Problem 25 - College Algebra

Question

Find: a. \((f \circ g)(x)\) b. \(\left(g^{\circ} f\right)(x)\) c. \((f \circ g)(2)\) $$f(x)=\sqrt{x}, g(x)=x-1$$

Step-by-Step Solution

Answer

a. \((f \circ g)(x)= \sqrt{x-1}\), b. \((g \circ f)(x)= \sqrt{x}-1\), c. \((f \circ g)(2)= 1.\)

Step 1: Compute \((f \circ g)(x)\)

To find \((f \circ g)(x)\), substitute \(g(x)\) into \(f(x)\). Here, \(f(x)=\sqrt{x}\) and \(g(x)=x-1\). This gives \((f \circ g)(x) = \sqrt{g(x)} = \sqrt{x-1}.\)

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

To find \((g \circ f)(x)\), substitute \(f(x)\) into \(g(x)\). Here, \(f(x)=\sqrt{x}\) and \(g(x)=x-1\). This gives \((g \circ f)(x) = g(f(x)) = (\sqrt{x})-1.\)

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

To find \((f \circ g)(2)\), substitute 2 into \((f \circ g)(x)\). Here \((f \circ g)(2)= \sqrt{2-1}= \sqrt{1} =1.\)

Key Concepts

Function Operations

Understanding how to manipulate and operate with functions is fundamental in algebra and calculus. Function operations include addition, subtraction, multiplication, division, and composition of functions. For example, if you have two functions, let's say \( f(x) = 2x + 3 \) and \( g(x) = x^2 \), you can create new functions by performing operations with \( f \) and \( g \).The sum of \( f \) and \( g \) is \( (f + g)(x) = f(x) + g(x) = (2x + 3) + x^2 \). Similarly, the product of \( f \) and \( g \) is \( (f \times g)(x) = f(x) \times g(x) = (2x + 3) \times x^2 \).
What makes these operations particularly interesting is how they combine properties of two different functions into a new, unique function, which can be analyzed in terms of domain, range, and behavior. We could even explore composite functions, which are a special type of function operation.

Composite Functions

A composite function is formed when one function is applied to the result of another function. Not to be confused with simply multiplying or adding functions together, composing functions is about applying one function to the outputs of another.

Forming a Composite Function

To form a composite function, usually denoted as \( (f \underline{\phantom{xxx}}\circ\ \underline{\phantom{xxx}} g) \), we would take an input value, perform \( g \) on it, and then use the output to perform \( f \). The notation \( f(g(x)) \) or \( f \underline{\phantom{xxx}}\circ\ \underline{\phantom{xxx}} g(x) \) means exactly that.For instance, if \( f(x) = \sqrt{x} \) and \( g(x) = x - 1 \), then the composite function \( (f \underline{\phantom{xxx}}\circ\ \underline{\phantom{xxx}} g)(x) \) would be \( f(g(x)) = \sqrt{x - 1} \). Importantly, the domain of the composite function is determined by the domains of both \( f \) and \( g \), and where \( g(x) \) is within the domain of \( f \).
When creating composite functions, it's essential to understand the order does matter—a quintessential example of how function operations are not necessarily commutative, which means \( f \underline{\phantom{xxx}}\circ\ \underline{\phantom{xxx}} g \) is not the same as \( g \underline{\phantom{xxx}}\circ\ \underline{\phantom{xxx}} f \). This can lead to different outputs and even entirely different domains for the composite functions.

Function Evaluation

Function evaluation might seem straightforward, but it's where you actually get to see what functions and their compositions do with specific values. Evaluating a function \( f(x) \) at a given value involves replacing the variable \( x \) with that value and performing the necessary operations.

Evaluating at a Specific Point

For example, given \( f(x) = \sqrt{x} \) and \( g(x) = x - 1 \), to evaluate \( (f \underline{\phantom{xxx}}\circ\ \underline{\phantom{xxx}} g)(2) \), you would find the output of \( g \) when \( x = 2 \), which is \( 1 \), and then find what \( f \) does to that output, which in turn gives you \( f(1) \), and thus the final answer is \( \sqrt{1} = 1 \).
It's important for students to practice evaluating functions at different points because it builds understanding of how a function behaves and can also reveal patterns or special properties of the function, such as symmetry or period. It's not enough to know the operations theoretically; evaluating functions consolidates the understanding of their practical application.