Problem 57 - College Algebra Essentials

Question

Find a. \((f \circ g)(x) \quad \) b. \((g \circ f)(x) \quad \) c. \((f \circ g)(2) \quad \) d. \((g \circ f)(2)\) $$f(x)=x^{2}+2, g(x)=x^{2}-2$$

Step-by-Step Solution

Answer

\(a. (f \circ g)(x) = (x^{2}-2)^{2}+2 \ b. (g \circ f)(x) = (x^{2}+2)^{2}-2 \ c. (f \circ g)(2) = 4 \ d. (g \circ f)(2) = 16\)

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

To compute \(f \circ g\), substitute \(g(x)\) into \(f\). We get \(f(g(x))=(x^{2}-2)^{2}+2\).

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

Similarly, to compute \(g \circ f\), substitute \(f(x)\) into \(g\). We get \(g(f(x))=(x^{2}+2)^{2}-2\).

Step 3: Compute \((f \circ g)(2)\)

Plug \(x=2\) into \(f(g(x))\). We get \((f \circ g)(2)=(2^{2}-2)^{2}+2=4\).

Step 4: Compute \((g \circ f)(2)\)

Similarly, plug \(x=2\) into \(g(f(x))\). We get \((g \circ f)(2)=(2^{2}+2)^{2}-2=16\).

Key Concepts

Composition of Functions

The composition of functions is a fundamental concept where one function is applied to the results of another function. This creates a new function. For instance, if we have two functions, say \(f\) and \(g\), their composition can be either \(f \circ g\) or \(g \circ f\). The notation \(f \circ g(x)\) means that you first evaluate \(g(x)\) and then plug the result into \(f(x)\). Conversely, \(g \circ f(x)\) involves evaluating \(f(x)\) first and then using that output as the input for \(g(x)\). This can be very useful when dealing with multiple transformations or processes in mathematics.

Here are quick steps to compose functions:

• Identify which function to evaluate first based on the notation, i.e., \(f \circ g\) or \(g \circ f\).
• Calculate the output of the first function.
• Use that output as the input of the second function to find the final result.

By practicing this approach, you'll be crafting new and dynamic functions from those you already know.

Evaluating Functions

Evaluating a function involves calculating the output of a function given a specific input. Let's see how this works with composition. In our exercise, you were given \(f(x) = x^2 + 2\) and \(g(x) = x^2 - 2\).

To evaluate \((f \circ g)(x)\), substitute the entire function \(g(x) = x^2 - 2\) into \(f\). So, you get \(f(g(x)) = (x^2 - 2)^2 + 2\). Remember, the outer function takes the result of the inner function as its input.

Similarly, to evaluate \((g \circ f)(x)\), substitute \(f(x) = x^2 + 2\) into \(g\). This means \(g(f(x)) = (x^2 + 2)^2 - 2\).

With numerical evaluation, you can substitute actual numbers for \(x\), such as \((f \circ g)(2)\) or \((g \circ f)(2)\). This means you do the calculations with \(x = 2\), providing concrete output results. Understanding this process is crucial as evaluating functions is a core part of algebra.

Quadratic Functions

Quadratic functions are a type of polynomial function where the variable is squared, which means they have the standard form \(ax^2 + bx + c\). In the given \(f(x) = x^2 + 2\) and \(g(x) = x^2 - 2\), both are quadratic functions.

Quadratic functions generate parabolas when graphed, which can open upwards or downwards depending on the sign of the leading coefficient. In these forms, the coefficient of the \(x^2\) term is positive, so the parabolas open upwards.

Here's why quadratic functions matter in this context:

• They provide a straightforward example for function composition as the squared terms result in higher degree polynomials when composed.
• The complexity of their algebra increases, which helps visualize and understand the effects of composition on function structure and graph shape.

These qualities make quadratics a helpful tool when diving deeper into understanding how different functions interact through composition.