/*! 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 51 For each function \(f,\) constru... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For each function \(f,\) construct and simplify the difference quotient $$ \frac{f(x+h)-f(x)}{h} $$ $$f(x)=6 x+2$$

Short Answer

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Step by step solution

01

Understand the Function

Identify the given function. Here, the function is given as \(f(x) = 6x + 2\).
02

Substitute Into the Difference Quotient Formula

The difference quotient formula is \( \frac{f(x+h) - f(x)}{h} \). Substitute the given function into this formula to get \( \frac{(6(x+h) + 2) - (6x + 2)}{h} \).
03

Expand the Numerator

Expand the numerator by distributing the 6 inside the parentheses: \( 6(x + h) + 2 \) becomes \( 6x + 6h + 2 \). Now the expression is \( \frac{6x + 6h + 2 - (6x + 2)}{h} \).
04

Simplify the Numerator

Combine like terms in the numerator: \( 6x + 6h + 2 - 6x - 2 \). This simplifies to \( 6h \). Now the expression is \( \frac{6h}{h} \).
05

Simplify the Fraction

Cancel the common factor of \(h\) in the numerator and the denominator: \( \frac{6h}{h} = 6 \).

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Key Concepts

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

simplifying fractions
Simplifying fractions is a fundamental skill in mathematics. When we simplify a fraction, our goal is to make it as simple as possible without changing its value. In the context of the difference quotient, simplifying fractions becomes particularly important to arrive at a clear, understandable result.
After you've combined like terms in the numerator, you often end up with a term that includes a common factor with the denominator. For instance, if you have the fraction \(\frac{6h}{h}\), both the numerator and the denominator share a common factor of \(h\). You can cancel out this common factor:
\[ \frac{6h}{h} = 6 \]
This step is crucial as it simplifies the expression significantly. Remember, when you cancel out a term, always make sure that the term you're canceling isn't equal to zero, as division by zero is undefined.
Practice simplifying fractions by looking for common factors and canceling only when it's mathematically valid.
function evaluation
Function evaluation is the process of finding the value of a function for a particular input. When given a function like \(f(x) = 6x + 2\), you can find the function's value by substituting the input for \(x\).
For example, to find \(f(x + h) \), you replace \(x\) with \(x + h\) in the function:
\[ f(x + h) = 6(x + h) + 2 \]
This is helpful when working with the difference quotient, where we need to evaluate the function at \((x + h)\) and \(x\). So, keeping this method of substitution in mind, we substitute \(x + h\) into the given function.
Understanding function evaluation helps simplify complex expressions and is key in calculus, particularly in understanding rate changes and derivative concepts.
precalculus
Precalculus forms the foundation for understanding calculus, encompassing a range of topics including functions, algebra, and trigonometry. The difference quotient, which we are dealing with here, is a key concept in precalculus. It provides a way to examine how a function's value changes as the input changes slightly.
The difference quotient formula is:
\[ \frac{f(x + h) - f(x)}{h} \]
This formula helps you understand the average rate of change of the function over an interval. In precalculus, you'll deal with various functions like linear, polynomial, rational, and trigonometric functions, and learning to manipulate and simplify expressions using these functions. This builds your skills for tackling more advanced calculus problems.
By mastering precalculus concepts like simplifying fractions and evaluating functions, you'll be well-prepared to handle calculus topics such as limits, derivatives, and integrals.

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Most popular questions from this chapter

A power line is constructed from a power station at point A to an island at point I, which is 1 mi directly out in the water from a point \(\mathrm{B}\) on the shore. Point \(\mathrm{B}\) is \(4 \mathrm{mi}\) downshore from the power station at A. It costs 5000 dollar per mile to lay the power line under water and 3000 dollar per mile to lay the power line under ground. The line comes to the shore at point S downshore from A. Let \(x=\) the distance from B to S. (IMAGE CAN NOT COPY) a) Express the cost \(C\) of laying the line as a function of \(x\) b) At what distance \(x\) from point \(B\) should the line come to shore in order to minimize cost?

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