/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 $$ y=e^{x / 2}+e^{-x / 2}, \qu... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 36

$$ y=e^{x / 2}+e^{-x / 2}, \quad y=0, \quad x=-1, \quad x=2 $$

Short Answer

Expert verified
The function \(y=e^{x / 2}+e^{-x / 2}\) evaluates to approximately 1.649 at \(x=-1\) and approximately 3.683 at \(x=2\). Therefore, the given boundary condition \(y=0\) is not in the range of this function. The function does not meet the boundary conditions at the given points.

Step by step solution

01

Understand the function

The function \(y=e^{x / 2}+e^{-x / 2}\) is composed of two exponential terms. The first term, \(e^{x / 2}\), grows as \(x\) becomes larger, while the second term, \(e^{-x / 2}\), decreases as \(x\) gets bigger. It seems like this function doesn't cross the \(x\) axis, which means \(y=0\) is not in the range of this function.
02

Evaluating the function at \(x=-1\)

We now plug \(x=-1\) into the function. Therefore, \(y=(-1)=e^{-1 / 2}+e^{1 / 2}\). After computation, we get \(y\approx1.649\).
03

Evaluating the function at \(x=2\)

Now we substitute \(x=2\) into the function. Therefore, \(y=e^{2 / 2}+e^{-2 / 2}=e+e^{-1}\). After computation, we get \(y\approx3.683\).

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Key Concepts

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

Function Evaluation
Function evaluation is a fundamental concept in mathematics that involves determining the output of a function for a specific input value. To solve such problems, you substitute the given input value, often represented as "\( x \)," into the function's equation.
In the exercise provided, the function is \( y = e^{x/2} + e^{-x/2} \). Evaluation requires substituting specific \( x \) values into this equation:
  • For \( x = -1 \), the function becomes \( y = e^{-1/2} + e^{1/2} \). By approximating this using a calculator, you get \( y \approx 1.649 \).
  • For \( x = 2 \), substituting gives \( y = e^{1} + e^{-1} \). The approximate calculation here yields \( y \approx 3.683 \).
Function evaluation involves both understanding the functional form and performing calculations to obtain numeric results. It is an essential skill in algebra and calculus, enabling the exploration of how functions behave for different input values.
Exponential Growth and Decay
Exponential functions, like the components of the given function \( y = e^{x/2} + e^{-x/2} \), exhibit patterns of growth and decay depending on the sign and magnitude of the exponent. **Exponential growth** occurs when there is a positive exponent; conversely, **exponential decay** happens when the exponent is negative.
In the exercise:
  • \( e^{x/2} \) represents exponential growth because as \( x \) increases, \( e^{x/2} \) becomes larger.
  • \( e^{-x/2} \) represents exponential decay, as increasing \( x \) makes \( e^{-x/2} \) smaller.
These functions illustrate classic exponential behavior, evident in real-world phenomena such as population growth (exponential growth) or radioactive decay (exponential decay). Recognizing the behavior of exponential functions helps in predicting and understanding complex dynamic systems, which often display these kinds of patterns.
Mathematical Range and Domain
In mathematics, the **domain** of a function is the complete set of possible input values (\( x \) values) that the function can accept, while the **range** is the set of all possible output values (\( y \) values).
  • For the function \( y = e^{x/2} + e^{-x/2} \), the domain is all real numbers \( \mathbb{R} \). This is because there are no restrictions on \( x \) for the expressions \( e^{x/2} \) and \( e^{-x/2} \) 鈥� you can substitute any real number into the exponent.
  • The range, however, is restricted. Exponential functions are always positive, meaning \( e^{x/2} \) and \( e^{-x/2} \) are never zero, hence \( y = 0 \) is not achievable.
In simple terms, while you can input any \( x \) into the equation, the output \( y \) will always be greater than zero. Understanding domain and range is crucial for solving equations and graphing functions, as it establishes the behavior of the function across all possible values.

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