Problem 9
Question
\(7-14\) . Graph the function, not by plotting points, but by starting from the graph of \(y=e^{x}\) in Figure \(1 .\) State the domain, range, and asymptote. $$ y=e^{-x}-1 $$
Step-by-Step Solution
Verified Answer
Domain: \((-\infty, \infty)\), Range: \((-1, \infty)\), Asymptote: \(y=-1\).
1Step 1: Identify the base function
The given function is a transformation of the base function \( y = e^x \). Recognizing the base function helps us understand how transformations will affect it.
2Step 2: Apply transformation to the base function
The function \( y = e^{-x} - 1 \) involves two transformations to \( y = e^x \):1. Reflect \( y = e^x \) over the y-axis to get \( y = e^{-x} \).2. Shift the result downward by 1 unit to get \( y = e^{-x} - 1 \).
3Step 3: Sketch the transformed graph
Start with \( y = e^x \) which is an increasing curve. Reflect it over the y-axis to get \( y = e^{-x} \), making it a decreasing curve. Move this entire curve down by 1 unit to obtain the graph of \( y = e^{-x} - 1 \).
4Step 4: Determine the domain
The domain of \( y = e^{-x} - 1 \) is the same as that of \( y = e^x \), which is all real numbers. Therefore, the domain is \( (-\infty, \infty) \).
5Step 5: Determine the range
Since \( y = e^{-x} \) ranges from (0, \( \infty \)), shifting it down by 1 unit changes the range to \( (-1, \infty) \).
6Step 6: Identify the horizontal asymptote
For the function \( y = e^{-x} - 1 \), as \( x \to \infty \), \( e^{-x} \to 0 \). Thus, \( y \to -1 \). The horizontal asymptote is \( y = -1 \).
Key Concepts
Domain and RangeTransformations of Exponential FunctionsAsymptotes
Domain and Range
Understanding the domain and range of a function is essential in graphing. The domain refers to all possible input values (or x-values) that a function can accept. For the exponential function \( y = e^{-x} - 1 \), its domain is all real numbers \((-\infty, \infty)\). This is because exponential functions are defined everywhere on the real number line:
- There are no restrictions on the x-values for \( e^{-x} \).
- No operations like division by zero or square roots of negative numbers occur.
- The whole graph moves one unit down.
- This adjustment changes the range to \((-1, \infty)\).
Transformations of Exponential Functions
Transformations change the appearance of functions without altering their basic nature. For the function \( y = e^{-x} - 1 \), two transformations are applied to the base exponential function \( y = e^x \):
- First, an exponential reflection occurs over the y-axis, transforming \( y = e^x \) into \( y = e^{-x} \). A reflection across the y-axis affects the x-values by making them negative, which turns the increasing graph of \( e^x \) into the decreasing graph of \( e^{-x} \).
- Second, the function is shifted down by 1 unit, modifying it to \( y = e^{-x} - 1 \). Vertical shifts adjust the y-values directly by adding or subtracting from them. Here, subtracting 1 lowers every point on the graph by one unit, impacting the range and aligning the asymptote at \( y = -1 \).
Asymptotes
An asymptote is a line that a graph approaches but never actually reaches. In the function \( y = e^{-x} - 1 \), the horizontal asymptote is crucial for understanding the behavior of the graph as \( x \) becomes very large. As \( x \to \infty \), the expression \( e^{-x} \to 0 \), causing the entire function \( y = e^{-x} - 1 \) to approach \( -1 \).
- The line \( y = -1 \) acts like a barrier the function draws near to but never intersects.
- This horizontal asymptote illustrates that despite the diminishing value of \( e^{-x} \), \( y \) will always be slightly greater than \(-1\).
Other exercises in this chapter
Problem 9
\(7-18\) Evaluate the expression. $$ \log 4+\log 25 $$
View solution Problem 9
Find the solution of the exponential equation, rounded to four decimal places. \(3 e^{x}=10\)
View solution Problem 9
Sketch the graph of the function by making a table of values. Use a calculator if necessary. $$ f(x)=2^{x} $$
View solution Problem 10
Express the equation in exponential form. $$ \begin{array}{ll}{\text { (a) } \log _{3} 81=4} & {\text { (b) } \log _{8} 4=\frac{2}{3}}\end{array} $$
View solution