Problem 59 - Contemporary Precalculus

Question

Find the difference quotient of the given function. Then rationalize its numerator and simplify. $$f(x)=\sqrt{x^{2}+1}$$

Step-by-Step Solution

Answer

Answer: The simplified difference quotient for the given function is $$\frac{2x+h}{\sqrt{(x+h)^{2}+1}+\sqrt{x^{2}+1}}$$.

Step 1: Write down the difference quotient formula

Recall that the difference quotient is given by: $$\frac{f(x+h)-f(x)}{h}$$

Step 2: Plug in the given function

Substitute the function $$f(x)=\sqrt{x^{2}+1}$$ into the difference quotient formula: $$\frac{\sqrt{(x+h)^{2}+1}-\sqrt{x^{2}+1}}{h}$$

Step 3: Rationalize the numerator

To rationalize the numerator, we multiply both the numerator and denominator by the conjugate of the numerator: $$\frac{\sqrt{(x+h)^{2}+1}-\sqrt{x^{2}+1}}{h} \cdot \frac{\sqrt{(x+h)^{2}+1}+\sqrt{x^{2}+1}}{\sqrt{(x+h)^{2}+1}+\sqrt{x^{2}+1}}$$

Step 4: Simplify the numerator

Multiply the numerators together and use the difference of squares formula, \((a-b)(a+b)=a^{2}-b^{2}\): $$\frac{((x+h)^{2}+1)-(x^{2}+1)}{h(\sqrt{(x+h)^{2}+1}+\sqrt{x^{2}+1})}$$

Step 5: Simplify the expression

Simplify the terms in the numerator: $$\frac{x^{2}+2xh+h^{2}+1-x^{2}-1}{h(\sqrt{(x+h)^{2}+1}+\sqrt{x^{2}+1})}$$ Now, cancel out the terms that are identical in the numerator: $$\frac{2xh+h^{2}}{h(\sqrt{(x+h)^{2}+1}+\sqrt{x^{2}+1})}$$

Step 6: Factor out h from the numerator

Factor out the common factor, h, from the numerator: $$\frac{h(2x+h)}{h(\sqrt{(x+h)^{2}+1}+\sqrt{x^{2}+1})}$$

Step 7: Cancel the common factor h

Cancel the common factor h from the numerator and denominator: $$\frac{2x+h}{\sqrt{(x+h)^{2}+1}+\sqrt{x^{2}+1}}$$ The simplified difference quotient for the given function is: $$\frac{2x+h}{\sqrt{(x+h)^{2}+1}+\sqrt{x^{2}+1}}$$

Key Concepts

Rationalizing the Numerator

This process of rationalizing the numerator is a crucial step when dealing with radicals in an expression. By rationalizing, we can simplify complex expressions involving square roots to a form that is easier to handle.

In the context of the difference quotient, when we have a radical difference where one term is subtracted from another, it often helps to multiply by a conjugate. The conjugate is found by changing the sign between two terms. For instance, if you have \( \sqrt{(x+h)^2+1} - \sqrt{x^2+1} \), its conjugate would be \( \sqrt{(x+h)^2+1} + \sqrt{x^2+1} \). This technique leverages the difference of squares forumla, \( (a-b)(a+b)=a^2-b^2 \), to eliminate the radicals.

By doing this, when we multiply the numerator and denominator by this conjugate \( \sqrt{(x+h)^2+1} + \sqrt{x^2+1} \), radicals in the numerator are eliminated. This makes further simplification possible, enhancing our ability to solve and understand the function behavior.

Simplifying Expressions

Simplifying expressions comes after rationalizing the numerator. In our context, the primary aim is to make the expression as clean and concise as possible.

Once you have rationalized the numerator, simplifying the expression is a methodical process. Breaking down complex numerators or denominators into simpler terms is a well-achieved by looking for common factors or terms. In this instance, after rationalizing, the difference of squares formula has been applied to produce \( ((x+h)^2+1)-(x^2+1) \).

Further expansion and simplification remove square terms, constant terms, and collect like terms. You'll notice how we ended up with \( 2xh + h^2 \) in the numerator. Finding common factors shared between the numerator and each term in the expression helps us to simplify even further.

Good simplification requires breaking down until we find opportunities to cancel terms, leading to a function that communicates core features of the function more understandably.

Difference of Squares

The difference of squares concept is a powerful algebraic tool used in various math topics. Specifically, in this exercise, it pairs well with rationalizing as it helps manage radials.

The difference of squares formula is given as \( (a-b)(a+b) = a^2-b^2 \). This formula allows for the quick multiplication of conjugate pairs by eliminating the middle terms. It's evident in our algebraic manipulation how this method expertly clears roots from the expression.

In our example, applying the formula results in subtracting squares of each term, which gives us a simpler combined expression.

• First, identify the two terms that form your squares
• Then, apply \( a^2-b^2 \) which helps combine and simplify radicals effectively.

Utilizing the difference of squares gives clarity to handling expressions that may seem overwhelming with radicals. This effective trick is indispensable in advancing through algebra problems elegantly.