Problem 14 - 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) \(g(f(3))\) $$ f(x)=3 x-1, \quad g(x)=4 x^{2} $$

Step-by-Step Solution

Answer

(a) 12x^2 - 1; (b) 36x^2 - 24x + 4; (c) 47; (d) 256.

Step 1: Understand Functions and Composition

We have two functions: \( f(x) = 3x - 1 \) and \( g(x) = 4x^2 \). Composition of functions involves plugging one function into the other. Here, we need to find four different compositions.

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

For \((f \circ g)(x)\), we need to substitute \(g(x)\) into \(f(x)\). Start by replacing \(x\) in \(f(x) = 3x - 1\) with \(g(x) = 4x^2\).\[(f \circ g)(x) = f(g(x)) = f(4x^2) = 3(4x^2) - 1 = 12x^2 - 1\]

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

For \((g \circ f)(x)\), we substitute \(f(x)\) into \(g(x)\). Replace \(x\) in \(g(x) = 4x^2\) with \(f(x) = 3x - 1\).\[(g \circ f)(x) = g(f(x)) = g(3x - 1) = 4(3x - 1)^2\]Expand and simplify:\[4(3x - 1)^2 = 4(9x^2 - 6x + 1) = 36x^2 - 24x + 4\]

Step 4: Evaluate \(f(g(-2))\)

First, compute \(g(-2)\):\[g(-2) = 4(-2)^2 = 4(4) = 16\]Now, substitute \(g(-2) = 16\) into \(f(x)\):\[f(16) = 3(16) - 1 = 48 - 1 = 47\]

Step 5: Evaluate \(g(f(3))\)

First, find \(f(3)\):\[f(3) = 3(3) - 1 = 9 - 1 = 8\]Then, substitute \(f(3) = 8\) into \(g(x)\):\[g(8) = 4(8)^2 = 4(64) = 256\]

Step 6: Conclusion

Now we have the solutions for all parts of the exercise:(a) \((f \circ g)(x) = 12x^2 - 1\)(b) \((g \circ f)(x) = 36x^2 - 24x + 4\)(c) \(f(g(-2)) = 47\)(d) \(g(f(3)) = 256\)

Key Concepts

Algebra

Algebra is a branch of mathematics that helps us describe relationships using symbols and operations. It allows us to solve equations and understand mathematical concepts in a clear and structured way.
In algebra, we work with variables, which can represent numbers in expressions or equations. This flexibility is key in solving problems. Algebraic expressions involve numbers, variables, and operators like addition, subtraction, multiplication, and division.
For example, in the functions given, \( f(x) = 3x - 1 \) and \( g(x) = 4x^2 \), we see algebra at play through operations involving variables and constants. Understanding the syntax and structure of algebraic expressions is essential when performing operations like composition.

Functions

Functions in mathematics are like rules or machines that take an input, perform a calculation, and provide an output. They are typically expressed as \( f(x) \), where \( x \) is the input variable.
In our case, we have two functions: \( f(x) = 3x - 1 \) and \( g(x) = 4x^2 \). Each of these functions has its own rule:

• For \( f(x) \), you multiply \( x \) by 3 and subtract 1.
• For \( g(x) \), you square \( x \) and then multiply by 4.

These functions demonstrate how different equations outline processes to transform an input into a specific output. Understanding how functions operate is foundational for exploring more complex operations like composition.

Composition of Functions

Composition of functions is an advanced topic in algebra where the output of one function becomes the input of another. This process allows us to build more complex functions from simpler ones.
For example, in this exercise, we look at \((f \circ g)(x)\) and \((g \circ f)(x)\). This is read as "\(f\) composed with \(g\)" and "\(g\) composed with \(f\)", respectively. The notation often appears as \((f \circ g)(x)\), showing that we first apply \(g(x)\) and then \(f(x)\).

• \((f \circ g)(x) = f(g(x)) = 12x^2 - 1\) means that you take \(g(x)\), apply it first, and then use the resulting value as the input for \(f(x)\).
• \((g \circ f)(x) = g(f(x)) = 36x^2 - 24x + 4\) shows the reverse order, where \(f(x)\)'s output becomes the input for \(g(x)\).

Understanding these operations helps illustrate how outputs and processes can influence further calculations and results.

Substitution in Functions

Substitution in functions involves replacing the variable in a function's equation with a specific value or another expression. This is a key technique used in evaluating functions at specific points as well as in function composition.
For example, when evaluating \( f(g(-2)) \), you substitute \(-2\) into \( g(x) \):

• First, find \( g(-2) = 16 \), then use this result in \( f(x): f(16) = 47 \).

Similarly, for \( g(f(3)) \), you substitute \(3\) into \(f(x)\):

• Calculate \( f(3) = 8 \), and then find \( g(8) = 256 \).

Substitution simplifies function evaluations and is immensely helpful when 'plugging in' values, whether this is for specific calculations or transferring outputs for further use. This is integral in solving complex algebraic problems involving multiple layers of operations.