Problem 21 - Beginning and Intermediate Algebra

Question

Let \(h(x)=-2 x+9\) and \(k(x)=3 x-1 .\) Find a) \(\quad(k \circ h)(x)\) b) \(\quad(h \circ k)(x)\) c) \((k \circ h)(-1)\)

Step-by-Step Solution

Answer

a) \((k \circ h)(x) = -6x + 26\) b) \((h \circ k)(x) = -6x + 11\) c) \((k \circ h)(-1) = 32\)

Step 1: Write down the given functions

We are given the functions: \(h(x) = -2x + 9\) and \(k(x) = 3x - 1\)

Step 2: Find (k ⨁ h)(x)

To find (k ⨁ h)(x), we substitute the expression for h(x) into the function k: \((k ⨁ h)(x) = k(h(x)) = k(-2x + 9)\) Now, we will replace x in the function k with the expression for h(x): \((k ⨁ h)(x) = 3(-2x + 9) - 1\) Now, let's simplify the expression: \((k ⨁ h)(x) = -6x + 27 - 1\) \((k ⨁ h)(x) = -6x + 26\)

Step 3: Find (h ⨁ k)(x)

To find (h ⨁ k)(x), we substitute the expression for k(x) into the function h: \((h ⨁ k)(x) = h(k(x)) = h(3x - 1)\) Now, we will replace x in the function h with the expression for k(x): \((h ⨁ k)(x) = -2(3x - 1) + 9\) Now, let's simplify the expression: \((h ⨁ k)(x) = -6x + 2 + 9\) \((h ⨁ k)(x) = -6x + 11\)

Step 4: Evaluate (k ⨁ h)(-1)

Now, we need to find the value of (k ⨁ h)(x) at x = -1: \((k ⨁ h)(-1) = -6(-1) + 26\) \((k ⨁ h)(-1) = 6 + 26\) \((k ⨁ h)(-1) = 32\) Now, let's write down our results. a) \((k ⨁ h)(x) = -6x + 26\) b) \((h ⨁ k)(x) = -6x + 11\) c) \((k ⨁ h)(-1) = 32\)

Key Concepts

Composite Functions

Imagine you have two different functions, like two separate tasks or processes. A composite function is like combining these two functions to create a brand new one. Here's how it works: **You take one function and use its output as the input for the other function.** Think of it as a machine with two parts, where the same object (or number) passes through both parts. In this exercise:

• We have two functions:
  • h(x) = -2x + 9
  • k(x) = 3x - 1
• We're looking for composite functions such as
  • (k ∘ h)(x), which means plugging the result of h(x) into k(x)
  • (h ∘ k)(x), which means plugging the result of k(x) into h(x)

This process allows us to combine functions to see how one function impacts another, making them very useful for solving specific problems.

Algebraic Functions

Algebraic functions are those that use algebraic expressions to relate inputs to outputs. They involve operations such as addition, subtraction, multiplication, and division. Let's break down the functions in this exercise: - **h(x) = -2x + 9**: This is a linear function because it only involves a simple multiplication and addition. - **k(x) = 3x - 1**: This is another linear function, with its own unique multiplication and subtraction aspect. When creating composite functions like

• (k ∘ h)(x) = k(h(x))
• The value that h(x) calculates is then plugged into k(x), creating a new algebraic expression:
  • (k ∘ h)(x) = 3(-2x + 9) - 1 = -6x + 26

These functions are foundational in algebra and help in understanding relationships in mathematical problems. Knowing how to manipulate algebraic functions helps in forming new expressions and solving complex equations.

Function Evaluation

Function evaluation involves finding the output of a function for specific values of the input. It means you're checking what a function "says" about a number when you plug it in. Using our example from this exercise, we're using function evaluation to find:

• (k ∘ h)(-1)
• This means:
  • First finding h(-1), which gives the result from the function h(x) when x = -1.
  • Next, using this result to find the output of k, or k(h(-1)).
• Calculating gives: (k ∘ h)(-1) = 32

This step is crucial because it shows the real outcome when these functions are applied in sequence. By evaluating composite functions for specific values, you understand not only how they are structured but also how they operate on specific numbers. Evaluation builds familiarity and develops skills in handling functions in various mathematical applications.