Problem 74 - College Algebra Essentials

Question

Find a. \((f \circ g)(x) \qquad\) b. the domain of \(f \circ g\) $$f(x)=x^{2}+1, g(x)=\sqrt{2-x}$$

Step-by-Step Solution

Answer

a. The composition of the functions is \( (f \circ g)(x) = 3 - x \).\nb. The domain of \( f \circ g \) is \( -\infty <= x <=2 \).

Step 1: Calculate the composition of functions

The composition of two functions, \( f \) and \( g \), denoted as \( f \circ g \), is defined as \( (f \circ g)(x) = f(g(x)) \). By simply replacing every instance of \( x \) in \( f \) with the function \( g(x) \), we get \( (f \circ g)(x) \). Therefore, substituting \( g(x) =\sqrt{2-x} \) into \( f(x)=x^{2}+1 \), we get: \((f \circ g)(x)= (\sqrt{2-x})^2 + 1\).

Step 2: Simplify the expression

Simplifying the above expression, we get: \((f \circ g)(x) = 2-x+1 = 3-x\).

Step 3: Calculate the domain of the composite function

The domain of \( f \circ g \) is the set of all real numbers \( x \) such that \( g(x) \) is defined and \( g(x) \) lies in the domain of \( f \). The domain of \( g(x)=\sqrt{2-x} \) is \( -\infty <= x <=2 \), because the expression under the square root must be greater than or equal to zero. The result \( g(x) \) must also lie in the domain of \( f \), but since \( f \) is a parabola shifted one unit upward, its domain is all real numbers. Therefore, the domain of \( f \circ g \) is \( -\infty <= x <=2 \).

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial for any student delving into higher mathematics. The domain refers to the complete set of possible values of the independent variable, often denoted as 'x', for which the function is defined. In simpler terms, it’s the range of 'input' or 'x' values that you can plug into the function and get a valid output.

When calculating the domain of a composition of functions, such as \(f \circ g\), you must consider the domain of the inner function, \(g(x)\), and ensure that its outputs are suitable for the domain of the outer function, \(f(x)\). For \(f(x) = x^2 + 1\) and \(g(x) = \sqrt{2-x}\), the domain of \(g\) is restricted because square roots require non-negative arguments. Therefore, the domain of \(g\) is \(x \leq 2\). Since \(f\) is a quadratic function with no restrictions, its domain is all real numbers. However, a composition like \(f\circ g\) can only accommodate inputs from the domain of \(g\), as \(g\)’s output serves as \(f\)’s input.

Thus, when calculating domains for compositions, doing so step-wise—first finding the domain of \(g\), then assessing whether the range of \(g\) fits within the domain of \(f\)—ensures accuracy. In the given problem, the domain of the composition \(f \circ g\) is, therefore, all real numbers \(x\) such that \(x\leq 2\), or put differently, \( -\infty \leq x \leq2 \).

Function Operations

Function operations, including the composition of functions, are operations where functions are combined in various ways to produce new functions. In the case of \(f \circ g\), the function \(g\) is applied first, and then the function \(f\) is applied to the result of \(g\). This is what it means to 'compose' functions—putting one function into another.

There are other types of function operations such as addition, subtraction, multiplication, and division of functions, which are respectively represented as \( f + g \), \( f - g \), \( f \cdot g \), and \( \frac{f}{g} \).

Each has its rules and patterns. For example, for addition, we simply add the outputs of each function (\(f(x) + g(x)\) for all \(x\) in the domain common to both functions). But composition is unique—it’s not about performing a simple arithmetic operation on the outputs, it’s about applying one function to the output of another, creating a ripple effect of transformation.

Mastering these operations is fundamental to progressing in mathematics, as they form the building blocks for more complex concepts. The exercise showcases the composition function operation and its distinctive impact on both the resulting function and its domain.

Simplifying Expressions

Simplifying expressions is a vital skill that clears up complexity and reveals the core aspect of an algebraic expression. To simplify means to alter an expression to make it easier to understand or work with, without changing its value. This process often involves combining like terms, applying the distributive property, and reducing fractions.

In the context of function composition, simplification can make it easier to identify the domain and understand the nature of the composite function. For instance, the solution of the exercise involved simplifying \( (f \circ g)(x) = (\sqrt{2-x})^2 + 1 \) to \(3 - x\). This involves squaring the square root (which cancels out) and combining like terms. After simplification, it’s easier to see that the composed function, in essence, is a straight line, which is simple compared to the initial expression.

Simplification does not just stop at combining terms; it can include factoring expressions, canceling common factors, and rationalizing denominators. Each of these methods strives to convert expressions into their least complex form, thus laying a clear path for further operations or evaluations.