Problem 29 - Algebra 2

Question

Find \([g \circ h](x)\) and \([h \circ g](x)\) $$ \begin{array}{l}{g(x)=-5 x} \\ {h(x)=-3 x+1}\end{array} $$

Step-by-Step Solution

Answer

\([g \circ h](x) = 15x - 5\) and \([h \circ g](x) = 15x + 1\).

Step 1: Identify the functions

First, identify the given functions: \(g(x) = -5x\) and \(h(x) = -3x + 1\). These are the functions we'll be using in our compositions \([g \circ h](x)\) and \([h \circ g](x)\).

Step 2: Compute \([g \circ h](x)\)

Composition \([g \circ h](x)\) means we need to substitute \(h(x)\) into \(g(x)\). This means we calculate \(g(h(x))\). Substituting \(h(x) = -3x + 1\) into \(g(x) = -5x\) gives us \(g(h(x)) = g(-3x + 1) = -5(-3x + 1)\). Now, distribute to simplify: \(-5 \times -3x = 15x\) and \(-5 \times 1 = -5\). Therefore, \([g \circ h](x) = 15x - 5\).

Step 3: Compute \([h \circ g](x)\)

Composition \([h \circ g](x)\) means we substitute \(g(x)\) into \(h(x)\). Therefore, calculate \(h(g(x))\). Substituting \(g(x) = -5x\) into \(h(x) = -3x + 1\) gives us \(h(g(x)) = h(-5x) = -3(-5x) + 1\). Simplify by distributing: \(-3 \times -5x = 15x\) and add \(+1\) to get \([h \circ g](x) = 15x + 1\).

Key Concepts

Composite Functions

Composite functions are an essential concept in algebra that involve combining two or more functions. This process allows one to evaluate the outcome by substituting one function into another. In algebra, this is represented typically as \(f \circ g\), which is read as "f of g" or "f composed with g." This process helps in understanding how multiple processes can be combined to arrive at a final output.
To compute composite functions, you substitute the output of one function into the input of another function. For instance, if you have two functions, \(g(x) = -5x\) and \(h(x) = -3x + 1\), you can create composites like \([g \circ h](x)\) and \([h \circ g](x)\).

• For \(g \circ h\), substitute \(h(x)\) into \(g(x)\), resulting in \(g(h(x)) = 15x - 5\).
• For \(h \circ g\), substitute \(g(x)\) into \(h(x)\), giving \(h(g(x)) = 15x + 1\).

Composites allow you to evaluate complex combinations by simplifying calculations into sequential steps.

Algebraic Functions

Algebraic functions are equations composed of polynomials. They involve operations such as addition, subtraction, multiplication, and division integrated into variable expressions. For the given functions, \(g(x) = -5x\) is a linear algebraic function as it represents a straightforward polynomial (a line) with a slope and passes through the origin.
Larger algebraic functions can be manipulated similarly by substituting variables, constants, and other function expressions.
Algebraically:

• \(g(x) = -5x\) showcases a simple linear function with only a multiplication operation impacting the variable.
• \(h(x) = -3x + 1\) presents another linear polynomial with both multiplication and an additional constant shift (+1).

This manipulation of algebraic functions through substitution and composite formation exemplifies their flexibility and power in algebra.

Function Operations

Function operations refer to the various ways you can manipulate and work with functions, such as through addition, subtraction, multiplication, division, and composition. These operations are foundational in algebra because they reveal how functions interact and how they can be transformed or combined.
In the context of the given exercise, we focus on the operation of composition seen when combining \(g(x)\) and \(h(x)\) into different compositions:

• Composition: This involves substituting the outcome of one function directly into another, allowing the operational properties of both to manifest in the result.
• The operations of distributing and commutating coefficients (such as \(-5\) and \(-3\) in these functions) highlight how different algebraic manipulations are released during substitution, creating new functions like \(15x - 5\) and \(15x + 1\).

Understanding function operations, particularly composite functions, provides insight into function interactions, helping solve more complex mathematical problems effectively.