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Simplify the difference quotient \(\frac{f(x+h)-f(x)}{h}\) for the following functions. $$f(x)=\frac{2}{x}$$

Short Answer

Expert verified
Question: Simplify the difference quotient for the function \(f(x) = \frac{2}{x}\). Answer: The simplified difference quotient for the function \(f(x) = \frac{2}{x}\) is \(\frac{-2}{(x+h)x}\).

Step by step solution

01

Write down the difference quotient

The difference quotient is given by: $$\frac{f(x+h)-f(x)}{h}$$
02

Find \(f(x+h)\) and \(f(x)\) for \(f(x)=\frac{2}{x}\)

To do this, substitute \(x+h\) and \(x\) into \(f(x)=\frac{2}{x}\). For \(f(x+h)\): $$f(x+h) = \frac{2}{x+h}$$ For \(f(x)\): $$f(x) = \frac{2}{x}$$
03

Substitute the expressions above into the difference quotient

We get the expression: $$\frac{\frac{2}{x+h}-\frac{2}{x}}{h}$$
04

Simplify the fraction in the numerator

To do so, we can find a common denominator: $$\frac{\frac{2x-2(x+h)}{(x+h)x}}{h}$$
05

Simplify the expression in the numerator

Distribute the \(-2\) to both \(x\) and \(h\): $$\frac{\frac{2x-2x-2h}{(x+h)x}}{h}$$
06

Simplify the expression further

Notice that \(2x-2x\) is zero. We're left with: $$\frac{\frac{-2h}{(x+h)x}}{h}$$
07

Divide the fraction by \(h\)

To divide a fraction by a number, we can multiply the denominator of the fraction by the number: $$\frac{-2h}{(x+h)x \times h}$$
08

Simplify the expression to obtain the final result

Since the \(h\) terms cancel out, the simplified difference quotient is: $$\frac{-2}{(x+h)x}$$

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Key Concepts

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

Simplifying Expressions
Simplifying mathematical expressions is fundamental to understanding calculus problems, especially when dealing with difference quotients, which are expressions that measure the change in a function's value over a tiny change in input. In the exercise, we focus on transforming a complex-looking expression to a more manageable form by performing a series of algebraic operations.

The process involves finding common denominators, distributing terms, and combining like terms. As you have seen in the given solution, we combine the fractions by finding a common denominator, which allowed us to subtract one fraction from another. The critical step was recognizing the subtraction of like terms, which simplified to zero, drastically reducing the complexity of the expression. Always carefully distribute negative signs and look for opportunities to cancel out terms, just as we've canceled out 'h' in the final steps. Simplification is about making the expression easier to understand and work with, a skill that is invaluable when you start tackling more advanced calculus problems.
Rational Functions
A rational function is a quotient of two polynomial functions. The function \(f(x) = \frac{2}{x}\) from our exercise is a classic example of a rational function, where the numerator is a constant and the denominator is a linear polynomial. Rational functions often appear in calculus problems, particularly when evaluating limits or finding derivatives and integrals.

In our exercise, we work with the rational function by altering the input (adding 'h') and carefully examining the changes in the function's output. A key strategy for dealing with rational functions is finding common denominators, as we saw in steps 4 and 5 of the solution. When simplifying a difference quotient involving a rational function, it is crucial to correctly manipulate the expressions to avoid errors, especially when the variables are in the denominator.
Limits in Calculus
The concept of limits is fundamental in calculus and is subtly introduced through the difference quotient. Limits describe the behavior of functions as they approach a specific point, either along the x-axis (input) or as the function value approaches a number (output).

The difference quotient \(\frac{f(x+h)-f(x)}{h}\) is a precursor to the derivative, which measures how a function changes at any point and is defined formally using limits. What we are essentially doing with the difference quotient is looking at the ratio of the changes in 'y' to changes in 'x' as 'h' approaches zero. In this exercise, we do not take the limit as 'h' approaches zero, but it's an important next step in many calculus problems. Understanding how to work with the difference quotient, as shown in the steps of the solution, prepares you for mastering limits, which define rates of change and areas under a curve -- core concepts in the study of calculus.

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