Problem 10 - Intermediate Algebra

Question

Determine \((f \circ g)(x)\) and \((g \circ f)(x)\) for each pair of functions. Also specify the domain of \((f \circ g)(x)\) and \((g \circ f)(x)\). (Objective 1\()\) \(f(x)=-\frac{2}{x}\) and \(g(x)=-3 x+6\)

Step-by-Step Solution

Answer

\((f \circ g)(x) = \frac{2}{3x - 6}, x \neq 2\). \((g \circ f)(x) = \frac{6}{x} + 6, x \neq 0\).

Step 1: Understanding Function Composition

Function composition involves plugging one function into another. For example, \((f \circ g)(x)\) means plugging \(g(x)\) into \(f(x)\). Similarly, \((g \circ f)(x)\) is plugging \(f(x)\) into \(g(x)\).

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

To find \((f \circ g)(x)\), substitute \(g(x)\) into \(f(x)\):\[f(g(x)) = f(-3x + 6) = -\frac{2}{-3x + 6}\]

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

Simplifying \(-\frac{2}{-3x + 6}\), we should recognize there is no further simplification possible without additional instructions. Thus:\[f(g(x)) = \frac{2}{3x - 6}\]

Step 4: Determine the Domain of \((f \circ g)(x)\)

The domain of a composed function is restricted by the domain of the inner function and any restrictions from the composition. \(g(x)\) is defined for all real \(x\), but \(f(g(x))\) must not have a denominator equal to zero.Set the denominator according to \(f(g(x))\): \[3x - 6 eq 0\] Solve: \[3x eq 6\,\,\,x eq 2\] Thus, the domain of \(f \circ g\) is all real numbers except \(x eq 2\).

Step 5: Calculate \((g \circ f)(x)\)

To calculate \((g \circ f)(x)\), substitute \(f(x)\) into \(g(x)\): \[g(f(x)) = g\left(-\frac{2}{x}\right) = -3\left(-\frac{2}{x}\right) + 6\]

Step 6: Simplify \((g \circ f)(x)\)

Simplifying \(-3\left(-\frac{2}{x}\right) + 6\) gives:\[g(f(x)) = \frac{6}{x} + 6\]

Step 7: Determine the Domain of \((g \circ f)(x)\)

The domain of a composed function requires that the inner function \(f(x)\) must be defined, which excludes \(x = 0\). Since the result \(\frac{6}{x} + 6\) has no other restrictions, the domain of \((g \circ f)(x)\) is all real numbers except \(x eq 0\).

Key Concepts

Domain of Functions

When considering functions, the domain is a crucial concept. It represents all the possible input values (x-values) that a function can accept without running into issues, such as division by zero or taking an even root of a negative number.

For the composition \((f \circ g)(x)\), the function \(f\) is applied to the output of \(g\). This means that the domain of \((f \circ g)(x)\) must account for two key factors:

• The domain of \(g(x)\): Since \(g(x) = -3x + 6\) is a linear function, it technically can accept all real numbers.
• Restrictions from \(f(g(x))\): After \(g(x)\) is calculated, it is then plugged into \(f(x)\). For \(f(x) = -\frac{2}{x}\), \(f\) itself is undefined if the denominator is zero. Hence, we set \(3x - 6 eq 0\), solving this gives \(x eq 2\).

Therefore, the domain of \((f \circ g)(x)\) excludes \(x = 2\). For \((g \circ f)(x)\), \(f(x)\) limits the domain since it cannot be divided by zero, hence \(x eq 0\). No further exclusions arise from \(g(f(x))\). Thus, \((g \circ f)(x)\) is defined for all \(x eq 0\).

Rational Functions

Rational functions are quotients of two polynomials. They are written as \(R(x) = \frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials.

In rational functions, the main issue to watch for is when the denominator, \(Q(x)\), becomes zero because division by zero is undefined. This defines the restrictions within the domain.

For our functions, \(f(x) = -\frac{2}{x}\) and part of the composition as \((f \circ g)(x) = \frac{2}{3x - 6}\), these are examples of rational functions. Notice how each function's denominator indicates where the function is undefined by equating \(Q(x) eq 0\):

• For \(f(x)\), \(x eq 0\), since \(0\) cannot be a denominator.
• For \((f \circ g)(x)\), \(3x - 6 eq 0\) simplifies to \(x eq 2\).

Understanding the nature of rational functions helps in predicting function behavior and any restrictions on input values.

Algebraic Simplification

Simplification in algebra involves rewriting expressions in their simplest form to make calculations easier and more comprehensible.

When dealing with rational functions like those in the exercise, simplification often involves manipulating both the numerator and the denominator.

For \((f \circ g)(x) = \frac{2}{3x - 6}\), look for common factors that can be canceled out. In our example, the expression is already simplified as there is nothing common in the numerator and denominator.

In simplifying \((g \circ f)(x)\) from \(-3\left(-\frac{2}{x}\right) + 6\) to \(\frac{6}{x} + 6\), perform the operations step-by-step: distribute \(-3\), then simplify any like terms.

Simplification often prepares functions for most practical applications, ensuring clearer understanding and ease in further operations or evaluations.