Problem 55 - College Algebra with Modeling and Visualization

Question

Exercises \(53-56:\) Use the given \(f(x)\) and \(g(x)\) to evaluate each expression. \(f(x)=5 x-2, g(x)=|x|\) (a) \((f \circ g)(-4)\) (b) \((g \circ f)(5)\)

Step-by-Step Solution

Answer

(f ∘ g)(-4) = 18, (g ∘ f)(5) = 23.

Step 1: Understand the Composition of Functions

The composition of functions notation \((f \circ g)(x)\) means that you first apply \(g(x)\) and then use the result as the input for \(f(x)\). Similarly, \((g \circ f)(x)\) means you first apply \(f(x)\) and then use the result as the input for \(g(x)\).

Step 2: Evaluate \((f \circ g)(-4)\)

First, find \(g(-4)\). Since \(g(x) = |x|\), we have \(g(-4) = |-4| = 4\). Next, use this result to evaluate \(f(g(-4)) = f(4)\). Since \(f(x) = 5x - 2\), \(f(4) = 5 \times 4 - 2 = 20 - 2 = 18\). Thus, \((f \circ g)(-4) = 18\).

Step 3: Evaluate \((g \circ f)(5)\)

First, evaluate \(f(5)\). Since \(f(x) = 5x - 2\), we have \(f(5) = 5 \times 5 - 2 = 25 - 2 = 23\). Next, find \(g(f(5)) = g(23)\). Since \(g(x) = |x|\), \(g(23) = |23| = 23\). Therefore, \((g \circ f)(5) = 23\).

Key Concepts

Absolute Value

The absolute value function, denoted as \(|x|\), is a fundamental concept in mathematics that measures the distance of a number from zero on a number line. It always produces a non-negative result. No matter if the input is negative or positive, the absolute value returns the positive magnitude of that number.

For example:

• If \(x = 4\), then \(|x| = 4\)
• If \(x = -4\), then \(|x| = 4\)

In exercises involving functions like \(g(x) = |x|\), it’s important to remember that this operation simplifies the input to its positive equivalent. This is straightforward but essential when dealing with function composition.

The function \(g(x)\) simplifies many calculations, as seen in the step-by-step solution. By using \(g(-4) = 4\), it makes the application of other functions easier.

Linear Functions

Linear functions are among the simplest types of functions in mathematics. They can be represented by the equation \(f(x) = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. These functions graph to form a straight line and have constant rates of change.

In our particular exercise, \(f(x) = 5x - 2\) is a linear function. Here:

• \(m = 5\), indicating a steep upward slope.
• \(b = -2\), which means the line crosses the y-axis at \(-2\).

Linear functions are straightforward to evaluate at any given point. You simply multiply the input by the slope and then add the intercept. Here’s an example from the exercise:

• For \(f(5)\), calculate \(5 \times 5 - 2 = 23\).
• For \(f(4)\), calculate \(5 \times 4 - 2 = 18\).

When working with compositions involving linear functions, ensure each step is handled accurately according to the order of operations.

Evaluating Functions

Evaluating functions is the process of determining the output value of a function given a specific input. This involves substituting the input value into the function's equation and performing the necessary arithmetic operations.

When dealing with function compositions like \((f \circ g)(x)\) or \((g \circ f)(x)\), the evaluation process becomes a sequence of evaluations. First, you evaluate the inside function using the given input, and then you take the result to evaluate the outer function.For example:

• To find \((f \circ g)(-4)\), first compute \(g(-4)\) as \(4\), and then use this result in \(f(x)\) to get \(18\).
• To compute \((g \circ f)(5)\), first evaluate \(f(5)\), which is \(23\), and then apply \(g(x)\) to get the final result \(23\).

Remember, the processing order is key. Any misstep might lead to incorrect conclusions. Ensure you always substitute precisely and calculate accurately with confidence in your operations.