Problem 20 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 11-20: Find (a) \((f \circ g)(x)\) (b) \((g \circ f)(x)\) (c) \(f(g(-2))\) (d) \(g(f(3))\) $$ f(x)=5, \quad g(x)=x^{2} $$

Step-by-Step Solution

Answer

(a) 5, (b) 25, (c) 5, (d) 25.

Step 1: Find \((f \\circ g)(x)\)

The composite function \((f \circ g)(x)\) means that we substitute \(g(x)\) into \(f(x)\). Since \(f(x) = 5\) and is a constant function, regardless of the input to \(f\), the output is always 5. Hence, \((f \circ g)(x) = 5\).

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

The composite function \((g \circ f)(x)\) requires substituting \(f(x)\) into \(g(x)\). Since \(f(x) = 5\), we replace \(x\) in \(g(x)\) with 5. Thus, \(g(5) = 5^2 = 25\). Therefore, \((g \circ f)(x) = 25\).

Step 3: Evaluate \(f(g(-2))\)

First, find \(g(-2)\). Since \(g(x) = x^2\), substitute \(-2\) to get \(g(-2) = (-2)^2 = 4\). Then, evaluate \(f(4)\). Since \(f(x) = 5\) for any \(x\), \(f(4) = 5\). Thus, \(f(g(-2)) = 5\).

Step 4: Evaluate \(g(f(3))\)

First, find \(f(3)\). Since \(f(x) = 5\) for any \(x\), \(f(3) = 5\). Then, substitute into \(g\), giving \(g(5) = 5^2 = 25\). Therefore, \(g(f(3)) = 25\).

Key Concepts

Constant Function

A constant function is one of the simplest types of functions you can encounter in mathematics. Regardless of the input value, the output of a constant function is always the same. For instance, when dealing with the function \( f(x) = 5 \), no matter what value \( x \) takes, the function will always output 5. This property makes constant functions unique since they do not depend on the input value as other types of functions might.

• Output is always the same constant, regardless of the input.
• Graphically, a constant function appears as a horizontal line on a coordinate plane.
• In our example: \( f(x) = 5 \) makes \( f(x) \) a constant function.

Constant functions are straightforward and serve as excellent examples to simplify the understanding of other concepts, particularly when learning about operations such as composition.

Function Evaluation

Function evaluation is the process of determining the output of a function for a specific input value. Simply put, it's substituting a given number for the variable in the function and calculating the result. Suppose you have a function \( g(x) = x^2 \). To evaluate this function at \( x = -2 \), you substitute \(-2\) into the function: \( g(-2) = (-2)^2 = 4 \). Let's break it down further:

• Identify the function and its rule (e.g., \( g(x) = x^2 \)).
• Substitute the given input value into the function (e.g., replace \( x \) with \(-2\)).
• Perform the necessary calculations to find the output.

Function evaluation is a key skill in mathematics, especially in calculus and algebra, because it helps determine how functions behave at specific points.

Substitution Method

The substitution method is a fundamental technique used when working with composite functions. It involves replacing one part of a function with another value or expression to make further calculations.Let's consider the composite function \( (f \circ g)(x) \). Here, substitution means plugging \( g(x) \) into \( f(x) \). Using the functions from our example, \( f(x) = 5 \) and \( g(x) = x^2 \):

• Determine \( g(x) \) and substitute it into \( f(x) \).
• Since \( f(x) \) is a constant \( 5 \), the output remains 5, no matter what \( g(x) \) might be.

The substitution method simplifies the process of evaluating expressions, allowing easier derivation of complex function relationships.

Composite Function Notation

Composite function notation is a way to express the operation of one function within another. The notation \( (f \circ g)(x) \) represents a composite function, meaning \( f \) composed with \( g(x) \). This reads as "f of g of x."Here's a step-by-step explanation:

• \( g(x) \) is computed first, taking the value \( x \) as its input.
• The result from \( g(x) \) is then used as the input for \( f \), resulting in \( f(g(x)) \).
• This process may be reversed, as in \( g(f(x)) \), where the output of \( f(x) \) becomes the input for \( g \).

Composite functions are extremely useful in mathematics, especially for creating complex functions built from simpler components. This method is widely used in calculus, functional analysis, and various applications across different fields.