Problem 62 - Intermediate Algebra

Question

Let \(f(x)=3 x-2\) and \(g(x)=x^{2}+x .\) Find each of the following. $$ (f \circ g)(0) $$

Step-by-Step Solution

Answer

The value of \((f \circ g)(0)\) is -2.

Step 1: Understanding Composition of Functions

To solve problems dealing with the composition of functions, we need to apply one function to the results of another. In this problem, we need to find \((f \circ g)(x)\), which is defined as \(f(g(x))\). This means we first apply \(g(x)\) and then apply \(f\) to the result.

Step 2: Calculate \(g(0)\)

We need to evaluate \(g(x)\) at \(x = 0\). Using the definition of \(g(x) = x^2 + x\), we substitute 0 for \(x\):\[g(0) = 0^2 + 0 = 0\]

Step 3: Calculate \(f(g(0))\)

Now we use the result from Step 2 for the function \(f(x) = 3x - 2\). Since \(g(0) = 0\), we need to find \(f(0)\):\[f(0) = 3(0) - 2 = -2\]

Step 4: Conclusion

We have the result from the above steps that \((f \circ g)(0) = f(g(0)) = -2\). This means when we apply \(g\) to 0 and then apply \(f\) to the result, we get -2 as the final answer.

Key Concepts

Algebraic Functions

Algebraic functions are defined by polynomials, which are expressions consisting of variables and coefficients. You might see such functions as simple expressions like linear or quadratic equations. For instance, in the exercise given, we have two algebraic functions:

• \( f(x) = 3x - 2 \) which is a linear function. It is characterized by the first-degree polynomial where "3" is the coefficient of \( x \) and "-2" is the constant term.
• \( g(x) = x^2 + x \) which is a quadratic function. This function involves \( x^2 \), making it a second-degree polynomial. "1" is the coefficient for both \( x^2 \) and \( x \).

Algebraic functions are essential in understanding relationships between variables and solving equations through substitution and manipulation of polynomials. In practice, understanding how these functions behave graphically and algebraically is crucial for breaking down complex problems into simpler, manageable parts. They form the basis of many real-world applications, from physics calculations to economic models.

Evaluating Functions

Evaluating functions means determining the output of a function for a particular input value. It's like asking, "What is \(f(x)\) when \(x\) is a specific number?" In the context of the problem, we first evaluate the function \(g(x)\) at \(x = 0\), which involves substituting 0 into the expression for \(g(x)\):

• \( g(0) = 0^2 + 0 = 0 \)

This step involves following the order of operations, often referred to as PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction). The output you get from this process can then be used in subsequent calculations.
Once we have \(g(0)\), we proceed to evaluate \(f(x)\) using the output from \(g(x)\). This gives us:

• \( f(g(0)) = f(0) = 3 \times 0 - 2 = -2 \)

Evaluating each function step by step simplifies the process and reduces errors, especially in more complex functions.

Composite Functions

Composite functions involve applying one function to the result of another, a fundamental concept in mathematics symbolized as \((f \circ g)(x)\), which means \(f(g(x))\). This approach allows us to create new functions by combining existing ones, seeing how one operation affects another.
In our problem, we find the composite function by plugging the output of \(g(x)\) into \(f(x)\):

• Start with \( g(x) = x^2 + x \).
• Calculate \( g(0) = 0 \).
• Using the result of \( g(0) \), compute \( f(g(0)) = f(0) = -2 \).

Composite functions allow us to explore relationships between different processes. They play a crucial role in advanced fields like calculus or physics, where you often deal with compositions of more intricate functions such as trigonometric, exponential, or logarithmic functions. Understanding composite functions is essential in simplifying, transforming, and solving equations efficiently. Just remember, composite operations must follow the inside-to-outside rule, solving the inner function first before applying the outer function.