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Ch. 1 - Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 1 - FunctionsProblem 1.1.98
Chapter 1, Problem 1.1.98

Simplify the difference quotients ƒ(x+h) - ƒ(x) / h and ƒ(x) - ƒ(a) / (x-a) by rationalizing the numerator.
ƒ(x) = √(1-2x)

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1
First, identify the function given: \( f(x) = \sqrt{1-2x} \). We need to simplify the difference quotients \( \frac{f(x+h) - f(x)}{h} \) and \( \frac{f(x) - f(a)}{x-a} \).
For the first difference quotient \( \frac{f(x+h) - f(x)}{h} \), substitute \( f(x+h) = \sqrt{1-2(x+h)} = \sqrt{1-2x-2h} \). The expression becomes \( \frac{\sqrt{1-2x-2h} - \sqrt{1-2x}}{h} \).
To rationalize the numerator, multiply the numerator and the denominator by the conjugate of the numerator: \( \frac{\sqrt{1-2x-2h} - \sqrt{1-2x}}{h} \times \frac{\sqrt{1-2x-2h} + \sqrt{1-2x}}{\sqrt{1-2x-2h} + \sqrt{1-2x}} \).
Simplify the numerator using the difference of squares: \( (\sqrt{1-2x-2h})^2 - (\sqrt{1-2x})^2 = (1-2x-2h) - (1-2x) = -2h \). The expression becomes \( \frac{-2h}{h(\sqrt{1-2x-2h} + \sqrt{1-2x})} \).
Cancel \( h \) in the numerator and denominator: \( \frac{-2}{\sqrt{1-2x-2h} + \sqrt{1-2x}} \). This is the simplified form of the first difference quotient. Repeat a similar process for the second difference quotient \( \frac{f(x) - f(a)}{x-a} \) by substituting \( f(a) = \sqrt{1-2a} \) and rationalizing the numerator.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Difference Quotient

The difference quotient is a fundamental concept in calculus that represents the average rate of change of a function over an interval. It is expressed as (ƒ(x+h) - ƒ(x)) / h, where h is the change in x. This concept is crucial for understanding derivatives, as the limit of the difference quotient as h approaches zero gives the derivative of the function.
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The Quotient Rule

Rationalizing the Numerator

Rationalizing the numerator involves manipulating an expression to eliminate any irrational numbers from the numerator. This is often done by multiplying the numerator and denominator by the conjugate of the numerator. This technique simplifies expressions, making it easier to evaluate limits or perform algebraic operations, especially in calculus.
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Function Composition

Function composition is the process of applying one function to the results of another function. In the context of the given question, understanding how to evaluate ƒ(x+h) involves substituting (x+h) into the function ƒ(x) = √(1-2x). This concept is essential for simplifying expressions and understanding how changes in input affect the output of a function.
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Related Practice
Textbook Question

Find the inverse f−1(x)f^{-1}\(\left\)(x\(\right\)) of each function (on the given interval, if specified).

f(x)=ln(3x+1)f\(\left\)(x\(\right\))=\(\ln\]\left\)(3x+1\(\right\))

Textbook Question

Simplify the difference quotients ƒ(x+h) - ƒ(x) / h and ƒ(x) - ƒ(a) / (x-a) by rationalizing the numerator.

ƒ(x) = - (3/√x)

Textbook Question

Intersection problems Find the following points of intersection.


The point(s) of intersection of the parabolas y= x² and y= -x² + 8x

Textbook Question

Simplify the difference quotient (ƒ(x)-ƒ(a)) / (x-a) for the following functions.

ƒ(x) = (1/x) - x²

Textbook Question

Solving equations Solve the following equations.


3(ˣ³⁻⁴) = 15

Textbook Question

Inverse sines and cosines Evaluate or simplify the following expressions without using a calculator.


cos (cos⁻¹ ( -1 ))

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