Problem 25 - Beginning and Intermediate Algebra

Question

Let \(f(x)=2 x^{2}+3 x-10\) and \(g(x)=3 x-5 .\) Find a) \((f \circ g)(x)\) b) \(\quad(g \circ f)(x)\) c) \((f \circ g)(1)\)

Step-by-Step Solution

Answer

a) \( (f \circ g)(x) = 2(3x - 5)^2 + 3(3x - 5) - 10\) b) \( (g \circ f)(x) = 3(2x^2 + 3x - 10) - 5\) c) \( (f \circ g)(1) = -7\)

Step 1: Find (f ∘ g)(x)

To find the function composition \((f \circ g)(x)\), we need to plug \(g(x)\) into \(f(x)\). So, we have: \[ (f \circ g)(x) = f(g(x)) = f(3x - 5) \] Now we'll plug \(3x - 5\) into \(f(x) = 2x^2 + 3x - 10\): \[ (f \circ g)(x) = 2(3x - 5)^2 + 3(3x - 5) - 10 \]

Step 2: Find (g ∘ f)(x)

To find the function composition \((g \circ f)(x)\), we need to plug \(f(x)\) into \(g(x)\). So, we have: \[ (g \circ f)(x) = g(f(x)) = g(2x^2 + 3x - 10) \] Now we'll plug \(2x^2 + 3x - 10\) into \(g(x) = 3x - 5\): \[ (g \circ f)(x) = 3(2x^2 + 3x - 10) - 5 \]

Step 3: Compute (f ∘ g)(1)

To find the value of \((f \circ g)(1)\), we'll plug in \(1\) into the expression we found in Step 1. So, we have: \[ (f \circ g)(1) = 2(3(1) - 5)^2 + 3(3(1) - 5) - 10 \] Now we'll compute the value. In conclusion, we have found: a) \((f \circ g)(x) = 2(3x - 5)^2 + 3(3x - 5) - 10\) b) \((g \circ f)(x) = 3(2x^2 + 3x - 10) - 5\) c) \((f \circ g)(1) = -7\)

Key Concepts

Algebraic Functions

Algebraic functions are mathematical expressions that involve variables and constants, using operations like addition, subtraction, multiplication, division, and powers. In the exercise given, the functions \(f(x) = 2x^2 + 3x - 10\) and \(g(x) = 3x - 5\) are examples of algebraic functions.
These functions can model real-world phenomena and are used extensively in algebra.
They can be combined or manipulated to form new functions, allowing for greater flexibility in mathematical modeling.

When working with algebraic functions in the context of composition, like \((f \circ g)(x)\) or \((g \circ f)(x)\), it involves using these basic operations to create new composite expressions.

Mathematical Notation

Mathematical notation is a system of symbols used to represent numbers, operations, and relationships in a clear and concise way.
In our exercise, we used notation like \(f(x)\) and \(g(x)\) to denote algebraic functions.
These symbols define the relationship between inputs and outputs of a function.

Function composition is shown using the notation \((f \circ g)(x)\), which means that you first apply \(g(x)\) and then apply \(f(x)\) to the result.
This notation is vital as it provides a standardized way to express complex mathematical ideas succinctly.

Understanding this notation is crucial, especially when substituting one function into another.

Substitution

Substitution is a mathematical technique used to simplify the process of evaluating functions or solving equations.
In function composition, substitution involves replacing the variable in one function with another function.

For instance, when finding \((f \circ g)(x)\), you replace every occurrence of \(x\) in \(f(x)\) with \(g(x)\) or \(3x - 5\), leading to:

• \((f \circ g)(x) = f(3x - 5)\)

This method allows for evaluating the outcome of two functions being performed in succession.

Similarly, substitution in \((g \circ f)(x)\) means inserting \(f(x)\) into each \(x\) of \(g(x)\). This step-by-step replacement provides a clear pathway to obtain the composite functions.

Evaluating Functions

Evaluating functions involves determining the value of a function at a specific point or simplifying a composite function.
This is particularly important in the context of function composition, where it allows us to find the result after multiple operations.

In the original exercise, evaluating \((f \circ g)(1)\) requires substituting \(1\) for \(x\) in the expression after having substituted \(g(x)\) into \(f(x)\), leading to the final calculation step:

• \((f \circ g)(1) = 2(3(1) - 5)^2 + 3(3(1) - 5) - 10 = -7\)

Understanding how to evaluate the functions properly is critical for solving algebraic problems and verifying solutions.

These evaluations not only confirm the accuracy of the solution but also strengthen your grasp of function interplay.