Problem 13
Question
From \(x^{2} e^{y}=0\) we have \(x=0 .\) since \(e^{x}-1=e^{0}-1=0,\) the second equation is satisfied for an arbitrary value of \(y .\) Therefore any point of the form \((0, y)\) is a critical point.
Step-by-Step Solution
Verified Answer
The critical points are any points of the form \((0, y)\).
1Step 1: Identify the given equation
We are given the equation \(x^2 e^y = 0\). Our goal is to solve for \(x\) and find the values of \(y\) that satisfy the equation.
2Step 2: Solve for x
To satisfy the equation \(x^2 e^y = 0\), either \(x^2 = 0\) or the exponential term must be zero. Since the exponential function \(e^y\) is never zero, it follows that \(x^2\) must be zero, which gives \(x = 0\).
3Step 3: Analyze the role of the exponential term
Given that \(e^x - 1 = e^0 - 1 = 0\), the equation is satisfied for any \(y\) when \(x=0\). The condition \(e^x - 1 = 0\) simply ensures that the function is zero for any arbitrary \(y\). This highlights that the value of \(y\) does not affect the solution when \(x = 0\).
4Step 4: Conclusion about the critical points
Since the equation \(x^2 e^y = 0\) is satisfied with \(x = 0\) for any \(y\), any point of the form \((0, y)\) is a critical point. Therefore, the set of critical points is all points on the y-axis.
Key Concepts
Exponential FunctionSolving EquationsMathematical Analysis
Exponential Function
An exponential function is a mathematical expression in which a constant base, typically the number \(e\) (approximately 2.718), is raised to a variable exponent. This type of function is symbolically expressed as \(f(x) = e^x\). Exponential functions are known for their unique property of maintaining a non-zero output for any real number input.
This means that no input value, positive or negative, will result in the function equating to zero. When analyzing the equation \(x^2 e^y = 0\) given in the problem, we see that the exponential function \(e^y\) remains non-zero. Therefore, the term \(x^2\) must account for the zero product, leading to the conclusion that \(x\) must be zero.
The behavior of exponential functions is crucial in various fields such as growth models, compounding interest calculations, and decay processes. They exhibit rapid changes and are inherently bound to the principles of calculus and mathematical analysis given their constant rate of growth or decay.
This means that no input value, positive or negative, will result in the function equating to zero. When analyzing the equation \(x^2 e^y = 0\) given in the problem, we see that the exponential function \(e^y\) remains non-zero. Therefore, the term \(x^2\) must account for the zero product, leading to the conclusion that \(x\) must be zero.
The behavior of exponential functions is crucial in various fields such as growth models, compounding interest calculations, and decay processes. They exhibit rapid changes and are inherently bound to the principles of calculus and mathematical analysis given their constant rate of growth or decay.
Solving Equations
Solving equations involves finding the values of variables that satisfy the given mathematical expression. In this context, it’s about isolating the variables to meet the conditions laid out by the equation. The core principle utilized here is the zero-product property, which states that if a product of two terms equals zero, then at least one of those terms must be zero.
In dealing with the equation \(x^2 e^y = 0\), we leverage the zero-product property to determine that \(x^2\) must be zero since \(e^y\) is never zero. Consequently, \(x = 0\) is derived from solving \(x^2 = 0\). With this simplification, the solution is accepted because the exponential term cannot drive the equation to zero on its own.
This methodology of solving equations is foundational in algebra and calculus, allowing students to logically deduce solutions and understand the underlying structure of complex expressions.
In dealing with the equation \(x^2 e^y = 0\), we leverage the zero-product property to determine that \(x^2\) must be zero since \(e^y\) is never zero. Consequently, \(x = 0\) is derived from solving \(x^2 = 0\). With this simplification, the solution is accepted because the exponential term cannot drive the equation to zero on its own.
This methodology of solving equations is foundational in algebra and calculus, allowing students to logically deduce solutions and understand the underlying structure of complex expressions.
Mathematical Analysis
Mathematical analysis is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It offers a rigorous framework to examine the behavior of mathematical functions and sets.
In the exercise, critical points are found using mathematical analysis, which helps in understanding the conditions where the function or its derivatives vanish or where certain changes in their nature occur. Here, analyzing \(x^2 e^y = 0\) leads us to find critical points at \((0, y)\), indicating any point along the y-axis fulfills the equation. Such points are where the 'action happens' – where significant changes or behaviours of the function manifest.
Mathematical analysis is pivotal in exploring the nuances of function behavior and provides the tools to solve advanced mathematical problems that include continuous functions and changes in rates, which are core to calculus and real-world applications.
In the exercise, critical points are found using mathematical analysis, which helps in understanding the conditions where the function or its derivatives vanish or where certain changes in their nature occur. Here, analyzing \(x^2 e^y = 0\) leads us to find critical points at \((0, y)\), indicating any point along the y-axis fulfills the equation. Such points are where the 'action happens' – where significant changes or behaviours of the function manifest.
Mathematical analysis is pivotal in exploring the nuances of function behavior and provides the tools to solve advanced mathematical problems that include continuous functions and changes in rates, which are core to calculus and real-world applications.
Other exercises in this chapter
Problem 13
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View solution Problem 15
since \(x^{2}-y^{2}=0, y^{2}=x^{2}\) and so \(x^{2}-3 x+2=(x-1)(x-2)=0 .\) It follows that the critical points are (1,1) \((1,-1),(2,2),\) and \((2,-2) .\) We n
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