Problem 61 - College Algebra Essentials

Question

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

Step-by-Step Solution

Answer

Thus, \( (f \circ g)(x) = \sqrt{x-1} \), \( (g \circ f)(x) = \sqrt{x} - 1 \), \( (f \circ g)(2) = 1 \), and \( (g \circ f)(2) = \sqrt{2} - 1 \).

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

We need to substitute \( g(x) = x - 1 \) into \( f(x) = \sqrt{x} \). Therefore, \( (f \circ g)(x) = f(g(x)) = \sqrt{x-1} \)

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

Now we substitute \( f(x) = \sqrt{x} \) into \( g(x) = x - 1 \). Thus, \( (g \circ f)(x) = g(f(x)) = \sqrt{x} - 1

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

Substitute \( x = 2 \) into the function \( (f \circ g)(x) = \sqrt{x-1} \). That gives us \( (f \circ g)(2) = \sqrt{2-1} = 1 \)

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

Substitute \( x = 2 \) into the function \( (g \circ f)(x) = \sqrt{x} - 1 \). We get \( (g \circ f)(2) = \sqrt{2} - 1 \)

Key Concepts

Composition of Functions Algebra

Understanding the composition of functions is a foundational topic in algebra. It involves combining two functions in such a way that the output of one function becomes the input for the other. In simpler terms, suppose we have two functions, f and g. The composition \(f \circ g\) is read as 'f composed with g' and it means you first apply g, and then apply f to the result of g.

To find \(f \circ g\)(x), you start with the innermost function. In the given exercise, \( g(x) = x - 1 \), so you would first subtract 1 from x. Next, you take that result and use it as your new input for the function \( f(x) = \sqrt{x} \), effectively taking the square root of the previous outcome. This cascade of operations forms the crux of function composition in algebra.

It's important to note that \(f \circ g\) is not the same as \(g \circ f\); the composition of functions is not commutative, meaning their order matters a great deal. When students recognize the procedures and correctly apply them, solving these algebraic expressions becomes much straightforward.

Evaluating Function Compositions

Evaluating function compositions is often a step-by-step process. To properly evaluate \(f \circ g\)(x) or \(g \circ f\)(x), follow the order of operations to ensure that you apply each function correctly. Using the given example with \(f(x) = \sqrt{x} \) and \(g(x) = x - 1\), we look to the step-by-step solution to understand the process.

For \(f \circ g\):

• We plug \(g(x)\) into \(f(x)\), giving us \(\sqrt{x-1}\).
• Then for a specific value, say when \(x = 2\), we substitute that in to get the final numerical answer: \(\sqrt{2-1} = 1\).

Similarly, for \(g \circ f\), one would perform the following steps:

• Insert \(f(x)\) into \(g(x)\), resulting in \(\sqrt{x} - 1\).
• With \(x = 2\), the expression simplifies to \(\sqrt{2} - 1\).

Evaluation of function compositions demands careful substitution and arithmetic to arrive at the right result. Starting from the inner function and working outwards, and remembering to feed the output of one function into the input of another, are key themes for success in these problems.

Function Operations

Function operations include tasks such as function composition, but they also extend to other operations like addition, subtraction, multiplication, and division of functions. While the textbook example focuses on composition, it's worthwhile to mention these other operations shortly.

If we were to add or subtract functions, for instance, we would simply add or subtract their outputs—for example, \(f(x) + g(x) = \sqrt{x} + (x - 1)\). These operations are done directly and don’t involve substitution as with composition.

Keep in mind that each of these function operations has its set of rules. With comprehension and practice in performing various function operations, students can expand their problem-solving toolkit, making it easier to tackle a wider range of algebraic challenges.