Question

In Problems 97-100,graph each function,Based on the graph,state the domain and the range and find any intercepts.

$f\left(x\right)=\left\{\begin{array}{l}{e}^{x}ifx<0\\ e^{-x}ifx\ge 0\end{array}\right.$

Short answer

Graph of the given function

Domain of the given function $(-\infty ,\infty )$.

Range of the given function $(0,\infty )$.

The y-intercept is$(0,1)$ . The two functions have the line$y=0$ as their horizontal asymptote. As the value of$x$ increases, $e^x$becomes smaller and the graph will approach the line$y=0$ . But, it will not intersect the line. Similar is the case with the graph of $e^{-x}$also. This implies that there is no x-intercept.

Step-by-step solution

Step 1.Given information

The given function$f\left(x\right)=\left\{\begin{array}{l}{e}^{x}ifx<0\\ e^{-x}ifx\ge 0\end{array}\right.$

Step 2.Finding the x and y values for f(x)=ex

Let us begin with the graph of $f\left(x\right)$or $y=e^x$ . Since this function is defined for the value of x less than zero, select some values for $x$that are less than 0 and find the corresponding values of$y$.

Create a table using the values.

x	y=ex
-3	0
-2	0.1
-1	0.4

Step 3.Finding the x and y values for y=e-x

Similarly, for the function $y=e^{-x}$, select some values for x that are greater than or equal to 0 .

x	y=e-x
0	1
1	0.4
2	0.1

Step 4.Plot the points for the two functions. Connect each set of points using a smooth curve. 

The domain of a function is the set of all x-values for which the function is defined. We can see that $f\left(x\right)$ is defined for all real numbers. The domain of $f\left(x\right)$ is the interval $(-\infty ,\infty )$.

The range is the set of values of the function. From the graph, it is possible to say that the range of $f\left(x\right)$ is the interval $(0,\infty )$.

The y-intercept is $(0,1)$. The two functions have the line $y=0$ as their horizontal asymptote. As the value of x increases, $e^x$becomes smaller and the graph will approach the line $y=0$. But, it will not intersect the line. Similar is the case with the graph of$e^{-x}$ also. This implies that there is no x-intercept.