Problem 58 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(f(x)=\frac{x}{\sqrt{x}-1}\). Find and simplify. (a) \(f\left(\frac{1}{x}\right)\) (b) \(f(f(x))\)

Step-by-Step Solution

Answer

(a) \(-\frac{1}{1 - \sqrt{x}}\); (b) \(1 + \frac{1}{\sqrt{x} - 1}\).

Step 1: Find \(f\left(\frac{1}{x}\right)\)

To find \(f\left(\frac{1}{x}\right)\), substitute \(x = \frac{1}{x}\) into \(f(x)\). The given function is \(f(x) = \frac{x}{\sqrt{x} - 1}\). Replacing \(x\) with \(\frac{1}{x}\), we get:\[f\left(\frac{1}{x}\right) = \frac{\frac{1}{x}}{\sqrt{\frac{1}{x}} - 1} = \frac{1}{x} \times \frac{1}{\frac{1}{\sqrt{x}} - 1}\]Simplifying the denominator:\[\frac{1}{\sqrt{x}} - 1 = \frac{1-\sqrt{x}}{\sqrt{x}}\]Thus, \(f\left(\frac{1}{x}\right)\) becomes:\[f\left(\frac{1}{x}\right) = \frac{1}{x} \times \frac{\sqrt{x}}{1-\sqrt{x}} = \frac{\sqrt{x}}{x(1-\sqrt{x})}\]

Step 2: Simplification of (a)

Let's further simplify the expression from Step 1.\[f\left(\frac{1}{x}\right) = \frac{\sqrt{x}}{x(1-\sqrt{x})}\]To simplify, multiply numerator and denominator by \(-(\sqrt{x} - 1)\) to rationalize:\[f\left(\frac{1}{x}\right) = -\frac{\sqrt{x}}{x(\sqrt{x} - 1)} = -\frac{1}{x}\left(\frac{1}{\sqrt{x} - 1}\right) = -\frac{1}{1 - \sqrt{x}}\]

Step 3: Find \(f(f(x))\)

Start by substituting \(f(x)\) into itself. Given \(f(x) = \frac{x}{\sqrt{x} - 1}\), we substitute \(x\) with \(f(x)\):\[f(f(x)) = f\left(\frac{x}{\sqrt{x} - 1}\right)\]Substitute into the function:\[f(f(x)) = \frac{\frac{x}{\sqrt{x} - 1}}{\sqrt{\frac{x}{\sqrt{x} - 1}} - 1}\]

Step 4: Simplification of \( f(f(x)) \)

Now simplify the expression from Step 3. First, simplify the new function:\[\sqrt{\frac{x}{\sqrt{x} - 1}} = \frac{\sqrt{x}}{(\sqrt{x} - 1)^{1/2}}\]Thus, the inner function simplifies as:\[f(f(x)) = \frac{x}{\sqrt{x} - 1} \div \left(\frac{\sqrt{x}}{(\sqrt{x} - 1)^{1/2}} - 1\right)\]Multiply by the conjugate to rationalize, and simplify, resulting in:\[f(f(x)) = \frac{x(\sqrt{x} - 1)^{1/2}}{x - \sqrt{x}(\sqrt{x} - 1)}\]This simplifies further to recognize that the x terms cancel, leading the reduced expression to:\[f(f(x)) = 1 + \frac{1}{\sqrt{x} - 1}\]when fully simplified.

Key Concepts

Rationalization

Rationalization in mathematics is a process often used to eliminate radicals, especially from the denominator of fractions. This process simplifies the expressions and makes them easier to manage. When you rationalize, you're essentially getting rid of the square root or other irrational elements in the denominator by multiplying the numerator and the denominator by a suitable expression.For instance, consider the step in the given solution where we rationalize the expression \(f\left(\frac{1}{x}\right) = \frac{\sqrt{x}}{x(1-\sqrt{x})}\). Here, rationalization is achieved by multiplying by \(-1(-\sqrt{x} + 1)\), which effectively gets rid of the square root in the denominator.

• Why rationalize: It smoothens calculations and comparisons by providing a standard form.
• Process: Multiply top and bottom by the conjugate of the denominator.

This strategy is quite common in simplifying expressions that involve square roots or other irrational numbers.

Simplification

Simplifying expressions is key to making mathematical problems more accessible and solvable. Simplification involves reducing an expression to its simplest form, which often requires rationalization and combining like terms or reducing fractions.In the original solution, simplification occurs in multiple steps such as transforming \(f\left(\frac{1}{x}\right) = -\frac{1}{1 - \sqrt{x}}\) after rationalizing the previous form. This step highlights the essence of simplification as it breaks down the expression to a simpler structure without altering its value.

• Combining like terms: Look for opportunities to merge terms with similar variables or powers.
• Reducing fractions: Break down both numerator and denominator then cancel common factors.

Simplification is necessary to derive conclusions from complex algebraic expressions more accurately and swiftly.

Substitution

Substitution is a crucial technique in algebra, allowing you to replace one element of an expression with another to either simplify or solve it. It helps to transform a complex problem into a potentially easier version by temporarily or permanently replacing variables.In our task, one example is substituting \(x\) with \(\frac{1}{x}\) in \(f(x)\) for \(f\left(\frac{1}{x}\right)\). This substitution changes the perspective from the original function \(f(x)\), manipulating it to explore behavior at different variable forms.Additionally, in the case of finding \(f(f(x))\), substitution plays a pivotal role again. Here, one substitutes the entire function \(f(x) = \frac{x}{\sqrt{x} - 1}\) back into itself. This process illustrates how substitution enables complex compositions and entire function layers to be addressed effortlessly by altering component parts.

• Approach: Fit a simpler or temporary expression to reframe the problem.
• Benefit: Simplifies decision-making and potentially reveals patterns or simplifies solutions.