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. \(f(x)=\sqrt{x+1}\)

Short answer

\(\frac{1}{\sqrt{x+h+1} + \sqrt{x+1}}\)

Step-by-step solution

Identify the Function and Difference Quotient Formula

The given function is \(f(x) = \sqrt{x+1}\). The difference quotient formula is \(\frac{f(x+h) - f(x)}{h}\), for \(h eq 0\).

Substitute \(f(x)\) and \(f(x+h)\)

Calculate \(f(x+h)\): \(f(x+h) = \sqrt{(x+h)+1} = \sqrt{x+h+1}\). Then, the difference quotient becomes \(\frac{\sqrt{x+h+1} - \sqrt{x+1}}{h}\).

Rationalize the Numerator

To simplify, multiply the numerator and denominator by the conjugate of the numerator: \(\frac{\sqrt{x+h+1} - \sqrt{x+1}}{h} \cdot \frac{\sqrt{x+h+1} + \sqrt{x+1}}{\sqrt{x+h+1} + \sqrt{x+1}} = \frac{(\sqrt{x+h+1} - \sqrt{x+1})(\sqrt{x+h+1} + \sqrt{x+1})}{h(\sqrt{x+h+1} + \sqrt{x+1})}\).

Simplify Using the Difference of Squares

Simplify the numerator using the difference of squares formula: \( (a - b)(a + b) = a^2 - b^2 \). Hence, \((\sqrt{x+h+1})^2 - (\sqrt{x+1})^2 = (x+h+1) - (x+1) = h\). The expression becomes \(\frac{h}{h(\sqrt{x+h+1} + \sqrt{x+1})} = \frac{1}{\sqrt{x+h+1} + \sqrt{x+1}}\).

Write the Final Result

The simplified difference quotient of the function \(f(x) = \sqrt{x+1}\) is \(\frac{1}{\sqrt{x+h+1} + \sqrt{x+1}}\).

Key concepts

Algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating those symbols. In this exercise, we encounter algebraic expressions like \(\sqrt{x+1}\) and \(\sqrt{x+h+1}\). Understanding how to handle these symbols is key.
One fundamental aspect of algebra here is the use of the difference quotient formula. Calculating \(f(x+h)\) by substituting \(x+h\) into the function requires skills in algebraic manipulation.
This is an essential step before any further simplification or rationalization can be done.

Function

A function is a relation that uniquely assigns an output to each input. In this exercise, our function is \(f(x) = \sqrt{x+1}\).
To use the difference quotient formula, we need to evaluate \(f(x)\) and \(f(x+h)\). Here, substituting \(x+h\) for the argument of the function is vital. This gives us \(f(x+h) = \sqrt{x+h+1}\).
The function defines the relationship between \(x\) and the value derived from inserting \(x\) into the function expression. Recognizing this helps us understand various transformations and simplifications that follow.

Simplification

Simplification makes expressions easier to handle and understand. In algebra, this often involves reducing expressions to simpler forms.
In this exercise, the difference quotient initially looks complex. Calculating the difference between \(f(x+h)\) and \(f(x)\) results in \(\frac{\sqrt{x+h+1} - \sqrt{x+1}}{h}\).
However, this can be simplified significantly using algebraic techniques like rationalizing the numerator. This transformation is crucial to making the expression manageable, laying the groundwork for further analysis or application.

Rationalization

Rationalization is a method used to eliminate radicals from the denominator or numerator of an expression. It simplifies computations and makes expressions easier to interpret.
Here, the difference quotient \(\frac{\sqrt{x+h+1} - \sqrt{x+1}}{h}\) is rationalized by multiplying by its conjugate: \(\frac{\sqrt{x+h+1} + \sqrt{x+1}}{\sqrt{x+h+1} + \sqrt{x+1}}\).
This process leverages the difference of squares formula, turning the radical expression into a simpler form. Rationalizing the numerator transforms the expression significantly, showing the power and utility of this algebraic technique.

Differences of Squares

The differences of squares formula states that \(a^2 - b^2 = (a - b)(a + b)\). It's used to simplify various algebraic expressions.
In this problem, when we rationalize the numerator, we use this formula: \( (\sqrt{x+h+1})^2 - (\sqrt{x+1})^2 = x+h+1 - x-1 = h\).
This simplification is a critical step, turning a complex radical expression into a straightforward form. Understanding this key concept is instrumental in mastering algebraic manipulations such as those found in the difference quotient.