Problem 88 - A Graphical Approach to College Algebra

Question

Determine the difference quotient \(\frac{f(x+h)-f(x)}{h}\) (where \(h \neq 0\) ) for each function \(f\). Simplify completely. \(f(x)=\sqrt{x}\) (Hint: Rationalize the numerator.)

Step-by-Step Solution

Answer

The difference quotient is \(\frac{1}{\sqrt{x+h} + \sqrt{x}}\).

Step 1: Write Down the Difference Quotient Formula

Start with applying the difference quotient formula for the function \(f(x) = \sqrt{x}\). The formula is given by:\[\frac{f(x+h) - f(x)}{h} = \frac{\sqrt{x+h} - \sqrt{x}}{h}\] Here we will compute \(f(x+h) = \sqrt{x+h}\) and \(f(x) = \sqrt{x}\), and then substitute them into the formula.

Step 2: Rationalize the Numerator

To simplify the expression, we need to rationalize the numerator:\[\frac{\sqrt{x+h} - \sqrt{x}}{h}\]Multiply the numerator and the denominator by the conjugate of the numerator:\[\frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}\]

Step 3: Simplify the Expression

Multiply out the terms in the numerator using the difference of squares identity:\[(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x}) = (x+h) - x = h\]Substitute back into the difference quotient:\[\frac{h}{h(\sqrt{x+h} + \sqrt{x})} = \frac{1}{\sqrt{x+h} + \sqrt{x}}\]

Step 4: Final Simplified Answer

Cancel out \(h\) from the numerator and the denominator, which leaves us with the final simplified form:\[\frac{1}{\sqrt{x+h} + \sqrt{x}}\]This is the simplified difference quotient for \(f(x) = \sqrt{x}\).

Key Concepts

Rationalizing the Numerator

Rationalizing the numerator is a clever trick used to eliminate radicals from a fraction's numerator. This is particularly important in calculus, where it helps simplify difference quotients.

To rationalize the numerator, we multiply the numerator and the denominator by the conjugate of the original numerator. The conjugate is the same as the original expression but with the opposite sign in the middle. For example, if we have \(\sqrt{x+h} - \sqrt{x}\), the conjugate is \(\sqrt{x+h} + \sqrt{x}\).

This method is effective because it transforms the numerator into a difference of squares formula:

• \((a-b)(a+b) = a^2 - b^2\)

By multiplying by the conjugate, the radicals are eliminated, which is what we wish to achieve before moving to simplify further.

Algebraic Simplification

Algebraic simplification is crucial for making complex expressions easier to work with. When dealing with difference quotients, simplification often hinges on seeing patterns and recognizing identities.

Once we rationalized the numerator by multiplying it with its conjugate, we applied the difference of squares formula:

• \((\sqrt{x+h})^2 - (\sqrt{x})^2 = (x+h) - x\)

This simplifies to just \(h\) in the numerator.

In the provided exercise, this simplification allows us to cancel \(h\) from the numerator and denominator. This is a common step in calculus to remove unnecessary terms and make the expression cleaner.

By simplifying expressions algebraically, we ensure that calculations become straightforward and easy to interpret.

Square Roots

Square roots appear frequently in mathematics, and understanding how to manipulate them is indispensable. In a fraction, square roots can often complicate simplification. When part of the numerator, as seen in this exercise, they require special techniques for simplification.

Square roots are represented as \(\sqrt{x}\), and denote a number which, when multiplied by itself, gives \(x\). The algebraic property of square roots that is most useful is their behavior under multiplication: \(\sqrt{a} \times \sqrt{b} = \sqrt{ab}\).

Knowing how to rationalize square root expressions by using conjugates can reveal hidden simplifications like the difference of squares.

This manipulation not only aids in calculus problems but also enhances problem-solving skills in algebraic contexts.