Problem 15 - Contemporary Precalculus

Question

Find the indicated values, where $$ g(t)=t^{2}-t \text { and } f(x)=1+x B$$ $$g(f(2)+3)$$

Step-by-Step Solution

Answer

Answer: The value of g(f(2)+3) is $$4B^{2} + 14B + 12$$.

Step 1: Find f(2)

Substitute the value of x with 2 in f(x): $$f(2) = 1 + 2B$$

Step 2: Calculate g(f(2)+3)

Now substitute the value of f(2) into g(t) combined with +3: $$g(f(2)+3) = g(1 + 2B + 3)$$

Step 3: Simplify the argument

Combine like terms inside the function g(t): $$g(4+2B)$$

Step 4: Find g(4+2B)

Substitute the value of t with 4+2B in g(t): $$g(4+2B) = (4+2B)^{2} - (4+2B)$$

Step 5: Expand and simplify

Expand the squared term and simplify the expression: $$(4+2B)^{2} - (4+2B) = 16 + 16B + 4B^{2} - 4 - 2B$$

Step 6: Combine like terms

Combine the like terms to get the final expression: $$g(f(2)+3) = 4B^{2} + 14B + 12$$ The value of g(f(2)+3) is $$4B^{2} + 14B + 12$$.

Key Concepts

Function Evaluation

Function evaluation is the process of finding the output of a function given a specific input. In mathematical terms, if you have a function, say \( f(x) \), and you want to evaluate it at \( x = a \), you replace every \( x \) in the function with \( a \). This process helps to find what the function equals when \( x \) takes on the value \( a \).
For example, to evaluate the function \( f(x) = 1 + xB \) at \( x = 2 \), we substitute 2 in place of \( x \):

• \( f(2) = 1 + 2B \)

This substitution helps us understand the behavior of the function at specific points. To solve more complex expressions, we must combine this evaluated function with other expressions, like in our exercise, where \( f(x) \) is used as a part of another function \( g(t) \).

Polynomial Expansion

Polynomial expansion is the process of expressing a polynomial that is given in a condensed form, like \( (4+2B)^2 \), as a sum of its individual terms. This includes distributing powers through multiplication and then combining terms that behave similarly. This method is particularly useful when you need to simplify equations.
Let's take the expression \( (4+2B)^2 \) and expand it:

• \( (4+2B)^2 = (4+2B)(4+2B) \)
• Use the distributive property: \( = 4 imes 4 + 4 imes 2B + 2B imes 4 + 2B imes 2B \)
• Simplify each term: \( = 16 + 8B + 8B + 4B^2 \)
• Combine the like terms: \( = 16 + 16B + 4B^2 \)

In our exercise, this expansion is crucial because it helps us simplify \( g(t) \) by expressing it in terms of basic algebraic operations.

Simplification of Expressions

The simplification of algebraic expressions involves organizing an expression into the simplest form possible. This means combining like terms and reducing any unnecessary complexity.
After expanding polynomials or evaluating functions, simplification becomes the next step, which improves clarity and makes it easier to interpret or solve the expression.
In our exercise, after expanding \( (4+2B)^2 \), we found:

• \( 16 + 16B + 4B^2 \)

We are also subtracting \( (4+2B) \) from it, performed as follows:

• Initial equation: \( 16 + 16B + 4B^2 - 4 - 2B \)
• Simplify it further by combining like terms: \( 16 - 4 + 16B - 2B + 4B^2 \)
• Resulting simplified expression: \( 12 + 14B + 4B^2 \)

This step makes the expression more manageable and is essential for ensuring the expression's accuracy.

Algebraic Substitution

Algebraic substitution is the practice of replacing variables in an expression with another value or expression. This technique is pivotal in solving equations and evaluating functions because it allows us to manipulate and explore formulas practically.
In our problem, substantial use of substitution takes place:

• First, in substituting \( x = 2 \) into \( f(x) \) as \( f(2) = 1 + 2B \)
• Then, using this result in \( g(t) \), where \( t = f(2) + 3 = 4 + 2B \)

This follows directly into computing \( g(4 + 2B) \) where another substitution is made as \( t = 4 + 2B \) in the function \( g(t) = t^2 - t \), hence helping us progress through the exercise to reach an answer.
This process is invaluable in mathematics as it provides clear, step-by-step methods to approach and solve complex expressions.