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Chapter 2: Problem 62

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

Short Answer

Expert verified
\( -2x - h \).

Step by step solution

01

- Write Down the Function and Difference Quotient

Given the function \( f(x) = 2 - x^2 \), the difference quotient is \[ \frac{f(x+h) - f(x)}{h} \].
02

- Substitute \( f(x) \) and \( f(x+h) \)

First, find \( f(x+h) \). Using the function \( f(x) = 2 - x^2 \), substitute \( x+h \) for \( x \):\[ f(x+h) = 2 - (x+h)^2 \].Next, substitute \( f(x+h) \) and \( f(x) \) into the difference quotient:\[ \frac{(2 - (x+h)^2) - (2 - x^2)}{h} \].
03

- Expand and Simplify the Expression

Expand \( (x+h)^2 \):\[ (x+h)^2 = x^2 + 2xh + h^2 \].Replace it in the expression:\[ \frac{2 - x^2 - 2xh - h^2 - 2 + x^2}{h} \].Combine like terms:\[ \frac{-2xh - h^2}{h} \].
04

- Factor and Cancel \( h \)

Factor \( h \) out of the numerator:\[ \frac{h(-2x - h)}{h} \].Cancel the \( h \):\[ -2x - h \].

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Key Concepts

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

Function Substitution
Function substitution is simply replacing one variable with another in a given function. For example, given a function like \( f(x) = 2 - x^2 \), if we want to find \( f(x+h) \), we substitute \( x+h \) wherever there鈥檚 an \( x \) in the original function.

Here鈥檚 how:
  • Original function: \( f(x) = 2 - x^2 \).
  • Substitute \( x+h \): \( f(x+h) = 2 - (x+h)^2 \).

This step is crucial because it sets up the difference quotient for later calculations. Always remember to substitute accurately to avoid mistakes in further steps.
Binomial Expansion
Binomial expansion is used to expand expressions that are raised to a power, like \( (x+h)^2 \). Using the binomial theorem for expansion simplifies the calculations.
  • Start with \( (x+h)^2 \).
  • Expand it as \( x^2 + 2xh + h^2 \).
This expansion helps in breaking down the terms easily for further simplification.
Once the terms are expanded correctly, it becomes easier to combine and simplify in the following steps.
Simplifying Fractions
Simplifying fractions involves reducing them to their lowest terms. After expanding and combining like terms in the expression, you might end up with a fraction like \( \frac{-2xh - h^2}{h} \).
  • Factor out the common term in the numerator: \( \frac{h(-2x - h)}{h} \).
  • Cancel the common factor: \( -2x - h \).

This is where simplifying fractions helps in breaking down complex terms into straightforward expressions. It's an essential step in many algebra problems.
Polynomials
Polynomials are algebraic expressions that consist of variables and coefficients. In the given exercise, you encounter polynomials like \( 2 - x^2 \) and its variations.
  • A polynomial like \( f(x) = 2 - x^2 \) consists of terms of varying degrees.
  • During substitution and expansion, you create new polynomials that still follow algebraic rules.
Understanding polynomials helps in identifying like terms and combining them correctly.
Recognizing the structure of polynomials is crucial for expanding, simplifying, and solving algebraic expressions efficiently.

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Most popular questions from this chapter

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