Problem 40 - Contemporary Precalculus

Question

Expand and simplify the rule of the function \(g .\) $$f(x)=2 x^{3}+3 x^{2}+x \text { and } g(x)=f(x-1)$$

Step-by-Step Solution

Answer

Answer: The expanded and simplified expression for g(x) is g(x) = 2x^3 - 3x^2 + x.

Step 1: Substituting (x-1) into f(x)

To find the function \(g(x)\), we need to substitute \((x-1)\) into \(f(x)\). That means wherever we see an \(x\) in the function \(f(x)\), we will replace it with \((x-1)\). So, $$g(x) = f(x-1) = 2(x-1)^3 + 3(x-1)^2 + x - 1$$

Step 2: Expanding (x-1) terms

Now, we need to expand the cubes and squares of \((x-1)\) terms. Using the binomial theorem, we can expand \((x-1)^3\) and \((x-1)^2\) as follows: $$(x-1)^3 = x^3 - 3x^2 + 3x - 1$$ $$(x-1)^2 = x^2 - 2x + 1$$

Step 3: Substitute the expanded terms into g(x)

Now that we have the expanded forms of \((x-1)^3\) and \((x-1)^2\), we can substitute them back into the expression for \(g(x)\): $$g(x) = 2(x^3 - 3x^2 + 3x - 1) + 3(x^2 - 2x + 1) + x - 1$$

Step 4: Distribute the constants in g(x)

Now, we need to distribute the constants 2 and 3 in the expression of \(g(x)\). This will give us: $$g(x) = 2x^3 - 6x^2 + 6x - 2 + 3x^2 - 6x + 3 + x - 1$$

Step 5: Combine like terms

Finally, let's combine the like terms in the expression of \(g(x)\) to simplify it: $$g(x) = 2x^3 - 6x^2 + 3x^2 + 6x - 6x + x - 2 + 3 - 1$$ $$g(x) = 2x^3 - 3x^2 + x$$ So, the expanded and simplified expression for \(g(x)\) is: $$g(x) = 2x^3 - 3x^2 + x$$

Key Concepts

Polynomial Functions

A polynomial function is a mathematical expression consisting of variables and coefficients, structured in a form where non-negative integer powers of the variable are used. In the function \( f(x) = 2x^3 + 3x^2 + x \), we have a cubic polynomial, which means the highest degree of any term is 3. Each term in a polynomial can be characterized by:

• its coefficient - for instance, 2 in \(2x^3\),
• its base - which is the variable \(x\), and
• its exponent - which represents the power or degree of the term, such as 3 in \(x^3\).

Polynomials are foundational in algebra because they allow for the modeling and solving of real-world problems. They provide both the flexibility required to describe diverse functional relationships and the simplicity needed to be manageable.

Binomial Theorem

The Binomial Theorem is a powerful algebraic tool that provides a method for expanding expressions that are raised to a power, such as \((x - 1)^3\). It states that \((a + b)^n\) can be expanded as:\[\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]This formula incorporates binomial coefficients \(\binom{n}{k}\), which can be found using Pascal's Triangle or calculated as\(\frac{n!}{k!(n-k)!}\). For example, when applying the binomial theorem to expand \((x-1)^3\), we calculate:

• \(x^3\) \((k=0)\)
• \(-3x^2\) \((k=1)\)
• \(+3x\) \((k=2)\)
• \(-1\) \((k=3)\)

Thus, the expansion becomes \(x^3 - 3x^2 + 3x - 1\), following the calculated sequence.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations that represent mathematical relationships. In this exercise, we worked with the expression \[g(x) = 2x^3 - 6x^2 + 3x^2 + 6x - 6x + x - 2 + 3 - 1\], which was derived during the process of expanding \(f(x-1)\). When simplifying an algebraic expression:

• Identify like terms - terms that have the same variable part and exponent.
• Combine like terms by adding or subtracting coefficients.

This simplification process helps to reduce complex expressions into their simplest form, making them easier to understand and manipulate solving problems.

Function Composition

Function composition involves creating a new function by applying one function to the results of another. In our exercise, the function \(g(x)\) is defined as \(f(x-1)\). This is an example of function composition, where we input \((x-1)\) instead of \(x\) into the function \(f\). Function composition allows:

• Modification of existing functions to shift, scale, or transform them.
• Building more complex functions from simpler ones.

It's an essential part of higher level mathematics, providing a framework for constructing solutions that build on simpler base functions. The process of substituting \(x-1\) for \(x\) in \(f(x)\) is precisely the means by which function composition is commonly performed.