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Chapter 3: Problem 5

Use the intermediate-value theorem to show that $$ e^{-x}=x $$ has a solution in \((0,1)\).

Short Answer

Expert verified
The function changes from positive at 0 to negative at 1, so a solution exists in (0, 1) by the intermediate-value theorem.

Step by step solution

01

Define the Function

To use the intermediate-value theorem, first define the function that needs to be analyzed. Consider the function \( f(x) = e^{-x} - x \). We need to find if there exists a solution to \( e^{-x} = x \), which translates to finding a root of \( f(x) \), i.e., where \( f(x) = 0 \).
02

Evaluate the Function at Endpoints

Next, evaluate the function \( f(x) \) at the endpoints of the interval \((0, 1)\). Calculate:\[ f(0) = e^{0} - 0 = 1 \]\[ f(1) = e^{-1} - 1 = \frac{1}{e} - 1 \]Approximately, \( \frac{1}{e} \approx 0.368 \), so \( f(1) \approx 0.368 - 1 = -0.632 \). Thus, \( f(0) = 1 \) and \( f(1) \approx -0.632 \).
03

Analyze the Sign Change

Observe the sign change in \( f(x) \) over the interval \((0, 1)\):\( f(0) = 1 > 0 \) (positive) and \( f(1) \approx -0.632 < 0 \) (negative). Since \( f(x) \) changes from positive to negative, there must be a point \( c \) in \((0, 1)\) where \( f(c) = 0 \), due to the intermediate-value theorem.

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Key Concepts

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

Root Finding
Root finding is an essential concept in calculus, focusing on identifying a value, known as a root, where a given function equals zero. This task is fundamentally important because it allows us to find solutions to equations, predict system behaviors, and interpret real-world data efficiently. For the equation \( e^{-x} = x \), we are tasked with finding a root within an interval.
When approaching such problems, it's helpful to:
  • Define a new function \( f(x) \) by rearranging the original equation. For our example, \( f(x) = e^{-x} - x \).
  • The goal is then to determine where this function equals zero, that is \( f(x) = 0 \).
Finding roots can help in creating mathematical models and solving real-life problems. Understanding the conditions under which roots exist not only aids in solving the given equations but builds a solid mathematical foundation.
Functions and Calculus
The field of calculus introduces the powerful concept of functions, describing relationships between quantities. In the context of our problem, we work with the exponential function and its important property where \( e^x \) is continuously differentiable.
The function \( f(x) = e^{-x} - x \) is continuous and smooth over any real number due to both components being continuous.
  • Using calculus, we understand that the behavior of the function \( f(x) \) can be analyzed through its derivative, \( f'(x) \).
  • While the function itself is significant, its continuity and differentiability allow us to apply the Intermediate-Value Theorem, a major principle in calculus.
Functions in calculus provide the tools to analyze behavior over specific intervals, as seen in our interval \((0, 1)\). By examining changes in sign and function values, the applicability of calculus in solving equations becomes evident.
Evaluation of Intervals
Evaluating intervals relates closely to determining where functions satisfy specific conditions, such as a change in sign which implies a root. For the problem \( e^{-x} = x \), explore the interval \((0, 1)\) to identify where the equation holds true.
In applying the Intermediate-Value Theorem:
  • Compute \( f(x) \) at both ends of the interval. We get \( f(0) = 1 \) and \( f(1) \approx -0.632 \).
  • A change in sign from positive to negative within this interval indicates a root at some point \( c \).
Effectively evaluating intervals not only pinpoints where solutions exist but also validates the presence of solutions through continuous function evaluation. This practice is vital in calculus as it provides a clear and structured approach to problem-solving.

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Most popular questions from this chapter

In Problems 1-32, use a table or a graph to investigate each limit. $$ \lim _{x \rightarrow 0^{+}}\left(1-e^{-x}\right) $$

Evaluate the trigonometric limits. $$ \lim _{x \rightarrow 0} \frac{\sin x(1-\cos x)}{x^{2}} $$

Use a graphing calculator to investigate $$ \lim _{x \rightarrow 1} \sin \frac{1}{x-1} $$

In Problems \(37-54\), use the limit laws to evaluate each limit. $$ \lim _{x \rightarrow 1 / 2} \frac{1-x-2 x^{2}}{1-2 x} $$

In Problems \(37-54\), use the limit laws to evaluate each limit. $$ \lim _{x \rightarrow-5}\left(4+2 x^{2}\right) $$
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