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Chapter 7: Problem 56

Write each phrase as an expression. See Sections 1.3 and 7.2 The reciprocal of \(x+1\)

Short Answer

Expert verified
The reciprocal of \(x+1\) is \(\frac{1}{x+1}\).

Step by step solution

01

Understand the Term 'Reciprocal'

The term 'reciprocal' of a number or expression refers to the multiplicative inverse. This means if you have a number or expression, the reciprocal is what you multiply it by to get 1.
02

Write the Expression for Reciprocal

Since we are given the expression \(x + 1\), its reciprocal would be calculated by taking the inverse of this expression. Thus, the reciprocal is \(\frac{1}{x+1}\). When \(x + 1\) is multiplied by \(\frac{1}{x+1}\), the result is 1.

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Key Concepts

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

Reciprocal
The concept of a reciprocal is an important term in algebraic expressions. Put simply, the reciprocal of a number or a mathematical expression is the quantity that, when multiplied by the original number or expression, results in the value of 1. For example:
  • The reciprocal of 3 is \( \frac{1}{3} \) because \( 3 \times \frac{1}{3} = 1 \).
  • Similarly, the reciprocal of \( \frac{5}{7} \) is \( \frac{7}{5} \), since \( \frac{5}{7} \times \frac{7}{5} = 1 \).
In working with algebraic expressions like \( x + 1 \), the reciprocal would be \( \frac{1}{x+1} \). This is because multiplying \( x+1 \) by \( \frac{1}{x+1} \) gives you 1. Understanding reciprocals can be crucial for solving equations and balancing expressions. It helps simplify expressions and is a key tool in the algebra toolbox.
Multiplicative Inverse
The multiplicative inverse is essentially a fancy term for the reciprocal. It's the number or expression that you multiply with, to make the product equal to one. Let's consider why it's called the multiplicative "inverse":
  • "Multiply" refers to the operation you'll perform.
  • "Inverse" signifies that it is the opposite or flipped version.
Take the number 2. The multiplicative inverse of 2 is \( \frac{1}{2} \), because \( 2 \times \frac{1}{2} = 1 \).
In algebra, when you are dealing with expressions like \( x + 1 \), its multiplicative inverse (or reciprocal) is \( \frac{1}{x+1} \). Applying these inverses is handy in solving fractions and algebraic equations as they allow you to isolate variables and cancel out terms efficiently.
Writing Expressions
Writing algebraic expressions involves converting verbal or textual descriptions into mathematical format. It鈥檚 like translating a language into the language of math. Here are some steps and tips to guide you in writing expressions:
  • Identify keywords in the problem. For example, "the sum of" suggests adding.
  • Use variables to represent unknowns or changing quantities, like \( x \).
  • When dealing with reciprocals, understand that it means the multiplicative inverse, which you write as \( \frac{1}{expression} \).
So, if we look at the phrase "the reciprocal of \( x + 1 \)," you translate it simply into \( \frac{1}{x+1} \). Writing expressions is fundamental in algebra as it is the first step in problem-solving. It allows you to visualize and then manipulate numbers and symbols to find solutions.

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Most popular questions from this chapter

In the study of calculus, the difference quotient \(\frac{f(a+h)-f(a)}{h}\) is often found and simplified. Find and simplify this quotient for each function \(f(x)\) by following steps a through \(d\). a. Find \((a+h)\) b. Find \(f(a)\) c. Use steps a and b to find \(\frac{f(a+h)-f(a)}{h}\) d. Simplify the result of step \(\mathbf{c}\). $$ \frac{3}{x+1} $$

Simplify each complex fraction. See Examples 1 and 2. $$ \frac{\frac{x+3}{x^{2}-9}}{1+\frac{1}{x-3}} $$

Simplify. See Sections 5.1 and 5.5. $$ \frac{-36 x b^{3}}{9 x b^{2}} $$

Perform each indicated operation. See Section 1.3. $$ \frac{7}{8} \div \frac{1}{2} $$

Perform each indicated operation. The dose of medicine prescribed for a child depends on the child's age \(A\) in years and the adult dose \(D\) for the medication. Two expressions that give a child's dose are Young's Rule, \(\frac{D A}{A+12},\) and Cowling's Rule, \(\frac{D(A+1)}{24} .\) Find an expression for the difference in the doses given by these expressions.
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