Problem 5

Question

Let \(f(x)=x^{2}\) and \(g(x)=2 x-5 .\) Match each function in Group I with the correct expression in Group II. (Group II) A. \(4 x^{2}-20 x+25\) B. \(x^{2}-2 x+5\) C. \(2 x^{2}-5\) D. \(\frac{x^{2}}{2 x-5}\) E. \(x^{2}+2 x-5\) F. \(2 x^{3}-5 x^{2}\) (Group I) $$(f \circ g)(x)$$

Step-by-Step Solution

Verified
Answer
(f \circ g)(x) matches with A: \(4x^2 - 20x + 25\).
1Step 1: Understanding Composition
The expression \((f \circ g)(x)\) means the composition of functions f and g, which is \(f(g(x))\). This implies substituting \(g(x)\) into \(f(x)\).
2Step 2: Substitute g(x) into f(x)
Replace \(x\) in \(f(x) = x^2\) with \(g(x) = 2x - 5\). This gives us \(f(g(x)) = (2x - 5)^2\).
3Step 3: Expand the Expression
Expand \((2x - 5)^2\) using the formula \((a - b)^2 = a^2 - 2ab + b^2\):1. \((2x)^2 = 4x^2\)2. \(-2(2x)(5) = -20x\)3. \(5^2 = 25\)This results in \(4x^2 - 20x + 25\).
4Step 4: Match with Group II
Compare the expanded expression with the options in Group II. We see that \(4x^2 - 20x + 25\) corresponds to option A.

Key Concepts

Quadratic FunctionFunction OperationsSubstitution Method
Quadratic Function
A quadratic function is a type of polynomial function where the highest degree of the variable is 2. It generally takes the form of \(f(x) = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). This function will graph as a parabola in the coordinate plane.
Quadratic functions like \(f(x) = x^2\) are some of the most fundamental in algebra. We often manipulate these functions to observe different outcomes, such as finding their vertex, axis of symmetry, and roots (x-intercepts).
  • **Vertex**: The highest or lowest point on the parabola, depending on whether it opens upwards or downwards.
  • **Axis of Symmetry**: A vertical line that passes through the vertex, dividing the parabola into two mirror-image halves.
  • **Roots**: The solutions to the equation \(f(x) = 0\), which are the points where the parabola intersects the x-axis.
Understanding how quadratic functions transform and interact with other functions, such as through composition, is crucial for mastering their behaviors in various mathematical contexts.
Function Operations
Function operations involve performing arithmetic operations, such as addition, subtraction, multiplication, and division, on two or more functions. However, one of the most important operations is function composition, denoted by \((f \circ g)(x)\), which means applying one function to the results of another function.
Through operations like composition, we can create more complex functions from simpler ones. Here are a few key points to remember:
  • **Function Addition**: \((f+g)(x) = f(x) + g(x)\)
  • **Function Subtraction**: \((f-g)(x) = f(x) - g(x)\)
  • **Function Multiplication**: \((f\cdot g)(x) = f(x)\cdot g(x)\)
  • **Function Division**: \((\frac{f}{g})(x) = \frac{f(x)}{g(x)}\), \( g(x) eq 0\)
  • **Function Composition**: \((f \circ g)(x) = f(g(x))\)
In the given exercise, we specifically work with function composition, which requires substituting \(g(x)\) into \(f(x)\), illustrating how complex expressions can be formed from separate functions.
Substitution Method
The substitution method is a powerful technique used to simplify complex problems by replacing variables with equivalent expressions. This often makes it easier to manipulate and solve equations.
In the context of function composition as seen in the exercise, substitution is used to replace \(x\) in the function \(f(x)\) with the entire function \(g(x)\). Here’s how it works in simple steps:
  • Identify the function to be substituted: \(g(x) = 2x - 5\).
  • Substitute \(g(x)\) into \(f(x) = x^2\), transforming it into \(f(g(x)) = (2x - 5)^2\).
  • Apply algebraic expansion techniques: Use formulas like \((a - b)^2 = a^2 - 2ab + b^2\) to simplify \((2x - 5)^2\) into \(4x^2 - 20x + 25\).
By substituting one function into another, we can solve for more complex expressions, determining how the combined function behaves. This is essential for deeper understanding in algebra and calculus as we analyze more intricate relationships between functions.