Problem 17 - 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) \(g(f(3))\) $$ f(x)=4 x, \quad g(x)=2 x^{3}-5 x $$

Step-by-Step Solution

Answer

(f∘g)(x)=8x^3−20x; (g∘f)(x)=128x^3−20x; f(g(−2))=−24; g(f(3))=3396.

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

To find \((f \circ g)(x)\), substitute \(g(x)\) into \(f(x)\):\[f(g(x)) = f(2x^3 - 5x)\]Substitute into \(f(x) = 4x\):\[f(g(x)) = 4(2x^3 - 5x) = 8x^3 - 20x\]So, \((f \circ g)(x) = 8x^3 - 20x\).

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

For \((g \circ f)(x)\), substitute \(f(x)\) into \(g(x)\):\[g(f(x)) = g(4x)\]Using \(g(x) = 2x^3 - 5x\), substitute to get:\[g(f(x)) = 2(4x)^3 - 5(4x)\]Calculate further:\[= 2(64x^3) - 20x = 128x^3 - 20x\]Thus, \((g \circ f)(x) = 128x^3 - 20x\).

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

First calculate \(g(-2)\):\[g(-2) = 2(-2)^3 - 5(-2) = 2(-8) + 10 = -16 + 10 = -6\]Now substitute into \(f(x)\):\[f(-6) = 4(-6) = -24\]So, \(f(g(-2)) = -24\).

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

First calculate \(f(3)\):\[f(3) = 4(3) = 12\]Now substitute into \(g(x)\):\[g(12) = 2(12)^3 - 5(12)\]Calculate further:\[= 2(1728) - 60 = 3456 - 60 = 3396\]Thus, \(g(f(3)) = 3396\).

Key Concepts

Algebraic Functions

Algebraic functions are expressions constructed using algebraic operations such as addition, subtraction, multiplication, division, and raising to whole number powers. These functions are defined by polynomials, rational expressions, or root functions. In general, algebraic functions can take different forms and are used widely in calculus, algebra, and many real-world applications.
For instance, consider the functions:

• \( f(x) = 4x \): This is a linear algebraic function, as it represents a straight line when plotted on a graph.
• \( g(x) = 2x^3 - 5x \): This function involves a cubic term, making it a polynomial of degree 3.

The complexity and behavior of these functions can vary greatly, from simple linear trends to more intricate curves. Understanding these basic forms is crucial for manipulating and analyzing functions.

Function Evaluation

Function evaluation is the method of determining the output of a function for a specific input value. This calculation provides insights into how the function behaves for different values and is essential for plotting its graph or modeling situations mathematically.
In our exercise, evaluating functions involves substituting a specific number into the function's equation. For example:

• To evaluate \( f(x) \) at \( x = 3 \), you substitute 3 into the function: \( f(3) = 4 imes 3 = 12 \).
• Similarly, for \( g(x) \) and \( x = -2 \), you find: \( g(-2) = 2(-2)^3 - 5(-2) = -16 + 10 = -6 \).

By performing these substitutions, you determine the value of the function output, a foundational skill in algebra and calculus that aids in comparing function outputs and predicting future values.

Composite Functions

Composite functions involve combining two functions such that the output of one function becomes the input of another. This process is denoted as \((f \circ g)(x)\) or \(f(g(x))\). The goal is to build more complex functions from simple ones to solve advanced problems or describe multiple-step processes.
For example:

• To find \( (f \circ g)(x) \), substitute \( g(x) \) into \( f(x) \), resulting in \( f(2x^3 - 5x) = 8x^3 - 20x \).
• Conversely, \( (g \circ f)(x) \) involves substituting \( f(x) \) into \( g(x) \), yielding \( g(4x) = 128x^3 - 20x \).

Combining functions in this way allows exploring more detailed relationships between variables, providing a powerful tool for modeling and problem-solving.

Substitution Method

The substitution method is a critical mathematical process for evaluating expressions within functions or solving equations. It involves replacing a variable with another expression or value to simplify or solve the problem. This technique is versatile and appears often in calculus, algebra, and other mathematical disciplines.
In our example, applying the substitution method is apparent when calculating composite functions or evaluating functions at certain points. For instance:

• Evaluating \( f(g(-2)) \): First, substitute \(-2\) into \( g(x) \), resulting in \( -6 \). Then, substitute this value into \( f(x) \) to find \( f(-6) = -24 \).
• Similarly, for \( g(f(3)) \), start by evaluating \( f(3) = 12 \) and then substitute 12 into \( g(x) \) to obtain \( g(12) = 3396 \).

Understanding the substitution method is key for simplifying complex expressions and solving multi-step functions effectively.