Problem 22 - Algebra and Trigonometry

Question

\(17-22=\) Use \(f(x)=3 x-5\) and \(g(x)=2-x^{2}\) to evaluate the expression. $$ \begin{array}{ll}{\text { (a) }(f \circ f)(x)} & {\text { (b) }(g \circ g)(x)}\end{array} $$

Step-by-Step Solution

Answer

(f ∘ f)(x) = 9x - 20; (g ∘ g)(x) = -x^4 + 4x^2 - 2.

Step 1: Solve 17-22

First, evaluate the expression \(17 - 22\). This is a simple subtraction problem. When you subtract a larger number from a smaller one, the result is negative. Thus, \(17 - 22 = -5\).

Step 2: Understand Composition Notation

The notation \((f \circ f)(x)\) means applying the function \(f\) to \(x\) and then applying \(f\) again to the result. Similarly, \((g \circ g)(x)\) means applying \(g\) to \(x\) and then applying \(g\) again to the result.

Step 3: Apply f(x) to itself

Start by finding \(f(x) = 3x - 5\). For \((f \circ f)(x)\), we first need \(f(x)\), and then substitute \(f(x)\) back into \(f\). Calculate \(f(f(x)) = f(3x - 5) = 3(3x-5) - 5 = 9x - 15 - 5 = 9x - 20\). Therefore, \((f \circ f)(x) = 9x - 20\).

Step 4: Apply g(x) to itself

Find \(g(x) = 2 - x^2\). For \((g \circ g)(x)\), compute \(g(g(x)) = g(2 - x^2) = 2 - (2 - x^2)^2\). Expand \((2 - x^2)^2\) as follows: \((2 - x^2)(2 - x^2) = 4 - 4x^2 + x^4\). Substitute back to get \(g(g(x)) = 2 - (4 - 4x^2 + x^4) = 2 - 4 + 4x^2 - x^4 = -2 + 4x^2 - x^4\). Thus, \((g \circ g)(x) = -x^4 + 4x^2 - 2\).

Key Concepts

Polynomial Functions

Polynomial functions are expressions composed of variables raised to whole number exponents. They consist of terms added or subtracted from each other. For example, in the function \( f(x) = 3x - 5 \), we have two terms: \(3x\) and \(-5\). The first term, \(3x\), is a linear term because it's raised to the first power. The constant term, \(-5\), doesn't have a variable attached. This linear equation is a form of polynomial function.Polynomial functions can also take more complicated forms. Consider \( g(x) = 2 - x^2 \). Here, we have a quadratic polynomial, as the highest power is 2. Polynomial functions can be classified based on the highest power of \(x\):

• Linear: The highest power of \(x\) is 1.
• Quadratic: The highest power of \(x\) is 2.
• Cubic, Quartic, etc.: The highest power of \(x\) is 3 or more.

Understanding polynomial functions is fundamental as they model many real-world situations, ranging from simple linear equations to complex quadratic or higher-degree functions.

Algebraic Subtraction

Algebraic subtraction involves subtracting one algebraic expression from another. In regular subtraction, we operate with numbers, but in algebraic subtraction, we often deal with variables and exponents too. In the given problem, we first needed to solve \( 17 - 22 \).This step is straightforward. Subtracting a larger number from a smaller one results in a negative value, leading to \( -5 \). This rule holds for algebraic subtraction as well, but instead of numbers alone, you will generally work with expressions like \( a - b \). When you subtract polynomials, distribute the negative sign to each term in the polynomial being subtracted. Consider subtracting two linear functions:

• Example 1: \((3x + 5) - (2x + 1)\)
• Distribute the negative: \(3x + 5 - 2x - 1\)
• Combine like terms: \( (3x - 2x) + (5 - 1) = x + 4 \)

Thus, handling algebraic subtraction requires these steps of dealing with coefficients, variables, and maintaining keen organization of terms.

Function Evaluation

Function evaluation is the process of finding outputs for various inputs into a function. Given a function, say \( f(x) = 3x - 5 \), to evaluate \( f(x) \), substitute the value of \( x \) into the function.Function composition, marked by \((f \circ f)(x)\) or \((g \circ g)(x)\), is a form of evaluation where a function is applied multiple times. For \( f(f(x)) \), we initially calculate \( f(x) \), then use this result as the input again. For our function, this translates to:

• Compute \( f(3x - 5) \)
• Straightforwardly evaluate the expression: \(3(3x - 5) - 5\)
• Resulting in: \(9x - 15 - 5 = 9x - 20\)

In many cases, breaking down complex functions step-by-step makes it easier. Likewise, \( g(g(x)) \) involves substituting \( g(x) = 2 - x^2 \) back into itself, eventually leading to \(-x^4 + 4x^2 - 2\). These steps might seem complex initially, but with practice, function evaluation becomes intuitive.