Simplify each complex fraction. [ 1/[(x+h)2 + 9] - 1/(x2+9)] / h

Ch. R - Review of Basic Concepts

Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook

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Lial 13th Edition

Ch. R - Review of Basic Concepts

Problem 87a

Chapter 1, Problem 87a

Simplify each complex fraction. [ 1/[(x+h)² + 9] - 1/(x²+9)] / h

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Identify the complex fraction: the numerator is the difference of two fractions, $\frac{1}{(x+h)^2 + 9} - \frac{1}{x^2 + 9}$, and the denominator is $h$.

Find a common denominator for the two fractions in the numerator, which is $[(x+h)^2 + 9][x^2 + 9]$.

Rewrite the numerator as a single fraction: $\frac{x^2 + 9 - ((x+h)^2 + 9)}{[(x+h)^2 + 9][x^2 + 9]}$.

Simplify the numerator inside the fraction by expanding and combining like terms: $x^2 + 9 - (x+h)^2 - 9$.

Rewrite the entire complex fraction as $\frac{\frac{\text{simplified numerator}}{[(x+h)^2 + 9][x^2 + 9]}}{h}$, then multiply by the reciprocal of $h$ to get $\frac{\text{simplified numerator}}{h \cdot [(x+h)^2 + 9][x^2 + 9]}$.

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Key Concepts

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

Complex Fractions

A complex fraction is a fraction where the numerator, denominator, or both contain fractions themselves. Simplifying involves rewriting the expression as a single fraction by finding common denominators or multiplying numerator and denominator by the least common denominator to eliminate the smaller fractions.

Difference of Fractions

Subtracting fractions requires a common denominator. To subtract 1/[(x+h)^2 + 9] and 1/(x^2 + 9), find the least common denominator, rewrite each fraction with this denominator, then subtract the numerators while keeping the denominator the same.

Simplifying Rational Expressions

After combining fractions, simplify the resulting rational expression by factoring, canceling common factors, and reducing the expression. This step is crucial to express the complex fraction in its simplest form, especially when variables and binomials are involved.

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