Question

Find the difference quotient of \(f\); that is, find \(\frac{f(x+h)-f(x)}{h}, h \neq 0,\) for each function. Be sure to simplify. \(f(x)=\frac{1}{\sqrt{x+2}}\)

Short answer

The difference quotient is \ \frac{-1}{(x+2) \cdot \sqrt{x+h+2}} \.

Step-by-step solution

- Set Up the Difference Quotient

Write the expression for the difference quotient: \[ \frac{f(x+h) - f(x)}{h} \].

- Find f(x+h)

Substitute \(x + h\) into the function \(f(x)\): \[ f(x+h) = \frac{1}{\sqrt{(x+h) + 2}} = \frac{1}{\sqrt{x+h+2}} \].

- Substitute f(x) and f(x+h) into the Difference Quotient

Substitute the expressions \(f(x)\) and \(f(x+h)\) into the difference quotient formula: \[ \frac{\frac{1}{\sqrt{x+h+2}} - \frac{1}{\sqrt{x+2}}}{h} \].

- Combine the Fractions

To combine the two fractions, find a common denominator: \[ \frac{\sqrt{x+2} \cdot \sqrt{x+2} - \sqrt{x+h+2} \cdot \sqrt{x+2}}{h \cdot \sqrt{x+h+2} \cdot \sqrt{x+2}} \], which simplifies to \[ \frac{\sqrt{x+2} - \sqrt{x+h+2}}{h \cdot \sqrt{x+2} \cdot \sqrt{x+h+2}} \].

- Simplify the Numerator

Rationalize the numerator by multiplying by the conjugate: \[ \frac{(\sqrt{x+2} - \sqrt{x+h+2})(\sqrt{x+2} + \sqrt{x+h+2})}{h \cdot \sqrt{x+2} \cdot \sqrt{x+h+2} \cdot (\sqrt{x+2} + \sqrt{x+h+2})} \].

- Simplify Further

Simplify the numerator and the denominator: Numerator: \[ (x+2) - (x+h+2) = -h \] Denominator: \[ h \cdot \sqrt{x+2} \cdot \sqrt{x+h+2} \cdot (\sqrt{x+2} + \sqrt{x+h+2}) \].

- Cancel Out h

Cancel the \(h\) from the numerator and denominator: \[ \frac{-1}{\sqrt{x+2} \cdot \sqrt{x+h+2} \cdot (\sqrt{x+2} + \sqrt{x+h+2})} \].

- Final Simplification

Simplify the fraction: \[ \frac{-1}{(x+2) \cdot \sqrt{x+h+2}} \].

Key concepts

Rationalizing the Numerator

To rationalize the numerator means to eliminate any square roots or cube roots from the numerator. It's similar to rationalizing the denominator, but this time we focus on the numerator.
In the original exercise, the fraction has a numerator that involves square roots. To simplify, we multiply both the numerator and the denominator by the conjugate of the numerator.
The conjugate of \( \sqrt{x+2} - \sqrt{x+h+2} \) is \( \sqrt{x+2} + \sqrt{x+h+2} \). By multiplying by this conjugate, square roots in the numerator will be converted into a simpler polynomial. This process is essential in making the expression simpler and more transparent.
For example: If you have \( \frac{ \sqrt{a} - \sqrt{b}}{c} \) and multiply by \( \sqrt{a} + \sqrt{b} \); you get:
- The numerator: \( ( \sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) \) which simplifies to \( a - b \)
- The denominator: stays \( c \).
It cleans up the square roots and results in a simpler form.

Finding Common Denominators

A crucial step when working with fractions in algebraic expressions is finding a common denominator.
Let's look at why it's essential in the context of difference quotients.
When working with the difference quotient \( \frac{f(x+h) - f(x)}{h} \), you often end up subtracting fractions.
To subtract two fractions, they need to have the same denominator. For example, in our exercise:

• The two fractions are \( \frac{1}{\sqrt{x+h+2}} \) and \( \frac{1}{\sqrt{x+2}} \)
• Their common denominator would be \( \sqrt{x+2} \cdot \sqrt{x+h+2} \)

Combining the fractions into one involves:

• Multiplying each fraction by the other's denominator.
• This ensures they share the same base and can be easily subtracted.

This standardizes the denominators and helps facilitate subsequent simplification steps.

Simplifying Algebraic Expressions

Simplifying algebraic expressions requires breaking down and combining terms to achieve the simplest form.
In the context of our exercise, simplifying involves several steps:

1. First, we combine fractions by finding common denominators.
2. Next, rationalize the numerator to eliminate square roots if necessary.
3. Then simplify the results by combining like terms or cancelling out common factors.

For instance, in our difference quotient:

• At one step, we encounter \( \frac{\frac{1}{\sqrt{x+h+2}} - \frac{1}{\sqrt{x+2}}}{h} \)
• After finding a common denominator and rationalizing the numerator, you get a simpler form.
• Finally, you cancel out common terms such as \( h \).

The ultimate goal is to make the expression as clear and manageable as possible, reducing it to its simplest form to understand the relationship it describes better.