Chapter 4, Problem 76RE

To determine

To describe: The behavior of the function as x increases without bound.

Explanation of Solution

Given:

The function is given as:

  $g\left(x\right)=e^x\cos x$

The function g ( x ) is multiplication of the two function, $y=e^x\text{and }y=\cos x$ .

As the function have the set of the real numbers as their domain. Hence, for any real number x :

  $\begin{array}{c}{e}^x>0\text{and }\cos x\le 1\\ e^x\left|\cos x\right|\le e^x\\ -e^x\le e^x\cos x\le e^x\end{array}$

Substituting $x=n\pi \text{in the function }g\left(x\right)$ :

  $\begin{array}{l}g\left(x\right)=e^x\cos n\pi \\ g\left(x\right)=\pm e^x\end{array}$

Substituting $x=\frac{\pi }{2}+n\pi$ into the equation 1 and solve for the g ( x ).

  $\begin{array}{l}g\left(x\right)=e^x\cos \left(\frac{\pi }{2}+n\pi \right)\\ g\left(x\right)=0\end{array}$

Therefore, the graph of the function g ( x ) touches the graph of the function $y=e^x$ or the graph of the function $y=-e^x\text{ at }x=n\pi$ and has x -intercepts at $x=\frac{\pi }{2}+n\pi$ .

The factor $e^x$ is known as the damping factor.

Now drawing the graph of the function: