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Chapter 10: Problem 4

\(f(x)=\pi-x, \quad 0

Short Answer

Expert verified
Every number x in the domain \(0 < x < \pi\) can be a solution for this equation within its given domain.

Step by step solution

01

Identify the function

The given function is \(f(x)=\pi-x\). Knowing what function we are dealing with is the first and the most straightforward step.
02

Solve for x

To solve for x, you need to isolate x in the function equation. This will involve taking \(\pi\) to the other side of the equation which gives you \(x = \pi - f(x)\). However, this result doesn't provide us with a particular solution to solve for x.
03

Identify the domain

Given that \(0 < x < \pi\), this is the domain we are working within in this task. This, however, doesn't yield a definitive answer for x as we would typically expect in a simple algebraic equation.
04

Conclude the solution

Given that the equation doesn't result in a specific value for x, and it's recognized that there's no additional information given in the problem, we can reason that all the values of x that satisfy \(0 < x < \pi\) are solutions for this equation, within its given domain. In other words, x can be any number between 0 and \(\pi\) (exclusive).

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Key Concepts

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

Function Solutions
Understanding how to find solutions to functions is a fundamental skill in calculus. A function solution essentially involves finding the range of values that a variable, typically x, can take such that the function is defined and satisfies any given conditions.

For the function in the example, denoted by f(x) = \(\pi - x\), we are looking for all possible values of x that make the function valid within the constraints of its domain. While in algebra you might expect a single solution for x, in calculus, functions can have multiple solutions that form an interval or a set of intervals. One of the main goals when dealing with function solutions is to explore the relationship between x and f(x) within the allowed range of values determined by the function's domain.
Domain of a Function
The domain of a function is the set of all possible input values (x-values) for which the function is defined. In other words, it represents all the x-values that you can safely plug into the function without causing any undefined behavior, such as division by zero or taking the square root of a negative number.

In our exercise, the domain is explicitly given as 0 < x < \(\pi\). This interval notation means that x can be any number greater than 0 and less than \(\pi\), but cannot include these endpoints. Knowing the domain is crucial because it tells us the scope within which we can search for function solutions, ensuring that we only consider valid inputs when analyzing our function.
Isolate Variables
To isolate variables in an equation is to rearrange the equation so that one variable stands alone on one side of the equation. This process often involves basic algebraic operations such as adding, subtracting, multiplying, or dividing both sides of the equation by the same number or expression.

In the provided example, the process of isolating x begins with the function definition, f(x) = \(\pi - x\), and leads to the rearrangement of x = \(\pi\) - f(x). Isolating variables is a critical step in solving equations because it simplifies the expression, making it easier to interpret the relationship between the variables and to ultimately find the solutions.
Step by Step Problem Solving
Effective step by step problem solving in mathematics helps break down complex tasks into manageable pieces. It's an approach that involves following a logical sequence of steps to tackle each aspect of a problem, ensuring nothing is overlooked.

Following this method, we first identify the function and its structure. Next, we attempt to isolate the variable of interest, in this case x. After that, we determine the domain, which is the valid range of x-values for the function. Lastly, we interpret these findings to conclude the set of possible solutions. This structured methodical approach aids in a clear understanding and reduces errors, ultimately empowering students to handle more complex problems with confidence.

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Most popular questions from this chapter

Verify d' Alembert's solution \((32)\) to the initial value problem \((26)-(28)\) when \(f(x)\) has a continuous second derivative and \(g(x)\) has a continuous first derivative by substituting it directly into the equations. $$ \frac{\partial^{2} u}{\partial t^{2}} $$ \(=\alpha^{2} \frac{\partial^{2} u}{\partial x^{2}}\), \(-\infty<\)x\(<\infty\), \(t>0\), $$ u(x, 0)=f(x) $$ \(-\infty<\)x\(<\infty\), $$ \frac{\partial u}{\partial t}(x, 0) $$, \(=g(x)\), \(-\infty<\)x\(<\infty\) for the given functions \(f(x)\) and \(g(x)\)

\(f(x, y)=\cos x \sin y+4 \cos 2 x \sin y-3 \cos 3 x \sin 4 y\)

Vibrating Drum. A vibrating circular membrane of unit radius whose edges are held fixed in a plane and whose displacement \(u(r, t)\) depends only on the radial distance \(r\) from the center and on the time \(t\) is governed by the initial boundary value problem. $$ \frac{\partial^{2} u}{\partial t^{2}} $$ \(\alpha^{2}\left(\frac{\partial^{2} u}{\partial r^{2}}+\frac{1}{r} \frac{\partial u}{\partial r}\right)\), \(0\)<\(r\)<1,\( \quad t>0\), $$ u(1, t)=0, \quad t>0 $$ $$ u(r, t) \quad \text { remains finite as } r \rightarrow 0^{+} $$ Show that there is a family of solutions of the form $$ u_{n}(r, t)=\left[a_{n} \cos \left(k_{\pi} \alpha t\right)+b_{n} \sin \left(k_{n} \alpha t\right)\right] J_{0}\left(k_{\pi} r\right) $$ where \(J_{0}\) is the Bessel function of the first kind of order zero (see page 478\()\) and \(0\)

\(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0, \quad 0< x <\pi, \quad 0< y <1,\) \(\frac{\partial u}{\partial x}(0, y)=\frac{\partial u}{\partial x}(\pi, y)=0, \quad 0 \leq y \leq 1,\) \(u(x, 0)=\cos x-\cos 3 x, \quad 0 \leq x \leq \pi,\) \(u(x, 1)=\cos 2 x, \quad 0 \leq x \leq \pi\)

$$f(x)=e^{x}$$ $$-\pi< x<\pi$$
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