/*! 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 56 Find the domain of the function.... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the domain of the function. $$h(x)=\frac{10}{x^{2}-2 x}$$

Short Answer

Expert verified
The domain of the function \(h(x)=\frac{10}{x^{2}-2 x}\) is (-∞, 0) union (0, 2) union (2, ∞).

Step by step solution

01

Set the denominator equal to zero

For the function \(h(x)=\frac{10}{x^{2}-2 x}\), set the denominator equal to zero. We have: \(x^{2}-2 x = 0\).
02

Solve for x

Next, solve the equation for x. This can be done by factoring out x in the equation \(x^{2}-2 x = 0\) which results in \(x(x-2) = 0\). Setting each factor equal to zero gives the solutions \(x = 0\) and \(x = 2\).
03

Define the domain

The solutions \(x = 0\) and \(x = 2\) are the values which \(x\) cannot take as they would make the denominator equal to zero. Therefore, the function is defined for all real numbers except for \(x = 0\) and \(x = 2\). The domain of function \(h(x)\) can therefore be expressed in interval notation as: (-∞, 0) union (0, 2) union (2, ∞).

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Key Concepts

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

Solving Rational Equations
Rational equations are equations involving at least one rational expression. A rational expression consists of a numerator and a denominator, where the denominator cannot be zero since division by zero is undefined. In solving rational equations such as the textbook example, \(h(x)=\frac{10}{x^{2}-2 x}\), the first goal is usually to find for what values of \(x\) the denominator equals to zero, making the equation undefined.

To solve the equation, you can factor the denominator (if possible) to find its zeros. In our example, the denominator is factored to \(x(x-2)\). This leads to the values \(x = 0\) and \(x = 2\) which cannot be solutions to the equation as they would make the denominator zero. In problems where the equation must be solved for a variable, you'd typically move all terms to one side to have zero on the other and then find a common denominator to combine the rational expressions if there are more than one.
Factorization in Algebra
Factorization is a cornerstone of algebra that involves breaking down numbers or expressions into a product of simpler factors. It is an invaluable tool in solving equations, simplifying expressions, and finding zeros of functions. When you factor the quadratic equation \(x^{2}-2x = 0\) as done in the textbook problem, you make the problem easier to solve by identifying its roots.

Remember, the first step in factorization is to look for a common factor. In our example, the common factor is \(x\), which leads to the factored form \(x(x-2)\). Other methods of factorization include grouping, using special products formulas (such as perfect square trinomials), and applying the quadratic formula. Understanding factorization allows students to solve quadratic equations and find the domain of functions, which is especially useful when dealing with rational expressions.
Interval Notation
Interval notation is a method used to describe sets of numbers that fall within a certain range. It is particularly useful in mathematics to denote domains, ranges, or sets of solutions in a concise and clear manner. In interval notation, open brackets (\()\) indicate that an endpoint is not included—this includes \( -\infty \) and \( +\infty \), as they are not real numbers and cannot be included in the set. Closed brackets \[ \], on the other hand, signify that an endpoint is included in the interval.

In the context of the domain of a function, say \(h(x)=\frac{10}{x^{2}-2 x}\), interval notation enables us to express the domain as a union of intervals that do not include the points where the denominator is zero (\(x = 0\) and \(x = 2\)). As shown in the example, the domain is represented as \( (-\infty, 0) \cup (0, 2) \cup (2, +\infty) \) which succinctly conveys that the function exists for all real numbers except 0 and 2.

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Most popular questions from this chapter

Determine whether the function has an inverse function. If it does, then find the inverse function. $$f(x)=\left\\{\begin{array}{ll}-x, & x \leq 0 \\\x^{2}-3 x, & x>0\end{array}\right.$$

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