Problem 36 - Intermediate Algebra

Question

Determine the indicated functional values. (Objective 2 ) If \(f(x)=-5 x+1\) and \(g(x)=\sqrt{4 x+1}\), find \((f \circ g)(6)\) and \((g \circ f)(-1)\).

Step-by-Step Solution

Answer

\((f \circ g)(6) = -24\) and \((g \circ f)(-1) = 5\).

Step 1: Understand Function Composition

Function composition is the process of applying one function to the results of another. It is denoted as \((f \circ g)(x)\), which means we first apply the function \(g\) and then apply \(f\) to the result. Similarly, \((g \circ f)(x)\) means we first apply \(f\) and then apply \(g\).

Step 2: Understanding the given functions

We have two functions: \(f(x) = -5x + 1\) and \(g(x) = \sqrt{4x + 1}\). We will use these functions to evaluate \((f \circ g)(6)\) and \((g \circ f)(-1)\).

Step 3: Find \(g(6)\) for \((f \circ g)(6)\)

To find \((f \circ g)(6)\), start by calculating \(g(6)\). Substitute \(x = 6\) into \(g(x)\): \[ g(6) = \sqrt{4(6) + 1} = \sqrt{25} = 5 \].

Step 4: Evaluate \(f(g(6))\)

Now that we have \(g(6) = 5\), substitute this into \(f(x)\) to find \(f(5)\): \[ f(5) = -5(5) + 1 = -25 + 1 = -24 \].

Step 5: Find \(f(-1)\) for \((g \circ f)(-1)\)

To find \((g \circ f)(-1)\), start by calculating \(f(-1)\). Substitute \(x = -1\) into \(f(x)\): \[ f(-1) = -5(-1) + 1 = 5 + 1 = 6 \].

Step 6: Evaluate \(g(f(-1))\)

Now that we have \(f(-1) = 6\), substitute this into \(g(x)\) to find \(g(6)\): \[ g(6) = \sqrt{4(6) + 1} = \sqrt{25} = 5 \].

Step 7: Summarize Results

Combining the results from Steps 4 and 6, we have \((f \circ g)(6) = -24\) and \((g \circ f)(-1) = 5\).

Key Concepts

Intermediate Algebra

Intermediate algebra is a crucial stepping stone in the world of mathematics that builds the bridge between basic algebraic principles and more advanced topics, such as calculus. It emphasizes a deeper understanding of functions and equations, like linear and radical ones, which are essential for solving real-world problems.
One fundamental aspect of intermediate algebra is the manipulation and simplification of algebraic expressions involving these equations. For instance, when dealing with the functions in the original exercise,

• we must understand how to evaluate and simplify functions like \(f(x) = -5x + 1\) and \(g(x) = \sqrt{4x + 1}\)
• and how they interact when composed together.

Through the exercise of function composition, intermediate algebra helps students practice working with complex operations and improves their problem-solving skills. This foundation prepares them for the more demanding calculations found in calculus and other higher-level courses.

Functional Values

Functional values are the outputs we get when we apply specific inputs to functions. They show the relationship between independent and dependent variables, helping us understand how changes in one variable affect another.
In the exercise, we work with the functions \(f(x) = -5x + 1\) and \(g(x) = \sqrt{4x + 1}\). Calculating functional values involves substituting inputs into these functions to get the outputs. For example, finding \(f(5)\) means substituting 5 into \(f(x)\) to get a result of \(-24\).
Similarly, finding \(g(6)\) gives us 5. This concept is fundamental in understanding how functions behave, allowing us to predict outcomes and analyze data effectively.

• Compute the output by direct substitution.
• Grasp the interplay between different functions through their outputs.

These calculations underpin much of what students will encounter in their future studies of mathematics and science.

Composite Functions

Composite functions combine two or more functions to create a new function, represented as \(f \circ g\) or \(g \circ f\). This means that you apply one function to an input, then use that result as the input for the next function.
Understanding this concept is critical for solving exercises like the one given, where we need to find \( (f \circ g)(6) \) and \( (g \circ f)(-1) \).
To break it down:

• For \(f \circ g(6)\), calculate \(g(6)\), getting 5, and then find \(f(5)\), resulting in \(-24\).
• For \(g \circ f(-1)\), find \(f(-1)\), yielding 6, and calculate \(g(6)\) to get 5.

This technique of function layering is incredibly useful and often seen in calculus, where functions' transformations and combinations are key. Practicing composite function problems enhances a student's logical reasoning and can be applied in various mathematical and scientific contexts.
By mastering composite functions, students develop a keen eye for detail and the ability to solve more complex mathematical problems confidently.