Problem 60 - College Algebra

Question

Find a. \((f \circ g)(x)\) b. \((g \circ f)(x)\) c. \((f \circ g)(2)\) d. \((g \circ f)(2)\) $$ f(x)-5 x-2, g(x)--x^{2}+4 x-1 $$

Step-by-Step Solution

Answer

a. \((f \circ g)(x) = 5x^2 + 20x - 7\), b. \((g \circ f)(x) = 25x^2 - 40x + 11\), c. \((f \circ g)(2) = 53\), d. \((g \circ f)(2) = 31\)

Step 1: Express Function Compositions

Start by expressing the function compositions:a. \((f \circ g)(x) = f(g(x))\)b. \((g \circ f)(x) = g(f(x))\)

Step 2: Substitute Value of g(x) into f(x)

For part a, replace \(g(x)\) with its given function in \(f(g(x))\): \(f(g(x)) = f(x^2 + 4x - 1)\)

Step 3: Substitute g(x) into f(x) Equation

Then, substitute \(x^2 + 4x - 1\) for \(x\) in \(5x - 2\): \((f \circ g)(x) = 5(x^2 + 4x - 1) - 2 = 5x^2 + 20x - 5 - 2 = 5x^2 + 20x - 7\)

Step 4: For part b, Substitute Value of f(x) into g(x)

Now, substitute \(f(x)\) with its function in \(g(f(x))\): \(g(f(x)) = g(5x - 2)\)

Step 5: Substitute f(x) into g(x) Equation

Replace \(5x - 2\) with \(x\) in \(x^2 + 4x - 1\): \((g \circ f)(x) = (5x - 2)^2 + 4(5x - 2) - 1 = 25x^2 -20x + 4 -20x + 8 - 1 = 25x^2 -40x + 11\)

Step 6: Compute the Function Compositions at x = 2

For parts c and d, substitute \(x = 2\) into the resulting function in step 3 for \((f \circ g)(x)\) and in step 5 for \((g \circ f)(x)\):c. \((f \circ g)(2) = 5(2)^2 + 20(2) - 7 = 20 + 40 - 7 = 53\)d. \((g \circ f)(2) = 25*2^2 - 40*2 + 11 = 100 - 80 + 11 = 31\)

Key Concepts

Function Operations

Function operations refer to the various ways in which two or more functions can interact with each other. One of the key operations is **function composition**, denoted by \(f \circ g\), which represents the function \(f(g(x))\). This means you first apply the function \(g\) to the variable \(x\), and then take the output of this operation and use it as the input for the function \(f\). This creates a new function based on the actions of the two original functions.

### Steps to Compose Functions

• Identify the inner function (the function that will be applied first). In the composition \(f \circ g\), \(g\) is the inner function.
• Identify the outer function (the function to be applied second). Here, \(f\) acts as the outer function.
• Substitute the entire inner function into every occurrence of the variable in the outer function's equation.

These steps help in simplifying the combination of functions and can also be used to determine the output for specific values once the composed function is formed.

Algebraic Expressions

Understanding algebraic expressions is crucial when working with function operations. An algebraic expression is a mathematical phrase that can include numbers, variables, and operation symbols. When dealing with function operations like composition, you'll often integrate one function's algebraic expression into another.

### Simplifying Algebraic Expressions

• Combine like terms: Ensure terms with the same variable raised to the same power are combined.
• Apply the distributive property to multiply terms as needed, such as \((a + b)c = ac + bc\).
• Simplify the equation step by step, keeping track of operations to avoid errors.

In the example problem, to find the composition \(f(g(x))\), you inserted the expression for \(g(x)\) where \(x\) appears in \(f(x)\), resulting in a longer algebraic expression that required careful simplification.

Substitution Method

The substitution method plays a vital role in solving equations involving function compositions. It involves replacing a variable with a specific value or another expression.

### How to Use Substitution

• Identify the target variable or expression to be replaced.
• Replace it with the given expression or value, such as substituting \(g(x)\) into \(f\) in \(f(g(x))\).
• Simplify the new expression to find the desired output.

For function compositions, this method simplifies the process of combining functions and finding specific numerical results. In our exercise, substituting \(x = 2\) into the expressions allowed the calculation of specific values of the composite functions, demonstrating how substitution transforms and simplifies the solution process.