Problem 19 - Calculus with Concepts in Calculus

Question

Find the domain and rule of \(g \circ f\) and \(f \circ g\). \(f(x)=1-x\) and \(g(x)=2 x+5\)

Step-by-Step Solution

Answer

The domain and rule of \(g \circ f\) are all real numbers and \(7 - 2x\); for \(f \circ g\), they are all real numbers and \(-2x - 4\).

Step 1: Understanding Composition

To find \( g \circ f \) and \( f \circ g \), we perform function composition. This means that for \( g \circ f \), you substitute \( f(x) \) into \( g(x) \), and for \( f \circ g \), you substitute \( g(x) \) into \( f(x) \). We also need to find the domains for both composed functions.

Step 2: Find \( g \circ f(x) \)

Substitute \( f(x) = 1 - x \) into \( g(x) = 2x + 5 \):\[g(f(x)) = g(1-x) = 2(1-x) + 5 = 2 - 2x + 5 = 7 - 2x\]

Step 3: Determine Domain of \( g \circ f \)

The domain of \( g \circ f \) is determined by the domain of \( f(x) \) since \( g(x) = 2x+5 \) can take any real number input. Thus, the domain of \( g \circ f \) is all real numbers \((-\infty, \infty)\) as \( f(x) = 1-x \) has no restrictions on \( x \).

Step 4: Find \( f \circ g(x) \)

Substitute \( g(x) = 2x + 5 \) into \( f(x) = 1-x \):\[f(g(x)) = f(2x+5) = 1 - (2x + 5) = 1 - 2x - 5 = -2x - 4\]

Step 5: Determine Domain of \( f \circ g \)

The domain of \( f \circ g \) is determined by the domain of \( g(x) \) since \( f(x) = 1-x \) can take any real number input. Thus, the domain of \( f \circ g \) is all real numbers \((-\infty, \infty)\) as \( g(x) = 2x+5 \) has no restrictions on \( x \).

Key Concepts

Domain of Functions

The domain of a function refers to all the possible input values (generally represented as \(x\)) that a function can accept without leading to any errors or undefined behavior. In our example, we have two functions, \(f(x) = 1 - x\) and \(g(x) = 2x + 5\). Both of these functions are linear, which means they can handle any real numbers as inputs. This results in both functions having the domain of all real numbers, expressed as \((-\infty, \infty)\). Understanding the domain becomes particularly important when working with composed functions like \(g \circ f\) and \(f \circ g\). In function composition, it’s crucial to ensure that the chosen input value for the first function in the composition can produce an output accepted by the second function. In our example, since both \(f(x)\) and \(g(x)\) can handle all real numbers, the composed functions \(g \circ f\) and \(f \circ g\) also have domains that cover all real numbers.

Combining Functions

Combining functions through composition involves plugging one function into another. It’s a straightforward way to combine operations that different functions perform on a given input. To create the composition \(g \circ f\), for example, you plug the entire function \(f(x) = 1 - x\) into the function \(g(x) = 2x + 5\). This results in a new function, \[g(f(x)) = g(1-x) = 2(1-x) + 5\], which simplifies down to \[7-2x\]. For the composition \(f \circ g\), you plug \(g(x)\) into \(f(x)\), yielding: \[f(g(x)) = f(2x+5) = 1 - (2x + 5)\], which further simplifies to \(-2x - 4\). The key here is to carefully substitute the entire expression of one function into every place where \(x\) appears in the other function. Composition allows you to explore how multiple transformations of an input can be combined into a single operation.

Algebraic Manipulation

Algebraic manipulation is crucial when you are working with function composition. It involves simplifying and rearranging expressions to find a more convenient or understandable form. Let's look at the example of composing \(g(f(x)) = 7 - 2x\). Here, after substituting \(f(x)\) into \(g(x)\), you perform arithmetic operations:

• Multiply the constant 2 by both 1 and \(-x\) to get \(2 - 2x\).
• Add 5 to get the final simplified form \(7 - 2x\).

Likewise, for \(f(g(x)) = -2x - 4\), you should:

• Subtract \(2x + 5\) from 1, which first expands to \(1 - 2x - 5\).
• Then simplify to \(-2x - 4\).

These steps of basic arithmetic—combining like terms, distributing constants, and rearranging terms—are the essence of algebraic manipulation.Performing these steps correctly ensures that your composed function provides an accurate representation of the transformations it represents.