Problem 55 - College Algebra Essentials

Question

Find a. \((f \circ g)(x) \quad \) b. \((g \circ f)(x) \quad \) c. \((f \circ g)(2) \quad \) d. \((g \circ f)(2)\) $$f(x)=4 x-3, g(x)=5 x^{2}-2$$

Step-by-Step Solution

Answer

The composed functions are: (a) \((f \circ g)(x) = 20x^{2} - 11\), (b) \((g \circ f)(x) = 20x^{2} - 30x + 23\), (c) \((f \circ g)(2) = 69\) and (d) \((g \circ f)(2) = 43\).

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

The function composition \(f \circ g\)(x) means to substitute the function g(x) in place of x in the function f(x). So we have, \[f(g(x))=f(5x^{2}-2) = 4(5x^{2}-2) - 3 = 20x^{2} - 8 - 3 = 20x^{2} - 11\]

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

The function composition \((g \circ f)(x)\) means to substitute the function f(x) in place of x in the function g(x). So we have, \[g(f(x))=g(4x-3)=5(4x-3)^{2} - 2 = 20x^{2} - 30x + 23\]

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

To find \((f \circ g)(2)\), substitute \(x=2\) into the function we got in step 1: \((f \circ g)(2) = 20(2^{2}) - 11 = 20(4) - 11 = 80 - 11 = 69\].

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

Similar to step 3, to find \((g \circ f)(2)\), substitute \(x=2\) into the function we found in step 2: \((g \circ f)(2) = 20 (2^{2}) - 30 (2) + 23 = 80 - 60 + 23 = 43\].

Key Concepts

Polynomial Functions

Polynomial functions are expressions that involve variables raised to whole number exponents. These functions can be simple, like a linear function, or more complex, such as a quadratic or cubic function. In general, a polynomial is expressed as:

• an equation of the form: \(a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0\)
• where \(a_n, a_{n-1}, ..., a_0\) are constants known as coefficients
• and \(n\) is a non-negative integer representing the degree of the polynomial.

In the given exercise, both \(f(x) = 4x - 3\) and \(g(x) = 5x^2 - 2\) are polynomial functions.
Function \(f(x)\) is a linear function as it's a first-degree polynomial, meaning its highest power of \(x\) is one. Whereas, \(g(x)\) is a quadratic function as it's a second-degree polynomial, with the highest power of \(x\) being two. These polynomial functions often appear in real-world situations, making their understanding crucial.

Substitution Method

The substitution method is a key technique used in mathematics, especially when dealing with function compositions. It involves replacing a variable with another expression or function. This method simplifies complex expressions and is essential for solving substitution problems in functions.
In the context of function composition, this method is pivotal. It refers to the process of substituting one function into another. If you have two functions, say \(f(x)\) and \(g(x)\), the composition \((f \circ g)(x)\) involves substituting \(g(x)\) into \(f(x)\).

• This means replacing \(x\) in \(f\) with the entire expression of \(g(x)\).
• Similarly, \((g \circ f)(x)\) involves inserting \(f(x)\) into \(g(x)\).

In the exercise, the substitution method is applied to find both \((f \circ g)(x)\) and \((g \circ f)(x)\). By carefully substituting and then simplifying, you develop a clearer understanding of how different functions interact in composition.

Evaluation of Functions

Once you understand the composition of functions, the next step is their evaluation. This process involves calculating the value of a function for a specific input. In algebra, evaluating functions is often done by substituting a specific value of the variable into the expression of the function.
For example, after finding the composite functions \((f \circ g)(x)\) and \((g \circ f)(x)\), evaluating these at a particular point, say \(x = 2\), helps determine specific numerical outcomes.

• In the exercise, \((f \circ g)(2)\) was calculated by substituting \(x = 2\) into the expression derived from \(f(g(x))\).
• Similarly, \((g \circ f)(2)\) involved substituting \(x = 2\) into the expression derived from \(g(f(x))\).

Such evaluations are crucial as they provide insights into the function's value at precise points, proving vital in numerous real-world applications.