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Chapter 8: Problem 18

(a) \(f(x) \equiv \frac{3 x}{(x-2)(x+1)}\) (b) \(f(x) \equiv \frac{2}{x-2}+\frac{1}{x+1}\)

Short Answer

Expert verified
Both representations are equivalent: \( \frac{3x}{(x-2)(x+1)} = \frac{2}{x-2}+\frac{1}{x+1} \)

Step by step solution

01

Simplify the expression for part (a)

Given: (a) \( f(x) \equiv\ \frac{3x}{(x-2)(x+1)} \)First, recognize that the initial fraction cannot be simplified further directly.
02

Compare the fractions in part (a) and part (b)

Compare the fraction from part (a) to the form presented in part (b): (b) \( f(x) \equiv\ \frac{2}{x-2} + \frac{1}{x+1} \)Now, let's check if they represent the same function by finding a common denominator.
03

Combine the terms in part (b) under a common denominator

For part (b), combine the terms: \( \frac{2}{x-2} + \frac{1}{x+1} \)Find a common denominator: \( \frac{2(x+1) + 1(x-2)}{(x-2)(x+1)} \)This results in: \( \frac{2x + 2 + x - 2}{(x-2)(x+1)} \)
04

Simplify the combined expression

Simplify the numerator: \( 2x + 2 + x - 2 = 3x \)Thus,\( \frac{2(x+1) + 1(x-2)}{(x-2)(x+1)} = \frac{3x}{(x-2)(x+1)} \)
05

Conclusion

Since both (a) and (b) simplify to the same expression: Part (a): \(\frac{3x}{(x-2)(x+1)} \)Part (b): \( \frac{2(x+1) + 1(x-2)}{(x-2)(x+1)} \)Both representations are equivalent.

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Key Concepts

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

partial fractions
Partial fractions are a way of breaking down a complex fraction into simpler parts. This technique is especially useful in integration and solving differential equations.
Imagine you have a fraction like: \(\frac{3x}{(x-2)(x+1)}\)
It looks complicated, right? By using partial fractions, we can rewrite this expression in a simpler form, like: \(\frac{2}{x-2} + \frac{1}{x+1}\)
This transformation makes it easier to work with the fraction in various mathematical scenarios.
simplification
Simplification involves making a mathematical expression more manageable while keeping it equivalent to the original.
In the context of our exercise, we have: \(\frac{3x}{(x-2)(x+1)}\)
The goal is to transform it into a simpler form without changing its value. By finding a common denominator for the fractions in part (b), we combine them into: \(\frac{2(x+1) + 1(x-2)}{(x-2)(x+1)}\)
This ultimately simplifies to: \(\frac{3x}{(x-2)(x+1)}\)
Both forms are simpler to handle in different scenarios.
common denominator
Finding a common denominator is crucial when adding or subtracting fractions.
For instance, to combine \(\frac{2}{x-2} + \frac{1}{x+1}\), we need a common denominator, which is \( (x-2)(x+1)\).
To achieve this, we rewrite each fraction: \(\frac{2(x+1) + 1(x-2)}{(x-2)(x+1)}\)
This common denominator allows us to add the numerators together:
  • 2(x+1) becomes 2x + 2
  • 1(x-2) becomes x - 2
Combining these, we get: 2x + 2 + x - 2 = 3x.
This helps create a single fraction that's easier to manipulate.
algebraic expression
An algebraic expression consists of numbers, variables, and operators (like +, -, *, /).
For example, \(\frac{3x}{(x-2)(x+1)}\) is an algebraic expression.
Breaking down this complex expression can make it easier to work with.
  • In part (a), we have a straightforward form of the expression.
  • In part (b), we divided it into simpler terms, showcasing partial fractions.
Understanding how to convert between these forms is key in algebra, helping you solve equations and simplify problems effectively.

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Most popular questions from this chapter

If \(\tan y=x\) find the value of \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}\) when \(y=\frac{\pi}{4}\).

Given that \(u\) and \(v\) are functions of \(x\), prove from first principles that $$ \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{u}{v}\right)=\frac{v \mathrm{~d} u / \mathrm{d} x-u \mathrm{~d} v / \mathrm{d} x}{v^{2}} $$ Find, in a simplified form, the derivative of the function $$ \frac{2+\ln (1+x)^{2}}{2-\ln (1-x)^{2}} $$

\(\frac{\mathrm{d}}{\mathrm{d} x}\left[\ln \frac{x}{1+x}\right]=\frac{\mathrm{d}}{\mathrm{d} x}[\ln x]-\frac{\mathrm{d}}{\mathrm{d} x}[\ln (1+x)]\).

If \(y=x \arctan x\), show that: (a) \(x\left(1+x^{2}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=x^{2}+\left(1+x^{2}\right) y\) (b) \(\left(1+x^{2}\right) \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}+2 x \frac{\mathrm{d} y}{\mathrm{~d} x}-2 y=2\).

\(\frac{\mathrm{d}}{\mathrm{d} x} a^{x}\) is: (a) \(x a^{x-1}\) (b) \(a^{x}\) (c) \(x \ln a\) (d) \(a^{x} \ln a\) (e) none of these.
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