/*! 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 31 Evaluate the difference quotient... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the difference quotient for the given function. Simplify your answer. \(f(x)=\frac{1}{x}, \quad \frac{f(x)-f(a)}{x-a}\)

Short Answer

Expert verified
The simplified difference quotient is \( -\frac{1}{xa} \).

Step by step solution

01

Substitute the function into the difference quotient

We begin by substituting the given function into the difference quotient formula. The formula is: \( \frac{f(x) - f(a)}{x - a} \). Since \( f(x) = \frac{1}{x} \), we replace \( f(x) \) and \( f(a) \) with their respective expressions: \( \frac{1}{x} \) and \( \frac{1}{a} \). The expression becomes: \( \frac{\frac{1}{x} - \frac{1}{a}}{x - a} \).
02

Simplify the numerator

Simplify the numerator \( \frac{1}{x} - \frac{1}{a} \) by finding a common denominator. The common denominator for \( x \) and \( a \) is \( xa \). Thus, \( \frac{1}{x} - \frac{1}{a} \) becomes \( \frac{a - x}{xa} \).
03

Simplify the entire expression

Substitute \( \frac{a - x}{xa} \) back into the expression: \( \frac{\frac{a - x}{xa}}{x - a} \). This can be rewritten as \( \frac{a - x}{xa(x-a)} \). Notice \( a - x \) can be factored as \( -(x - a) \). Thus the expression \( \frac{a - x}{xa(x-a)} \) becomes \( -\frac{1}{xa} \) by cancelling out the \( x-a \) terms.

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Key Concepts

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

Simplifying Algebraic Expressions
Simplifying algebraic expressions is crucial for solving complex mathematical problems. It involves reducing expressions to their simplest form by performing operations like addition, subtraction, multiplication, and division.
Consider an expression such as \[ \frac{1}{x} - \frac{1}{a} \]. To simplify it, we seek a common denominator, which in this case is \[ xa \]. This allows us to combine the terms into a single fraction: \[ \frac{a - x}{xa} \].
Simplifying expressions in this manner makes them easier to work with and can reveal hidden relationships between variables. So, whenever you encounter a complex expression, search for opportunities to simplify by finding common denominators or factoring.
Rational Functions
Rational functions are a key concept in calculus and algebra. These are functions formed by the ratio of two polynomials, such as \[ f(x) = \frac{1}{x} \]. They are integral in modeling real-world phenomena and in complex calculations.
A rational function can be represented as \[ \frac{p(x)}{q(x)} \], where \[ p(x) \] and \[ q(x) \] are polynomials. The domain of a rational function excludes any values that make the denominator zero, as this would make the function undefined.
  • Rational functions often feature vertical asymptotes, which occur at values that make the denominator zero.
  • They can also have horizontal or oblique asymptotes, which describe the function's behavior as \[ x \] approaches infinity.
These characteristics make rational functions versatile in mathematical modeling, from physics to economics.
Derivative Concept
The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. It is often introduced using the concept of the difference quotient.
The difference quotient \[ \frac{f(x) - f(a)}{x - a} \] computes the average rate of change of the function \[ f(x) \] over an interval. As \[ x \] approaches \[ a \], this quotient approaches the derivative \[ f'(a) \], giving us the instantaneous rate of change.
  • The derivative provides crucial information about the function, such as slope or gradient, and is used to find minima and maxima.
  • In practical applications, derivatives can help optimize processes and predict behavior of dynamic systems.
Knowing how to calculate and interpret derivatives is key to understanding the world through a mathematical lens.
Common Denominator in Fractions
Finding a common denominator is an essential skill when working with fractions, especially in algebra and calculus. It allows us to add, subtract, or simplify fractions effectively.
When you have \[ \frac{1}{x} \] and \[ \frac{1}{a} \], the common denominator is \[ xa \]. This makes it possible to rewrite the difference \[ \frac{1}{x} - \frac{1}{a} \] as \[ \frac{a - x}{xa} \].
  • Getting a common denominator involves multiplying each fraction by a form of 1 that will make the denominators the same. This transforms each fraction without changing its value.
  • Once fractions share a common denominator, they can be directly added, subtracted, or simplified as needed.
Mastering the technique of finding a common denominator is crucial for simplifying rational expressions and solving algebraic equations.

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Most popular questions from this chapter

Us population The table gives the population of theUnited States, in millions, for the years \(1900-2010 .\) Use a calculator with exponential regression capability to model the US population since \(1900 .\) Use the model to estimate the population in 1925 and to predict the population in the year \(2020 .\) $$\begin{array}{|c|c|}\hline \text { Year } & {\text { Population }} \\\ \hline 1900 & {76} \\ \hline 1910 & {92} \\ \hline 1920 & {106} \\ \hline 1940 & {131} \\ {1950} & {150} \\ \hline\end{array}$$$$ \begin{array}{|c|c|}\hline \text { Year } & {\text { Population }} \\ \hline 1960 & {179} \\ {1970} & {203} \\ \hline 1990 & {220} \\ {2000} & {281} \\\ {2010} & {310} \\ \hline\end{array}$$

(a) Find the domain of \(f(x)=\ln \left(e^{x}-3\right)\) (b) Find \(f^{-1}\) and its domain.

Data points \((x, y)\) are given. (a) Draw a scatter plot of the data points. (b) Make semilog and log-log plots of the data. (c) Is a linear, power, or exponential function appropriate for modeling these data? (d) Find an appropriate model for the data and then graph the model together with a scatter plot of the data. $$\begin{array}{|c|c|c|c|c|c|c|}\hline x & {2} & {4} & {6} & {8} & {10} & {12} \\\ \hline y & {0.08} & {0.12} & {0.18} & {0.26} & {0.35} & {0.53} \\\ \hline\end{array}$$ \(\begin{array}{|c|c|c|c|c|c|c|}\hline x & {10} & {20} & {30} & {40} & {50} & {60} \\ \hline y & {29} & {82} & {150} & {236} & {330} & {430} \\\ \hline\end{array}\)

Express the given quantity as a single logarithm. $$ \ln 5+5 \ln 3 $$

40\. Find the first 40 terms of the sequence defined by $$a_{n+1}=\left\\{\begin{array}{ll}{\frac{1}{2} a_{n}} & {\text { if } a_{n} \text { is an even number }} \\ {3 a_{n}+1} & {\text { if } a_{n} \text { is an odd number }}\end{array}\right.$$ and \(a_{1}=11 .\) Do the same if \(a_{1}=25 .\) Make a conjecture about this type of sequence.

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