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Chapter 2: Problem 51

Simplify the difference quotient \(\frac{f(x+h)-f(x)}{h}\) if \(h \neq 0\). \(f(x)=\sqrt{x-3}\) (Hint: Rationalize the numerator.)

Short Answer

Expert verified
\( \frac{1}{\sqrt{x+h-3} + \sqrt{x-3}} \)

Step by step solution

01

Define the difference quotient

The difference quotient for the function \( f(x) = \sqrt{x-3} \) is given by \( \frac{f(x+h) - f(x)}{h} \). We need to simplify this expression.
02

Find \( f(x+h) \)

Substitute \( x+h \) into the function: \( f(x+h) = \sqrt{(x+h) - 3} = \sqrt{x+h-3} \).
03

Substitute into the difference quotient

Substitute \( f(x+h) \) and \( f(x) \) into the difference quotient: \[ \frac{\sqrt{x+h-3} - \sqrt{x-3}}{h} \].
04

Rationalize the numerator

Multiply the numerator and the denominator by the conjugate of the numerator: \( \frac{(\sqrt{x+h-3} - \sqrt{x-3})(\sqrt{x+h-3} + \sqrt{x-3})}{h(\sqrt{x+h-3} + \sqrt{x-3})} \).
05

Simplify the numerator

The numerator becomes \((\sqrt{x+h-3})^2 - (\sqrt{x-3})^2 = (x+h-3) - (x-3)\). Simplify this to \(h\).
06

Cancel \( h \) in the fraction

The expression becomes \( \frac{h}{h(\sqrt{x+h-3} + \sqrt{x-3})} = \frac{1}{\sqrt{x+h-3} + \sqrt{x-3}} \), provided \( h eq 0 \).
07

Final simplified form

The simplified form of the difference quotient is \( \frac{1}{\sqrt{x+h-3} + \sqrt{x-3}} \).

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Key Concepts

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

Rationalizing the numerator
Rationalizing the numerator involves removing radicals (square roots) from the numerator of a fraction. This process makes the expression easier to simplify and work with.
In the problem, we have the expression \( \frac{\sqrt{x+h-3} - \sqrt{x-3}}{h} \). To rationalize the numerator, we multiply both the numerator and the denominator by the conjugate, which is \( \sqrt{x+h-3} + \sqrt{x-3} \).
  • The conjugate is simply the same expression as the numerator with the sign between the terms flipped.
  • This multiplication eliminates the radical in the numerator by using the identity \((a-b)(a+b) = a^2 - b^2\).
This technique helps tidy up radical expressions and is a standard procedure in algebra.
Function simplification
Function simplification is the process of rewriting a function more simply. This often involves algebraic manipulations to reduce complexity.
When simplifying the difference quotient with radicals, such as in our example, eliminating common terms was key. After rationalizing, the numerator simplified to \(h\) because the expression became \((x+h-3)-(x-3)\), which subtracts the constants and variable terms.
  • After simplifying the numerator to \(h\), we could cancel it with the \(h\) in the denominator.
  • This simplification was only possible because \(h eq 0\).
Ultimately, function simplification made our expression much simpler: \( \frac{1}{\sqrt{x+h-3} + \sqrt{x-3}} \).
Radical expressions
Radical expressions include roots, like square roots. They are common in many areas of mathematics, including precalculus.
In our task, we dealt with radical expressions when simplifying the function \( f(x) = \sqrt{x-3} \).
  • Radical expressions can appear difficult to handle initially because they're not as straightforward to manipulate as other algebraic forms.
  • Using techniques like rationalizing the numerator helps in simplifying these expressions.
Mastering radical expressions is crucial, as they often come up in various calculus problems later. Understanding their properties helps simplify and solve equations involving them.
Precalculus
Precalculus serves as a foundation for understanding calculus. It includes a wide range of mathematical concepts, including functions and radicals.
The difference quotient is one such concept that originates in precalculus. It is vital to understand this as it lays the groundwork for calculating derivatives in calculus.
  • By working through simplification, we see how these precalculus skills are directly applied.
  • Achieving proficiency in these topics ensures readiness for tackling calculus in the future.
Precalculus bridges high school algebra and calculus, preparing students with the necessary skills and understanding for more advanced study.

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Most popular questions from this chapter

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