Problem 26 - Beginning and Intermediate Algebra

Question

Let \(h(x)=3 x^{2}-8 x+2\) and \(k(x)=2 x-3 .\) Finc a) \(\quad(h \circ k)(x)\) b) \(\quad(k \circ h)(x)\) c) \(\quad(k \circ h)(0)\)

Step-by-Step Solution

Answer

a) \((h \circ k)(x) = 12x^2 - 52x + 53\), b) \((k \circ h)(x) = 6x^2 - 16x + 1\), c) \((k \circ h)(0) = 1\).

Step 1: Define the functions h(x) and k(x)

Given, h(x) = 3x^2 - 8x + 2 and k(x) = 2x - 3.

Step 2: Find (h◦k)(x), i.e., h(k(x))

We need to substitute k(x) in place of x in the function h(x). \(h(k(x)) = 3(2x - 3)^2 - 8(2x - 3) + 2\)

Step 3: Simplify and find (h◦k)(x)

Now, we will simplify the expression to find (h◦k)(x): \((h \circ k)(x) = 3(4x^2 - 12x + 9) - 16x + 24 + 2\) \((h \circ k)(x) = 12x^2 - 36x + 27 - 16x + 24 + 2\) \((h \circ k)(x) = 12x^2 - 52x + 53\)

Step 4: Find (k◦h)(x), i.e., k(h(x))

We need to substitute h(x) in place of x in the function k(x). \(k(h(x)) = 2(3x^2 - 8x + 2) - 3\)

Step 5: Simplify and find (k◦h)(x)

Now, we will simplify the expression to find (k◦h)(x): \((k \circ h)(x) = 6x^2 - 16x + 4 - 3\) \((k \circ h)(x) = 6x^2 - 16x + 1\)

Step 6: Calculate (k◦h)(0), i.e., k(h(0))

Now, we will substitute x = 0 in the expression for (k◦h)(x) to find (k◦h)(0): \((k \circ h)(0) = 6(0)^2 - 16(0) + 1\) \((k \circ h)(0) = 0 - 0 + 1\) \((k \circ h)(0) = 1\) The answers are: a) (h◦k)(x) = 12x^2 - 52x + 53, b) (k◦h)(x) = 6x^2 - 16x + 1, c) (k◦h)(0) = 1.

Key Concepts

Polynomial Functions

Polynomial functions are expressions that involve variables raised to positive integer powers, along with their corresponding coefficients. They are considered some of the simplest forms of mathematical expressions. An example of a polynomial function is the quadratic function, such as \( h(x) = 3x^2 - 8x + 2 \). This function includes terms like \(3x^2\), \(-8x\), and \(2\), each representing a part of the whole polynomial.
The characteristics of polynomial functions include:

• Degree: Determined by the highest power of the variable in the function. For \( h(x) \), the degree is 2, making it a quadratic polynomial.
• Coefficients: The constants that multiply the variable terms, like 3 and -8 in \( h(x) \).
• Constant Term: This term has no variable attached, such as \(2\) in \( h(x) \).

Polynomial functions can be combined, substituted into one another, and simplified to create more complex expressions, showcasing their versatility in calculus and algebra.

Substitution Method

The substitution method is a key mathematical technique used to solve problems by replacing a variable with a different expression. This technique is especially useful in function composition, where one function is applied to the result of another function.
For the given exercise, we look into compositional functions (h◦k)(x) and (k◦h)(x):

• To find \((h \circ k)(x)\), we substitute \(k(x) = 2x - 3\) into \(h(x)\). This means wherever you see \(x\) in \(h(x)\), replace it with \(2x - 3\), leading to \( h(k(x)) = 3(2x - 3)^2 - 8(2x - 3) + 2 \).
• To find \((k \circ h)(x)\), we substitute \(h(x) = 3x^2 - 8x + 2\) into \(k(x)\), so that \(x\) in \(k(x)\) gets replaced with \(h(x)\) resulting in \( k(h(x)) = 2(3x^2 - 8x + 2) - 3 \).

This method is invaluable for evaluating complex expressions and functions in both algebra and calculus, as it allows one to systematically approach problems by breaking them into manageable parts.

Simplification Process

Simplification is the process of reducing a complex expression to its most basic form, often making it easier to work with or evaluate. This process involves combining like terms, performing arithmetic operations, and applying algebraic identities.
In the exercise at hand, after substituting, we performed the following:

• For \((h \circ k)(x)\), after substituting \(k(x)\) into \(h(x)\), the expression \(3(2x - 3)^2 - 8(2x - 3) + 2\) was expanded and like terms were combined to finally obtain \(12x^2 - 52x + 53\).
• For \((k \circ h)(x)\), the expansion and combination of terms within \(2(3x^2 - 8x + 2) - 3\) led to \(6x^2 - 16x + 1\).

Simplification is crucial, as it transforms mathematical expressions, making them efficient for evaluations like \((k \circ h)(0)\), where substituting zero directly provides the result without unnecessary calculations. Simplifying expressions can reveal fundamental characteristics and behaviors of functions, being a key skill in mathematical problem-solving.