Problem 63 - Precalculus Essentials

Question

Find. a. \((f \circ g)(x)\) b. \((g \circ f)(x)\) c. \((f \circ g)(2)\) d. \((g \circ f)(2)\) $$f(x)=2 x-3, g(x)=\frac{x+3}{2}$$

Step-by-Step Solution

Answer

a. \((f \circ g)(x)=x\), b. \((g \circ f)(x)=x\), c. \((f \circ g)(2)=2\), d. \((g \circ f)(2)=2\)

Step 1: Compose \(f \circ g\)

To calculate \((f \circ g)(x)\), we need to substitute \(g(x)\) into \(f(x)\). So, wherever there is \(x\) in the \(f(x)\) function, we replace it with \(g(x)\), that leads to: \(f(g(x)) = 2 * g(x) - 3\). Substitute \(g(x) = \frac{x+3}{2}\) into \(f(g(x))\), we get \(f(g(x)) = 2 * (\frac{x+3}{2}) - 3 = x + 3 - 3 = x\)

Step 2: Compose \(g \circ f\)

Now, for \((g \circ f)(x)\), we substitute \(f(x)\) into \(g(x)\). Hence, wherever there is \(x\) in the \(g(x)\) function, we replace it with \(f(x)\), that leads to: \(g(f(x)) = \frac{f(x) + 3}{2}\). Substitute \(f(x) = 2x - 3\) into \(g(f(x))\), we get \(g(f(x)) = \frac{2x - 3 + 3}{2} = \frac{2x}{2} = x \)

Step 3: Compute \((f \circ g)(2)\)

After having the composite function \(f \circ g = x\), now substitute \(x = 2\) into it, we get \((f \circ g)(2) = 2 \)

Step 4: Compute \((g \circ f)(2)\)

Similarly, after having the composite function \(g \circ f = x\), now substitute \(x = 2\) into it, we get \((g \circ f)(2) = 2 \)

Key Concepts

Function Composition

Function composition is a powerful tool in mathematics where the output of one function becomes the input for another. Think of it as a sequence of operations where one function is applied to the results of another function. In the given exercise, composing functions means swapping each instance of the variable in one function with an entire other function.
For example, to find \((f \circ g)(x)\), we substitute the function \(g(x)\) into \(f(x)\). Seem difficult? It's just like replacing every \(x\) in \(f(x)\) with \(g(x)\), which makes our expression look like \(f(g(x))\). This substitution links the two functions, enabling us to explore how they interact.
Function composition helps streamline complicated expressions and provides a simpler framework to understand multi-step processes. It is crucial in breaking down complex mathematical problems into more manageable parts.

Mathematical Operations

When dealing with function composition, be ready to perform a series of mathematical operations. For the problem at hand, to compose \(f \circ g\), the multiplication, addition, and division are crucial. For instance, when we substitute \(g(x)\) into \(f(x)\) as in \(f(g(x)) = 2 * \left(\frac{x+3}{2}\right) - 3\), we first multiply \(2\) by \(\frac{x+3}{2}\).
Observe how we do these calculations step by step, ensuring each operation is clear:

• Multiply: Simplify \(2 * \left(\frac{x+3}{2}\right)\) which results in \(x+3\).
• Subtract: From \(x + 3\), subtract \(3\) to finally simplify and get \(x\).

This systematic progression demonstrates how different operations are applied sequentially to reduce more complicated strings of functions to manageable forms. Always pay close attention to the order of operations, as this can significantly change your results.

Substitution in Functions

In mathematics, substitution is like swapping out parts to see how they fit into other expressions. It transforms the known into something new by making strategic replacements. Here, when substituting \(g(x)\) into \(f(x)\), a simple substitution involves replacing every \(x\) in \(f(x)\) with the expression for \(g(x)\).
This technique helps us simplify or solve problems by expressing one function in terms of another. In the given exercise, the substitution process turns \(f(x) = 2x - 3\) into a new expression by slotting in \(g(x) = \frac{x+3}{2}\). Similarly, when we switch roles and put \(f(x)\) into \(g(x)\), we've practiced substitution again, showing how adaptable it is across varying scenarios.
Overall, substitution is a key skill in manipulating and understanding functions. Practicing it helps solidify your grasp of how functions interrelate, making even intimidating problems more approachable.