Problem 62 - A Graphical Approach to Precalculus with Limits

Question

Use \(f(x)\) and \(g(x)\) to find each composition. Identify is domain. (Use a calculator if necessary to find the domain.) \(\begin{array}{llll}\text { (a) }(f \circ g)(x) & \text { (b) }(g \circ f)(x) & \text { (c) }(f \circ f)(x)\end{array}\) $$f(x)=\frac{1}{x+1}, g(x)=3-6 x$$

Step-by-Step Solution

Answer

(a) \(\frac{1}{4 - 6x}\) with \( x \neq \frac{2}{3} \), (b) \(3 - \frac{6}{x+1}\) with \( x \neq -1 \), (c) \(\frac{1}{x+2}\) with \( x \neq -2 \).

Step 1: Understand Function Composition

Function composition involves plugging one function into another. For two functions \( f(x) \) and \( g(x) \): - \((f \circ g)(x)\) means substituting \( g(x) \) into \( f(x) \).- \((g \circ f)(x)\) means substituting \( f(x) \) into \( g(x) \).- \((f \circ f)(x)\) involves substituting \( f(x) \) into itself.

Step 2: Compute \((f \circ g)(x)\)

First, substitute \( g(x) = 3 - 6x \) into \( f(x) = \frac{1}{x+1} \). This gives: \[(f \circ g)(x) = f(g(x)) = \frac{1}{(3 - 6x) + 1} = \frac{1}{4 - 6x}\] The domain requires that the denominator is not zero: \[ 4 - 6x eq 0 \] Solving: \[ 6x eq 4 \quad \Rightarrow \quad x eq \frac{2}{3} \] Thus, the domain is all real numbers except \( x = \frac{2}{3} \).

Step 3: Compute \((g \circ f)(x)\)

Substitute \( f(x) = \frac{1}{x+1} \) into \( g(x) = 3 - 6x \). This yields: \[(g \circ f)(x) = g(f(x)) = 3 - 6\left(\frac{1}{x+1}\right) = 3 - \frac{6}{x+1}\]Ensure the denominator is not zero: \[ x+1 eq 0 \quad \Rightarrow \quad x eq -1 \] Hence, the domain is all real numbers except \( x = -1 \).

Step 4: Compute \((f \circ f)(x)\)

Substitute \( f(x) = \frac{1}{x+1} \) into itself: \[(f \circ f)(x) = f(f(x)) = \frac{1}{\left(\frac{1}{x+1}\right) + 1} = \frac{1}{\frac{1 + x + 1}{x+1}} = \frac{x+1}{x+2} = \frac{1}{x+2}.\] The domain condition involves setting the denominator not equal to zero: \[ x+2 eq 0 \quad \Rightarrow \quad x eq -2 \] Hence, the domain is all real numbers except \( x = -2 \).

Step 5: Conclusion

We have calculated all compositions:- \((f \circ g)(x) = \frac{1}{4 - 6x}\) with domain \( x eq \frac{2}{3} \)- \((g \circ f)(x) = 3 - \frac{6}{x+1}\) with domain \( x eq -1 \)- \((f \circ f)(x) = \frac{1}{x+2}\) with domain \( x eq -2 \)

Key Concepts

Domain of a Function

The domain of a function essentially refers to the collection of all possible inputs (usually represented as \(x\)) that a function can accept. For instance, when dealing with rational functions, you need to be cautious about the values of \(x\) that make the denominator zero.

When you perform operations on functions, like composing them, it's important to determine the domain for the resulting composition. This is done to ensure you don’t end up dividing by zero, which results in undefined values. Here's how you find the domain:

• Set the denominator of any rational expression to not equal zero (use \( eq \)).
• Solve for \(x\).
• The domain is all real numbers except for the value(s) that make the denominator zero.

For example, if after the function composition you have \(\frac{1}{4 - 6x}\), you would solve \(4 - 6x eq 0\), leading to \(x eq \frac{2}{3}\). Thus, the domain is all real numbers except \(\frac{2}{3}\).

Rational Functions

Rational functions are quotients of two polynomials, meaning they have the general form \(\frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials. What makes rational functions unique are their potential restrictions on the domain caused by the denominator.

These types of functions can have values of \(x\) that lead to a zero in the denominator, resulting in points where the function is undefined. To handle rational functions effectively:

• Always check the denominator for values that might make it zero.
• Examine the behavior of the function as it approaches these undefined points (e.g., asymptotic behavior).
• Perform simplifications cautiously, as cancelling terms affects domain restrictions.

For instance, for the original function \(f(x) = \frac{1}{x+1}\), the value that makes the fraction undefined is \(x = -1\), thus \(f\) has a domain of all real numbers except \(-1\). Incorporating this analysis into function operations helps avoid errors and ensures accurate calculations.

Function Operations

Function operations include addition, subtraction, multiplication, division, and composition of functions. These operations allow us to combine and modify functions in useful ways.

**Function Composition**
Composition of functions, often denoted by \( (f \circ g)(x)\), involves plugging one function into another. It's like stacking functions on top of each other.

• First, calculate the inner function, \(g(x)\).
• Then, feed the output of \(g(x)\) into \(f(x)\).
• Common issues arise when the inner function produces an output not in the domain of the outer function.

For example, with \((f \circ g)(x) = \frac{1}{4 - 6x}\), you must ensure that \(4 - 6x eq 0\) to avoid undefined values. This results in excluding \(x = \frac{2}{3}\) from the domain.

Being mindful of these operations allows for seamless function manipulation while maintaining accurate domains and avoiding undefined points. Understanding how these combine is essential in calculus and algebra for solving more complex problems.