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Chapter 3: Problem 38

Graph the exponential function using transformations. State the \(y\) -intercept, two additional points, the domain, the range, and the horizontal asymptote. $$f(x)=1+e^{-x}$$

Short Answer

Expert verified
The y-intercept is (0, 2), additional points are (1, 1.368) and (-1, 3.718), domain is all real numbers, range is (1, 鈭�), and the horizontal asymptote is y = 1.

Step by step solution

01

Identify the Parent Function

The parent function for the given function is the exponential function \( f(x) = e^{-x} \). This is a basic decreasing exponential function with a horizontal asymptote at \( y = 0 \).
02

Apply Transformations

The function \( f(x) = 1 + e^{-x} \) involves a vertical shift of the parent function \( e^{-x} \) upwards by 1 unit. This means that the graph will move 1 unit upward.
03

Determine the Horizontal Asymptote

The horizontal shift affects the horizontal asymptote. The transformation shifts the asymptote from \( y = 0 \) to \( y = 1 \). Therefore, the horizontal asymptote of the function is \( y = 1 \).
04

Find the y-Intercept

To find the \( y \)-intercept, substitute \( x = 0 \) into the function: \( f(0) = 1 + e^{0} = 2 \). Therefore, the \( y \)-intercept is \( (0, 2) \).
05

Calculate Additional Points

Choose two values for \( x \) to find additional points. For \( x = 1 \), \( f(1) = 1 + e^{-1} \approx 1.368 \). For \( x = -1 \), \( f(-1) = 1 + e^{1} \approx 3.718 \). Thus, the points are approximately \( (1, 1.368) \) and \( (-1, 3.718) \).
06

Determine Domain and Range

The domain of an exponential function like \( f(x) = 1 + e^{-x} \) is all real numbers, \( (-\infty, \infty) \). The range is determined by the vertical shift, so it is \( (1, \infty) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Graph Transformations
When dealing with exponential functions like \( f(x) = 1 + e^{-x} \), understanding how graph transformations work is crucial. Transformations can alter a graph's position or shape. Let's break it down:
  • Vertical Shifts: In the function \( f(x) = 1 + e^{-x} \), the "1" indicates a vertical shift of the entire graph upward by 1 unit. This means every point on the graph of \( e^{-x} \) is pushed 1 unit higher.
  • Reflection and Dilation: The \( e^{-x} \) suggests a reflection over the y-axis, compared to \( e^x \). However, since we are focusing mainly on vertical shifts here, our attention is primarily on how the graph is lifted upwards.
By applying these transformations, you can visualize how the original exponential graph transforms into the one described by the function.
Horizontal Asymptote
An exponential function often includes a horizontal asymptote, a line that the graph approaches but never quite touches. For \( f(x) = 1 + e^{-x} \), this is no exception.
  • Original Asymptote: The parent function \( e^{-x} \) has a horizontal asymptote at \( y = 0 \). This is where the graph levels off as \( x \rightarrow \, \infty \).
  • Transformed Asymptote: Due to the vertical shift upwards by 1, the new horizontal asymptote becomes \( y = 1 \). This means the transformed graph will approach the line \( y = 1 \) infinitely, but it will never actually touch or cross it.
Understanding the horizontal asymptote helps in predicting the behavior of the function's graph at extreme values of \( x \).
Y-Intercept
The \( y \)-intercept of a function is where the graph crosses the y-axis. For any function, setting \( x = 0 \) finds this intercept easily.
  • Finding the Intercept: For \( f(x) = 1 + e^{-x} \), substitute \( x = 0 \) to get \( f(0) = 1 + e^{0} = 2 \).
  • Intercept Point: This calculation shows the \( y \)-intercept is at the point \( (0, 2) \). It is a crucial point that helps in sketching the graph as it serves as the initial position of the graph on the y-axis.
The \( y \)-intercept is a key feature for quickly identifying where the graph starts on the vertical line through \( x = 0 \).
Domain and Range
For any function, particularly exponential ones, determining the domain and range is essential.
  • Domain: The domain of the function \( f(x) = 1 + e^{-x} \) is all real numbers. This means \( x \) can be any real number, represented as \((-\infty, \infty)\).
  • Range: The range is impacted by the vertical shift from transformations. Here, since \( e^{-x} \) can't be negative and we've shifted upwards by 1, the range is \( (1, \infty) \). This reflects that the outputs of the function are always above 1 but can grow infinitely larger.
The domain and range provide a full picture of where the function can exist and how it behaves across its graph.

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Most popular questions from this chapter

Wing Shan just graduated from dental school owing \(\$ 80,000\) in student loans. The annual interest is \(6 \% .\) Her time \(t\) to pay off the loan is given by $$t=-\frac{\ln \left[1-\frac{80,000(0.06)}{n R}\right]}{n \ln \left(1+\frac{0.06}{n}\right)}$$ where \(n\) is the number of payment periods per year and \(R\) is the periodic payment. a. Use a graphing utility to graph $$t_{1}=-\frac{\ln \left[1-\frac{80,000(0.06)}{12 x}\right]}{12 \ln \left(1+\frac{0.06}{12}\right)} \text { as } Y_{1} \text { and }$$ $$t_{2}=-\frac{\ln \left[1-\frac{80,000(0.06)}{26 x}\right]}{26 \ln \left(1+\frac{0.06}{26}\right)} \text { as } Y_{2}$$ Explain the difference in the two graphs. b. Use the \([\text { TRACE }]\) key to estimate the number of years that it will take Wing Shan to pay off her student loan if she can afford a monthly payment of \(\$ 800\) c. If she can make a biweekly payment of \(\$ 400\), estimate the number of years that it will take her to pay off the loan. d. If she adds \(\$ 200\) more to her monthly or \(\$ 100\) more to her biweekly payment, estimate the number of years that it will take her to pay off the loan.

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