Problem 13 - Thomas Calculus

Question

Copy and complete the following table. $$\begin{array}{ll}{g(x)} & {f(x)} & {(f \circ g)(x)} \\ \hline\end{array}$$ $$ \begin{array}{ll}{\text { a. } x-7} & {\sqrt{x}} \\ {\text { b. } x+2} & {3 x} \\ {\text { c. } ?} & {\sqrt{x-5}} & {\sqrt{x^{2}-5}}\end{array} $$ $$ \text { d. } \frac{x}{x-1} \quad \frac{x}{x-1} \quad ? $$ $$ \text { e. }? \quad 1+\frac{1}{x} \quad x $$ $$ \begin{array}{ll}{\text { f. } \frac{1}{x}} & {?} & {x}\end{array} $$

Step-by-Step Solution

Answer

c. g(x) = x^2; d. (f ∘ g)(x) = \frac{x}{x-2}; e. g(x) = \frac{1}{x-1}; f. f(x) = \frac{1}{x}.

Step 1: Composition of functions definition

The composition of functions \(f \circ g\)(x) means applying function \(g\) and then applying function \(f\) to the result. Hence \( (f \circ g)(x) = f(g(x)) \).

Step 2: Solving for c

In part (c), we have \(f(x) = \sqrt{x-5}\) and \( (f \circ g)(x) = \sqrt{x^2 - 5}\). To solve for \(g(x)\), we set \( \sqrt{g(x) - 5}= \sqrt{x^2 - 5} \), thus \( g(x) = x^2\).

Step 3: Solving for d

For part (d), we have both \(g(x) = \frac{x}{x-1} \) and \(f(x) = \frac{x}{x-1} \). Therefore, \( (f \circ g)(x) = f(g(x)) = \frac{\left(\frac{x}{x-1}\right)}{\left(\frac{x}{x-1}\right) - 1} = \frac{x}{x-2}\).

Step 4: Solving for e

In part (e), \(f(x) = 1 + \frac{1}{x}\) and \( (f \circ g)(x) = x \). To find \(g(x)\), note that if \(f(g(x)) = x\) and \(f(x) = 1 + \frac{1}{x}\), we need \((1 + \frac{1}{g(x)}) = x\). Solving gives \(g(x) = \frac{1}{x-1} \).

Step 5: Solving for f

For part (f), we have \(g(x) = \frac{1}{x}\) and \( (f \circ g)(x) = x \). Thus \(f(g(x)) = x\) becomes \(f\left(\frac{1}{x}\right) = x \). Hence we need to find \(f(x)\) such that \(f(x) = \frac{1}{x} \), resulting in the identity function requirement.

Key Concepts

Inverse Functions

Inverse functions are about reversing the effect of a function. Imagine you have a function that turns apples into apple juice. The inverse function would turn apple juice back into apples. Mathematically, if you have a function \( f(x) \) and its inverse is \( f^{-1}(x) \), then applying \( f \) and then \( f^{-1} \) will get you back to your starting point: \( f(f^{-1}(x)) = x \).

In our exercise, identifying inverse functions helps solve parts like "(e)" and "(f)". Consider "(e)": we want \( f(g(x)) = x \), which means \( g(x) \) should undo the effect of \( f(x) = 1 + \frac{1}{x} \). This leads us to find \( g(x) = \frac{1}{x-1} \), as substituting it into \( f \) brings us back to \( x \).

The pattern of working with inverse functions is essential in simplifying equations and solving problems, especially where you need to find a function that reverses another's effect.

Algebraic Manipulation

Algebraic manipulation is the art and science of rearranging equations to find unknown variables. Think of it as untangling a knot or reshuffling the pieces of a puzzle until everything fits perfectly.

Let's take part "(c)" of the exercise for example. Here, we start with \( f(x) = \sqrt{x-5} \) and need \( (f \circ g)(x) = \sqrt{x^2 - 5} \). To solve for \( g(x) \), you equate \( \sqrt{g(x) - 5} = \sqrt{x^2 - 5} \), and through algebraic manipulation, simplify the equation to find \( g(x) = x^2 \).

By carefully rearranging terms, making legal operations like adding, subtracting, multiplying, or using roots, we can isolate variables or simplify complex expressions. This manipulation ensures we arrive at the correct solution or function necessary for the given problem.

Square Root Function

The square root function is a particular type of function that involves finding a number, which when squared, gives the original number. For example, the square root of 16 is 4, because \( 4^2 = 16 \). The function itself is expressed as \( f(x) = \sqrt{x} \).

In the context of the exercise, the square root function is integral in sections like "(c)" and "(d)". Take "(c)": the function \( f(x) = \sqrt{x-5} \) transforms its input by removing 5, then finding the square root. When composed with \( g(x) = x^2 \), which previously we found, it ensures the requirement \( (f \circ g)(x) = \sqrt{x^2 - 5} \) is satisfied.

• Functions like \( f(x) = \sqrt{x} \) are crucial for expressing operations dealing with areas, geometry, and real-world scenarios like population models.
• The challenge lies in managing square roots in algebraic expressions, which sometimes require squaring both sides to remove the radicals.

Being familiar with how square root functions operate can greatly enhance your problem-solving toolkit, especially in algebra and calculus contexts.