Problem 58 - Contemporary Precalculus

Question

Find the difference quotient of the given function. Then rationalize its numerator and simplify. $$g(x)=2 \sqrt{x+3}$$

Step-by-Step Solution

Answer

Answer: The simplified difference quotient for the given function is $$\frac{2}{\sqrt{x + h + 3} + \sqrt{x + 3}}$$.

Step 1: Substitute g(x + h) and g(x)

Replace $$x$$ with $$(x + h)$$ in the formula to obtain $$g(x + h)$$, and leave $$g(x)$$ as-is: $$\frac{g(x + h) - g(x)}{h} = \frac{2 \sqrt{(x + h) + 3} - 2 \sqrt{x + 3}}{h}$$

Step 2: Simplify the expression

We can factor out $$2$$ from the numerator and simplify the expression: $$\frac{2(\sqrt{x + h + 3} - \sqrt{x + 3})}{h}$$

Step 3: Rationalize the numerator

To rationalize the numerator, multiply both numerator and denominator by the conjugate of the numerator, which is $$(\sqrt{x + h + 3} + \sqrt{x + 3})$$: $$\frac{2(\sqrt{x + h + 3} - \sqrt{x + 3})(\sqrt{x + h + 3} + \sqrt{x + 3})}{h(\sqrt{x + h + 3} + \sqrt{x + 3})}$$

Step 4: Simplify the final expression

Now, use the difference of squares formula $$(a - b)(a + b) = a^2 - b^2$$ to simplify the numerator: $$\frac{2((x + h + 3) - (x + 3))}{h(\sqrt{x + h + 3} + \sqrt{x + 3})}$$ Further simplification gives: $$\frac{2h}{h(\sqrt{x + h + 3} + \sqrt{x + 3})}$$ Finally, we can cancel the $$h$$ term in the numerator and denominator: $$\frac{2}{\sqrt{x + h + 3} + \sqrt{x + 3}}$$ Therefore, the simplified difference quotient for the given function is: $$\frac{2}{\sqrt{x + h + 3} + \sqrt{x + 3}}$$

Key Concepts

Rationalizing the Numerator

Rationalizing the numerator is a useful algebraic technique when dealing with expressions involving radicals. In this context, we aim to eliminate the radicals from the numerator of our expression. When radicals or square roots appear in the numerator, they can sometimes make it difficult to simplify expressions or take limits, a common step in calculus.
To achieve this, we utilize the conjugate of the expression in the numerator. A conjugate is formed by changing the sign between two terms. For example, the conjugate of \( \sqrt{a} - \sqrt{b} \) is \( \sqrt{a} + \sqrt{b} \). By multiplying the numerator and the denominator by this conjugate, the radical terms often cancel out, simplifying the expression.

• Multiply both numerator and denominator by the conjugate.
• This process helps in removing square roots from the numerator.
• Ultimately leads to further simplification of the expression.

In our exercise, the expression was \( \sqrt{x+h+3} - \sqrt{x+3} \). We multiplied it by its conjugate, \( \sqrt{x+h+3} + \sqrt{x+3} \), thereby rationalizing the numerator.

Difference of Squares Formula

The difference of squares formula is an algebraic identity that is fundamental in simplifying products involving conjugates. It states that \((a - b)(a + b) = a^2 - b^2\). When you multiply two conjugates, the result is a difference of squares, effectively getting rid of middle terms and leaving us with simple subtraction.
This formula is especially handy during the rationalization of numerators, where radicals meet their conjugates. In our example, by using this identity, the expression \((\sqrt{x + h + 3} - \sqrt{x + 3})(\sqrt{x + h + 3} + \sqrt{x + 3})\) simplifies to \((x + h + 3) - (x + 3)\). As a result, it gives this neat difference: \( h \).

• Works whenever you encounter expressions of form \((a-b)(a+b)\).
• Eliminates middle terms, simplifying calculations.
• Converts expressions with radicals into simple differences.

In each step of our original problem, using this formula effectively turned the radical terms into manageable linear terms.

Simplification Techniques

Simplification techniques are essential for reducing complex mathematical expressions to a more manageable form. Key steps usually involve factoring, canceling common terms, and simplifying fractions. In our exercise, after rationalizing the numerator, we proceeded to simplify the resulting expression further.
The primary technique used was canceling common terms. We noticed \( h \) appeared in both the numerator and denominator. By dividing both by \( h \), we simplified the fraction, which allowed cancellation. However, it's crucial that \( h \) does not equal zero, as division by zero is undefined.

• Always look for common factors in numerators and denominators.
• Cancel common factors when possible.
• Be cautious of division by zero.
• Simplification often reveals underlying structures of functions or solutions.

By using these strategies, we arrived at the final expression, clearly showing the simplified form of the initial difference quotient.