Function Composition And Domain Determination For F(x) = 1/x And G(x) = (x-3)/2

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Delving into Function Composition

In the realm of mathematics, function composition stands as a fundamental operation that combines two functions to create a new function. This operation, often denoted by the symbol "∘", essentially involves applying one function to the result of another. In this exploration, we embark on a journey to unravel the composition of two specific functions: $f(x) = \frac{1}{x}$ and $g(x) = \frac{x-3}{2}$. Our primary objective is to determine the formula for $(g \circ f)(x)$, which represents the composition of g with f, and subsequently simplify the resulting expression. Furthermore, we will delve into the intricacies of identifying the domain of this composite function, ensuring we comprehend the set of all possible input values for which the function is defined.

Function composition is a powerful tool that allows us to build complex functions from simpler ones. The notation $(g \circ f)(x)$ signifies that we first apply the function f to the input x, obtaining the result f(x). Then, we take this result and apply the function g to it, yielding g(f(x)). This sequential application of functions forms the core of function composition. Understanding the order in which functions are applied is crucial, as $(g \circ f)(x)$ is generally different from $(f \circ g)(x)$. In our case, we are specifically interested in the composition where f is applied first, followed by g.

The process of finding the formula for $(g \circ f)(x)$ involves substituting the function f(x) into the function g(x). This means that wherever we see x in the expression for g(x), we replace it with the entire expression for f(x). This substitution is the key to unraveling the composite function. Once we have performed this substitution, we often simplify the resulting expression to obtain a more concise and manageable formula. Simplification may involve algebraic manipulations such as combining like terms, canceling common factors, or applying algebraic identities. The goal is to express the composite function in its simplest and most understandable form.

Unveiling the Formula for (g ∘ f)(x)

To embark on our quest to find the formula for $(g \circ f)(x)$, we must first recall the definitions of our two functions:

f(x)=1xf(x) = \frac{1}{x}

g(x)=x−32g(x) = \frac{x-3}{2}

The notation $(g \circ f)(x)$ signifies that we first apply the function f to the input x, and then apply the function g to the result. In mathematical terms, this can be expressed as:

(g∘f)(x)=g(f(x))(g \circ f)(x) = g(f(x))

This equation tells us that to find $(g \circ f)(x)$, we need to substitute f(x) into g(x). In other words, we replace every instance of x in the expression for g(x) with the entire expression for f(x). Let's perform this substitution:

g(f(x))=g(1x)=(1x)−32g(f(x)) = g(\frac{1}{x}) = \frac{(\frac{1}{x}) - 3}{2}

Now, we have an expression for $(g \circ f)(x)$, but it is not yet in its simplest form. To simplify this expression, we need to perform some algebraic manipulations. Our goal is to eliminate the complex fraction, which is a fraction within a fraction. To do this, we can multiply both the numerator and denominator of the main fraction by x, the denominator of the inner fraction:

(1x)−32=x((1x)−3)2x=1−3x2x\frac{(\frac{1}{x}) - 3}{2} = \frac{x((\frac{1}{x}) - 3)}{2x} = \frac{1 - 3x}{2x}

Therefore, after simplification, we arrive at the formula for $(g \circ f)(x)$:

(g∘f)(x)=1−3x2x(g \circ f)(x) = \frac{1 - 3x}{2x}

This equation represents the composite function formed by applying f first and then g. It is a rational function, which means it is a ratio of two polynomials. The numerator is the polynomial 1 - 3x, and the denominator is the polynomial 2x. Understanding the form of this function is crucial for determining its domain, which is the set of all possible input values for which the function is defined.

Determining the Domain of (g ∘ f)(x)

Now that we have found the formula for $(g \circ f)(x)$, our next crucial step is to determine its domain. The domain of a function is the set of all possible input values (x-values) for which the function produces a valid output. In simpler terms, it's the set of numbers we can "plug in" to the function without causing any mathematical errors, such as division by zero or taking the square root of a negative number.

In the context of composite functions, determining the domain requires careful consideration of both the individual functions involved and the composite function itself. We must ensure that any value we input into the composite function is valid for both the inner function (f in our case) and the outer function (g in our case), as well as the final composite function. This means we need to identify any restrictions on the input values imposed by each function.

Let's start by examining the individual functions:

  • Function f(x) = 1/x: This function is a rational function, which means it has a denominator. The key restriction for rational functions is that the denominator cannot be equal to zero. If the denominator were zero, the function would be undefined because division by zero is not a mathematically valid operation. Therefore, for f(x) = 1/x, the domain is all real numbers except for x = 0. We can express this domain in interval notation as $(-\infty, 0) \cup (0, \infty)$.

  • Function g(x) = (x-3)/2: This function is also a rational function, but its denominator is a constant (2), which is never zero. Therefore, there are no restrictions on the input values for g(x). The domain of g(x) is all real numbers, which can be expressed in interval notation as $(-\infty, \infty)$.

Now, let's consider the composite function:

  • Composite Function $(g \circ f)(x) = \frac{1 - 3x}{2x}$

    This function is also a rational function, and again, we must ensure that the denominator is not zero. The denominator here is 2x, which is zero when x = 0. Therefore, similar to f(x), the composite function $(g \circ f)(x)$ is undefined when x = 0.

However, determining the domain of a composite function involves more than just looking at the final expression. We must also consider the domain restrictions imposed by the inner function, f(x), because the output of f(x) becomes the input for g(x). In our case, f(x) = 1/x, which, as we established earlier, has a domain of all real numbers except for x = 0. This means that even if the final expression for $(g \circ f)(x)$ were defined at x = 0, it would still be excluded from the domain because it is not in the domain of f(x).

Therefore, to find the domain of $(g \circ f)(x)$, we need to consider the following:

  1. Values of x that make the denominator of f(x) equal to zero.
  2. Values of x that make the denominator of $(g \circ f)(x)$ equal to zero.

In our case, both f(x) and $(g \circ f)(x)$ have a denominator of zero when x = 0. Thus, the domain of $(g \circ f)(x)$ is all real numbers except for x = 0. In interval notation, this is expressed as:

Domain of $(g \circ f)(x) = (-\infty, 0) \cup (0, \infty)$

This means that we can input any real number into the function $(g \circ f)(x)$ except for 0. Inputting 0 would lead to division by zero, making the function undefined.

Rounding the Answer (If Applicable)

In some cases, when dealing with domains that involve inequalities or specific numerical values, we might be asked to round our answer to a certain decimal place. However, in this particular problem, the domain of $(g \circ f)(x)$ is expressed as an interval, which represents a continuous range of values. There are no specific numerical values that need to be rounded. The domain is precisely defined as all real numbers except for 0, and the interval notation $(-\infty, 0) \cup (0, \infty)$ accurately represents this domain.

Therefore, no rounding is necessary in this case. The final answer for the domain of $(g \circ f)(x)$ is $(-\infty, 0) \cup (0, \infty)$.

Conclusion

In this exploration, we have successfully navigated the realm of function composition, specifically focusing on the functions $f(x) = \frac{1}{x}$ and $g(x) = \frac{x-3}{2}$. Our journey began with the determination of the formula for the composite function $(g \circ f)(x)$, which we found to be $\frac{1 - 3x}{2x}$. This process involved substituting f(x) into g(x) and simplifying the resulting expression.

Subsequently, we delved into the crucial aspect of identifying the domain of $(g \circ f)(x)$. We meticulously analyzed the domain restrictions imposed by both the individual functions, f(x) and g(x), and the composite function itself. This analysis revealed that the domain of $(g \circ f)(x)$ is all real numbers except for x = 0, which we expressed in interval notation as $(-\infty, 0) \cup (0, \infty)$.

Understanding function composition and domain determination is paramount in mathematics, as it allows us to build and analyze complex functions from simpler components. The ability to identify domain restrictions ensures that we are working within the valid input range of a function, preventing mathematical errors and ensuring meaningful results. This exploration has provided a comprehensive understanding of these concepts in the context of specific functions, laying a solid foundation for further mathematical endeavors.