Find
a. (fog) (x)
b. (go f) (x)
c. (fog) (2)
d. (go f) (2).
f(x) = 2x, g(x) = x+7

Verified step by step guidance

1. Step 1: Understand the problem. You are tasked with finding the compositions of two functions, f(x) = 2x and g(x) = x + 7. Specifically, you need to compute (f ∘ g)(x), (g ∘ f)(x), (f ∘ g)(2), and (g ∘ f)(2). The symbol ∘ represents function composition, meaning you substitute one function into the other.
2. Step 2: To find (f ∘ g)(x), substitute g(x) into f(x). This means replacing the input of f(x) with g(x). Start with f(x) = 2x, and replace x with g(x) = x + 7. The result is f(g(x)) = 2(x + 7). Simplify this expression.
3. Step 3: To find (g ∘ f)(x), substitute f(x) into g(x). This means replacing the input of g(x) with f(x). Start with g(x) = x + 7, and replace x with f(x) = 2x. The result is g(f(x)) = 2x + 7. Simplify this expression.
4. Step 4: To find (f ∘ g)(2), use the expression for (f ∘ g)(x) that you derived in Step 2. Substitute x = 2 into the expression f(g(x)) = 2(x + 7). Simplify the result.
5. Step 5: To find (g ∘ f)(2), use the expression for (g ∘ f)(x) that you derived in Step 3. Substitute x = 2 into the expression g(f(x)) = 2x + 7. Simplify the result.

Key Concepts

Function Composition

Function composition involves combining two functions to create a new function. If f(x) and g(x) are two functions, the composition (fog)(x) means applying g first and then f to the result, expressed as f(g(x)). Understanding this concept is crucial for solving problems that require evaluating composite functions.

Evaluating Functions

Evaluating functions means substituting a specific value into a function to find its output. For example, if f(x) = 2x, then f(2) = 2(2) = 4. This skill is essential for calculating the values of composite functions at given points, as required in the question.

Algebraic Manipulation

Algebraic manipulation refers to the process of rearranging and simplifying expressions using algebraic rules. This includes operations like addition, multiplication, and applying the distributive property. Mastery of these techniques is necessary to simplify the results of function compositions and to evaluate them accurately.