Problem 8 - A Graphical Approach to Precalculus with Limits

Question

Let \(f(x)=x^{2}+3 x\) and \(g(x)=2 x-1 .\) Perform the composition or operation indicated. $$(g \circ f)(-2)$$

Step-by-Step Solution

Answer

(g \circ f)(-2) = -5

Step 1: Understand the Composition

The composition \(g \circ f\) denotes the function \(g(f(x))\). This means we substitute \(f(x)\) into \(g(x)\) as the input.

Step 2: Evaluate f(-2)

Substitute \(-2\) into the function \(f(x) = x^2 + 3x\).\[ f(-2) = (-2)^2 + 3(-2) = 4 - 6 = -2 \]

Step 3: Use f(-2) as an Input for g(x)

Now that we have \(f(-2) = -2\), use this value as the input for \(g(x)\). Substitute it into \(g(x) = 2x - 1\).\[ g(f(-2)) = g(-2) = 2(-2) - 1 = -4 - 1 = -5 \]

Step 4: Conclusion

The composition \(g(f(-2))\) results in \(-5\). Therefore, \( (g \circ f)(-2) = -5 \).

Key Concepts

Algebraic Functions

Algebraic functions are mathematical expressions built using algebraic operations such as addition, subtraction, multiplication, division, and exponentiation. These functions are the building blocks of many mathematical models used in everyday life. In the context of function composition, algebraic functions like \(f(x) = x^2 + 3x\) and \(g(x) = 2x - 1\) are manipulated to form new functions by performing operations between them.

Algebraic functions can be evaluated by substituting values for \(x\) and simplifying. Understanding how these functions operate independently is crucial before attempting to compose them. For example, when evaluating \(f(x)\), each term in the function is addressed individually to reach the total expression as seen in \(f(-2) = (-2)^2 + 3(-2)\).

Composition of algebraic functions broadens the scope by allowing us to input one function into another, as demonstrated in the composition \((g \circ f)(-2)\). This requires substitution, where the entire expression of one function is used as the input for another.

Precalculus

Precalculus is a branch of mathematics that prepares students for calculus. It encompasses many foundational concepts, including algebraic functions and their compositions. The concept of function composition is essential in Precalculus since it sets the stage for understanding more complex relationships in calculus.

Function composition, such as \(g(f(-2))\), involves plugging one function into another. In a Precalculus context, this operation is vital because it illustrates how different functions interrelate, leading to the exploration of transformations and behavior of more advanced functions used in calculus.

By practicing function compositions in Precalculus, students build the necessary skills to manipulate and transform functions, which is crucial for solving real-world problems. Precalculus exercises often focus on understanding how compositions work numerically, graphically, and symbolically.

Step-by-Step Solutions

Step-by-step solutions are invaluable tools for understanding complex mathematical operations like function composition. When tackling a problem like \((g \circ f)(-2)\), breaking it down into manageable parts makes the task less daunting and the solution more comprehensible.

Each step follows logically from the last. For instance, in the given exercise, understanding \( f(x) \) is key before using its outcome as the input for \( g(x) \). By substituting \(-2\) into \(f(x )\) to find \( f(-2) \), you gather information needed for the next step.

• Evaluate the inner function: \( f(-2) = -2 \).
• Substitute this result into the outer function: \( g(-2) = -4 - 1 \).
• Conclude with the final answer: \((g \circ f)(-2) = -5\).

The clarity provided by step-by-step solutions makes them powerful for reinforcing the learning process, especially when students can directly see the effects of each operation on the final outcome.