1}\)tab.
2. Enter the equation\(f\left( x \right) = {e^x} - c{e^{ - x}}\)in the second tab.
3. Enter the “GRAPH” button in the graphing calculator.
4. Set the value of\(c\)as\(5,\frac{1}{5},0, - \frac{1}{5}, - 5\)and observe the graph of the function.

Visualization of the graph of the function as shown below:

It is observed that the minimum point \(\left( {\frac{1}{2}\ln c,2\sqrt c } \right)\) is parameterized by \(x = \frac{1}{2}\ln c,y = 2\sqrt c \). It is observed that after deleting the parameter, we obtain the minimum point lying on the graph of \(y = 2{e^x}\).