Problem 23 - Beginning and Intermediate Algebra

Question

Let \(g(x)=x^{2}-6 x+11\) and \(h(x)=x-4 .\) Find a) \(\quad(h \circ g)(x)\) b) \(\quad(g \circ h)(x)\) c) \(\quad(g \circ h)(4)\)

Step-by-Step Solution

Answer

a) \((h \circ g)(x) = x^2 - 6x + 7\) b) \((g \circ h)(x) = x^2 - 2x - 1\) c) \((g \circ h)(4) = 7\)

Step 1: Finding g(x)

We are given the function \(g(x) = x^2 - 6x + 11\).

Step 2: Find h(g(x))

Now we need to evaluate h at g(x), so we substitute g(x) into h(x). Since h(x) = x - 4, we get the following composition of functions: \[ (h \circ g)(x) = h(g(x)) = h(x^2 - 6x + 11). \] Now we need to replace x in h with g(x) expression that we found earlier: \[ (h \circ g)(x) = (x^2 - 6x + 11) - 4. \]

Step 3: Simplify (h°g)(x)

Simplifying the expression above yields: \[ (h \circ g)(x) = x^2 - 6x + 7. \] #b) Finding (g°h)(x)#

Step 1: Finding h(x)

We are given the function \(h(x) = x - 4\).

Step 2: Find g(h(x))

Now we need to evaluate g at h(x), so we substitute h(x) into g(x). Since g(x) = x^2 - 6x + 11, we get the following composition of functions: \[ (g \circ h)(x) = g(h(x)) = g(x - 4). \] Now we need to replace x in g with h(x) expression that we found earlier: \[ (g \circ h)(x) = (x - 4)^2 - 6(x - 4) + 11. \]

Step 3: Simplify (g°h)(x)

Simplifying the expression above yields: \[ (g \circ h)(x) = x^2 - 2x - 1. \] #c) Finding (g°h)(4)#

Step 1: Evaluate (g°h)(x) at x = 4

Now we need to find the value of the composed function at x = 4. So we substitute x = 4 into the expression we found for (g°h)(x): \[ (g \circ h)(4) = (4)^2 - 2(4) - 1. \]

Step 2: Simplify (g°h)(4)

Simplifying the expression above yields: \[ (g \circ h)(4) = 16 - 8 - 1 = 7. \] Results: a) (h°g)(x) = x^2 - 6x + 7 b) (g°h)(x) = x^2 - 2x - 1 c) (g°h)(4) = 7

Key Concepts

Polynomial Functions

Polynomial functions are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. These functions have the general form of a polynomial, such as:

• Monomial: A single-term polynomial, like \(3x^4\).
• Binomial: A two-term polynomial, like \(x^2 - 4x\).
• Trinomial: A three-term polynomial, like \(x^2 - 6x + 11\).

Polynomials can vary in degree, which is determined by the highest power of the variable in the expression. For example, \(x^2 - 6x + 11\) is a polynomial of degree 2. This means the graph of this function, \(g(x)\), will be a parabola. They are widely used in algebra to model various phenomena. Students often encounter polynomial functions in exercises involving function composition and simplification, similar to the exercise discussed here. Understanding polynomial functions is crucial for progress in algebra and calculus.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operators (like addition, subtraction, multiplication, etc.). In this exercise, both \(g(x)\) and \(h(x)\) are algebraic expressions. Let's break them down:

• \(g(x) = x^2 - 6x + 11\): This is a quadratic expression involving the square of a variable.
• \(h(x) = x - 4\): A linear expression that just subtracts a number from a variable.

When we work with algebraic expressions, we apply various algebraic operations, such as substitution, to simplify or modify them. For example, in function composition like \((h \circ g)(x)\), we substitute the entire algebraic expression of \(g(x)\) into \(h(x)\). Such manipulation requires understanding the principles of algebra to correctly evaluate and simplify the given expressions.

Evaluating Functions

Evaluating functions involves finding the value of a function for a particular input. It is an essential skill in mathematics, helping students to understand how functions behave. Let's see how it's done using the provided exercise:

• Function Composition: In problems like this, you need to substitute one function into another. For example, \((h \circ g)(x)\) means you plug \(g(x)\) into \(h(x)\). Ensure that you replace the variable in \(h(x)\) with \(g(x)\), simplifying to obtain: \(x^2 - 6x + 7\).
• Specific Values: Sometimes you need to evaluate a composite function for a specific value of \(x\). Using \((g \circ h)(x)\), substitute and simplify: \((g \circ h)(4) = 16 - 8 - 1 = 7\).

Using these methods allows you to develop a deeper understanding of how functions transform and how composite functions interact from one to another. Mastery of this concept is fundamental for solving more complex problems in advanced mathematics.