Problem 9 - Algebra 2 and Trigonometry

Question

In \(3-10\) , evaluate each composition for the given values if \(f(x)=3 x\) and \(g(x)=x-2\) $$ \mathrm{f}\left(\mathrm{g}\left(\frac{2}{3}\right)\right) $$

Step-by-Step Solution

Answer

The composition \( f(g(\frac{2}{3})) = -4 \).

Step 1: Evaluate Inner Function

First, evaluate the inner function \(g(x)\) by substituting \(x = \frac{2}{3}\). \[ g\left(\frac{2}{3}\right) = \frac{2}{3} - 2 = \frac{2}{3} - \frac{6}{3} = -\frac{4}{3} \] So \(g\left(\frac{2}{3}\right) = -\frac{4}{3}\).

Step 2: Substitute into Outer Function

Now, we take the result from Step 1 and substitute \(-\frac{4}{3}\) into the outer function \(f(x)\). \[ f(-\frac{4}{3}) = 3 \times \left(-\frac{4}{3}\right) = -4 \]

Step 3: Conclusion

The value of the composition \( f(g(\frac{2}{3})) \) is \(-4\).

Key Concepts

Inner Function Evaluation

In function composition, the first crucial step is to evaluate the inner function. An inner function is essentially the function that is applied first, before another function is applied to its output. So, for the exercise given, the inner function is denoted as \( g(x) \). When given a specific value to substitute into \( g(x) \), in this case \( x = \frac{2}{3} \), you systematically replace \( x \) with \( \frac{2}{3} \) in the formula for \( g(x) \).

Here, \( g(x) = x - 2 \). By substituting, we perform the following operation:

• Subtract \( 2 \) from \( \frac{2}{3} \), which involves converting \( 2 \) into \( \frac{6}{3} \) to have a common denominator.
• The result is \( \frac{2}{3} - \frac{6}{3} = -\frac{4}{3} \).

The evaluation of \( g\left(\frac{2}{3}\right) \) gives us \(-\frac{4}{3} \).

Outer Function Substitution

Once the inner function has been evaluated, the next step in function composition is substituting its result into the outer function. The outer function is the one that is applied after the inner function, often using the result from the inner function's evaluation as its input.

In our example, the outer function is \( f(x) = 3x \). You take the result from the previous step, which is \(-\frac{4}{3}\), and substitute it for \( x \) in \( f(x) \):

• Replace \( x \) with \(-\frac{4}{3} \) in \( 3x \).
• This becomes \( 3 \times \left(-\frac{4}{3}\right) \).

After carrying out the multiplication, you find that \( f\left(-\frac{4}{3}\right) = -4 \). This is how the outer function utilizes the output from the inner function.

Composition of Functions

The composition of functions is a fundamental concept where the output of one function becomes the input of another. Notation-wise, when you see \( f(g(x)) \), it signifies that function \( g \) is inside function \( f \). This composition process can involve several layers of functions, but in our example, we deal with only two - \( f \) and \( g \).

This concept is essential because it helps us deal with complex equations in an organized manner, where the solution of one function influences the outcome of the next. The main elements to consider in function compositions include:

• Identifying which function is inside or outside.
• Following the order of operations strictly, first dealing with the inner functions.

Hence, in the exercise, knowing that \( g(x) \) feeds into \( f(x) \) is key to processing the given expressions correctly.

Evaluation Process

The evaluation process for the composition of functions involves a strategic method. It's a series of well-defined steps that lead to an accurate answer. In this case, the exercise provided steps illustrate this flow well, ensuring clarity and understanding.

First, identify and evaluate the inner function as described earlier. Then proceed to substitute its output into the outer function. Carefully executing each step minimizes errors and solidifies the understanding of function relationships.

• Begin with the inner function, here \( g(x) \).
• Move to evaluating the outer function, \( f(x) \), using the output from the first step.

This systematic approach not only makes it easier to work through individual compositions but also develops a deeper mathematical intuition for more complex functions you might encounter.