Chapter 4.3, Problem 50E

Solution

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

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

Range: $y>-3$

Given:

A function $h\left(x\right)=y=e^{-2\left(x+1\right)}-3$ 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 is a set of all real numbers for any natural base functions.
3. The range is the set of 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 $y=e^{-2\left(x+1\right)}$ with $y=a{e}^{rx}$ gives $a=1$ and $r=-2$ so the function is exponential decay function because $a>0$ and $r<0$ .

Evaluate the function at different values of $x$ .

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

  $\begin{array}{l}y=e^{-2\left(0+1\right)}-3\\ y=e^{-2}-3\\ y=-2.86\end{array}$

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

  $\begin{array}{l}y=e^{-2\left(x+1\right)}-3\\ y=e^{-2\left(1+1\right)}-3\\ y=e^{-4}-3\\ y=-2.981\end{array}$

To sketch the graph, plot the points $\left(0,-2.86\right)$ and $\left(1,-2.981\right)$ and draw the curve.

  

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

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

It is clear from the graph that the horizontal asymptote is $y=-3$ . So, the range is,

  $y>-3$

Conclusion:

The domain and range of function is $\left(-\infty ,\infty \right)$ and $y>-3$ respectively.