Problem 92 - Precalculus Essentials

Question

Find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$for the given function. $$f(x)=\sqrt{x-1}$$

Step-by-Step Solution

Answer

The simplified difference quotient for the given function \( \sqrt{x-1} \) is \( \frac{1}{\sqrt{x + h - 1} + \sqrt{x - 1}} \)

Step 1: Identify the Function and Difference Quotient

f(x) is given as \( \sqrt{x-1} \). The difference quotient we need to find is \( \frac{f(x + h) - f(x)}{h} \), where \( h \neq 0 \).

Step 2: Apply the Function

First, apply the function to \( x+h \) and \( x \), we get \(f(x + h) = \sqrt{x + h - 1}\) and \(f(x) = \sqrt{x - 1}\).

Step 3: Substitute into the Difference Quotient

Substituting the above into the difference quotient yields \( \frac{\sqrt{x + h - 1} - \sqrt{x - 1}}{h} \).

Step 4: Rationalize the Numerator

Next, rationalize the numerator by multiplying the expression by its conjugate, which does not change its value but eliminates the square root in the numerator. The conjugate of \( \sqrt{x + h - 1} - \sqrt{x - 1} \) is \( \sqrt{x + h - 1} + \sqrt{x - 1} \). Thus the expression becomes \( \frac{\sqrt{x + h - 1} - \sqrt{x - 1}}{h} \cdot \frac{\sqrt{x + h - 1} + \sqrt{x - 1}}{\sqrt{x + h - 1} + \sqrt{x - 1}} \).

Step 5: Simplify the expression

Simplify the above expression using the difference of squares formula – \( (a - b)(a + b) = a^2 - b^2 \). Our expression becomes \( \frac{(x+h-1)-(x-1)}{h(\sqrt{x + h - 1} + \sqrt{x - 1})} \) which simplifies further to \( \frac{h}{h(\sqrt{x + h - 1} + \sqrt{x - 1})} \).

Step 6: Eliminate h from the numerator and denominator

Now, cancel out h from the numerator and the denominator to get the simplified difference quotient as \( \frac{1}{\sqrt{x + h - 1} + \sqrt{x - 1}} \).