Chapter 4.3, Problem 42E

Solution

To graph: The given function and determine its domain and range.

Domain $=\left(-\infty ,\infty \right)$

Range: $\left(0,\infty \right)$

Given:

A function $y=e^{-2x}$ is given.

Concept used:

1. A exponential function having form $y=a{e}^{rx}$ is said to growth function if $a>0$ and $r>0$ and it is said to be decay function if $a>0$ and $r<0$ .
2. The domain for any natural base exponential function is the set of all real numbers.
3. The range for any natural base exponential function is the set of all real numbers above or below the horizontal asymptote $y=h$ . It means if $a>0$ then $f\left(x\right)>h$ and if $a<0$ , then $f\left(x\right)<h$ .

Calculation:

The comparison of function $y=e^{-2x}$ with $y=a{e}^{rx}$ gives $a=1$ and $r=-2$ . As $a>0$ and $r<0$ ; therefore, the given function is exponential decay function.

Evaluate the values of the function for different values of $x$ :

When $x=0$ , then the value of $y$ is,

  $\begin{array}{l}y=e^{-2x}\\ y=e^0\\ y=1\end{array}$

When $x=1$ , then the value of $y$ is,

  $\begin{array}{l}y=e^{-2}\\ y=0.135\end{array}$

To sketch the graph of the given function, plot the points $\left(0,1\right)$ and $\left(1,0.135\right)$ in an Euclidean plane and connect them with a smooth curve:

  

The domain of the function is the set of all real numbers. So,

Domain $=\left(-\infty ,\infty \right)$

It is clear from the graph that the horizontal asymptote is $y=0$ . So, the range is $\left(0,\infty \right)$ .

Conclusion:

The domain and range of function is $\left(-\infty ,\infty \right)$ and $\left(0,\infty \right)$ , respectively.