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Given the function g(x) = 9 – x², evaluate IX + h) – g(x) h ± 0. h

Given the function g(x) = 9 – x², evaluate IX + h) – g(x) h ± 0. h

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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**Problem:** Given the function \( g(x) = 9 - x^2 \), evaluate \(\frac{g(x + h) - g(x)}{h}\), \( h \neq 0 \). **Explanation:** This problem asks you to find the difference quotient for the given function. The difference quotient is a fundamental concept in calculus that represents the slope of the secant line between two points on the graph of a function, often used as a step toward finding the derivative. 1. **Function Information:** - The function is \( g(x) = 9 - x^2 \). 2. **Difference Quotient:** - It is expressed as \(\frac{g(x + h) - g(x)}{h}\). - This formula is used to calculate the average rate of change of the function over a small interval \( h \). 3. **Objective:** - Evaluate the expression to find how the function changes as \( x \) changes by a small amount \( h \), assuming \( h \neq 0 \). This exercise helps illustrate how derivatives are calculated as \( h \) approaches zero.
Transcribed Image Text:**Problem:** Given the function \( g(x) = 9 - x^2 \), evaluate \(\frac{g(x + h) - g(x)}{h}\), \( h \neq 0 \). **Explanation:** This problem asks you to find the difference quotient for the given function. The difference quotient is a fundamental concept in calculus that represents the slope of the secant line between two points on the graph of a function, often used as a step toward finding the derivative. 1. **Function Information:** - The function is \( g(x) = 9 - x^2 \). 2. **Difference Quotient:** - It is expressed as \(\frac{g(x + h) - g(x)}{h}\). - This formula is used to calculate the average rate of change of the function over a small interval \( h \). 3. **Objective:** - Evaluate the expression to find how the function changes as \( x \) changes by a small amount \( h \), assuming \( h \neq 0 \). This exercise helps illustrate how derivatives are calculated as \( h \) approaches zero.
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