The Functions \[$ F \$\] And \[$ G \$\] Are Given. Evaluate \[$ F \circ G \$\] And Find The Domain Of The Composite Function \[$ F \circ G \$\].$\[ F(x) = \frac{5}{x+1} \\]$\[ G(x) = \frac{5}{x}

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The Functions of Composition: Evaluating and Finding the Domain of the Composite Function

In mathematics, the composition of functions is a fundamental concept that allows us to combine two or more functions to create a new function. Given two functions, { f $}$ and { g $}$, we can evaluate the composite function { f \circ g $}$ by substituting the expression for { g(x) $}$ into the function { f(x) $}$. In this article, we will evaluate the composite function { f \circ g $}$ and find the domain of the composite function { f \circ g $}$, where { f(x) = \frac{5}{x+1} $}$ and { g(x) = \frac{5}{x} $}$.

To evaluate the composite function { f \circ g $}$, we need to substitute the expression for { g(x) $}$ into the function { f(x) $}$. This means that we will replace { x $}$ in the function { f(x) $}$ with the expression { g(x) = \frac{5}{x} $}$.

{ f(g(x)) = f\left(\frac{5}{x}\right) = \frac{5}{\frac{5}{x}+1} $}$

To simplify the expression, we can multiply the numerator and denominator by { x $}$ to get rid of the fraction in the denominator.

{ f(g(x)) = \frac{5}{\frac{5}{x}+1} \cdot \frac{x}{x} = \frac{5x}{5+x} $}$

Therefore, the composite function { f \circ g $}$ is given by { f(g(x)) = \frac{5x}{5+x} $}$.

To find the domain of the composite function { f \circ g $}$, we need to consider the restrictions on the domain of both functions { f(x) $}$ and { g(x) $}$. The function { f(x) = \frac{5}{x+1} $}$ is defined for all real numbers except { x = -1 $}$, since the denominator cannot be zero. The function { g(x) = \frac{5}{x} $}$ is defined for all real numbers except { x = 0 $}$, since the denominator cannot be zero.

However, when we substitute the expression for { g(x) $}$ into the function { f(x) $}$, we get a new function { f(g(x)) = \frac{5x}{5+x} $}$. This function is defined for all real numbers except { x = -5 $}$, since the denominator cannot be zero.

Therefore, the domain of the composite function { f \circ g $}$ is all real numbers except { x = -5 $}$.

In conclusion, we have evaluated the composite function { f \circ g $}$ by substituting the expression for { g(x) $}$ into the function { f(x) $}$. We have also found the domain of the composite function { f \circ g $}$ by considering the restrictions on the domain of both functions { f(x) $}$ and { g(x) $}$. The composite function { f \circ g $}$ is given by { f(g(x)) = \frac{5x}{5+x} $}$, and its domain is all real numbers except { x = -5 $}$.

  1. Evaluate the composite function { f \circ g $}$ where { f(x) = \frac{1}{x-2} $}$ and { g(x) = \frac{1}{x+3} $}$.
  2. Find the domain of the composite function { f \circ g $}$ where { f(x) = \frac{1}{x-1} $}$ and { g(x) = \frac{1}{x+2} $}$.
  1. To evaluate the composite function { f \circ g $}$, we need to substitute the expression for { g(x) $}$ into the function { f(x) $}$.

{ f(g(x)) = f\left(\frac{1}{x+3}\right) = \frac{1}{\frac{1}{x+3}-2} $}$

To simplify the expression, we can multiply the numerator and denominator by { x+3 $}$ to get rid of the fraction in the denominator.

{ f(g(x)) = \frac{1}{\frac{1}{x+3}-2} \cdot \frac{x+3}{x+3} = \frac{x+3}{1-2(x+3)} $}$

Therefore, the composite function { f \circ g $}$ is given by { f(g(x)) = \frac{x+3}{1-2(x+3)} $}$.

  1. To find the domain of the composite function { f \circ g $}$, we need to consider the restrictions on the domain of both functions { f(x) $}$ and { g(x) $}$. The function { f(x) = \frac{1}{x-1} $}$ is defined for all real numbers except { x = 1 $}$, since the denominator cannot be zero. The function { g(x) = \frac{1}{x+2} $}$ is defined for all real numbers except { x = -2 $}$, since the denominator cannot be zero.

However, when we substitute the expression for { g(x) $}$ into the function { f(x) $}$, we get a new function { f(g(x)) = \frac{x+3}{1-2(x+3)} $}$. This function is defined for all real numbers except { x = -\frac{1}{2} $}$, since the denominator cannot be zero.

Therefore, the domain of the composite function { f \circ g $}$ is all real numbers except { x = -\frac{1}{2} $}$.

In conclusion, the composition of functions is a powerful tool in mathematics that allows us to combine two or more functions to create a new function. By evaluating the composite function and finding its domain, we can gain a deeper understanding of the behavior of the function and its restrictions. In this article, we have evaluated the composite function { f \circ g $}$ and found its domain, where { f(x) = \frac{5}{x+1} $}$ and { g(x) = \frac{5}{x} $}$. We have also provided example problems and solutions to help illustrate the concept.
Q&A: The Functions of Composition

In our previous article, we discussed the concept of composition of functions and evaluated the composite function { f \circ g $}$ where { f(x) = \frac{5}{x+1} $}$ and { g(x) = \frac{5}{x} $}$. We also found the domain of the composite function { f \circ g $}$. In this article, we will answer some frequently asked questions about the composition of functions and provide additional examples to help illustrate the concept.

A: The composition of functions is a way of combining two or more functions to create a new function. It is denoted by { f \circ g $}$ and is defined as { f(g(x)) $}$.

A: To evaluate the composite function, you need to substitute the expression for { g(x) $}$ into the function { f(x) $}$. This means that you will replace { x $}$ in the function { f(x) $}$ with the expression { g(x) $}$.

A: The domain of the composite function is the set of all possible input values for which the function is defined. To find the domain of the composite function, you need to consider the restrictions on the domain of both functions { f(x) $}$ and { g(x) $}$.

A: Yes, you can have multiple composite functions. For example, if you have three functions { f(x) $}$, { g(x) $}$, and { h(x) $}$, you can create a composite function { f \circ g \circ h $}$ by substituting the expression for { h(x) $}$ into the function { g(x) $}$, and then substituting the expression for { g(x) $}$ into the function { f(x) $}$.

A: A composite function is defined if the function { f(x) $}$ is defined for all values of { x $}$ in the domain of the function { g(x) $}$. If the function { f(x) $}$ is not defined for all values of { x $}$ in the domain of the function { g(x) $}$, then the composite function is not defined.

A: Yes, you can have a composite function with a variable in the denominator. For example, if you have two functions { f(x) = \frac{1}{x+1} $}$ and { g(x) = \frac{1}{x} $}$, you can create a composite function { f \circ g $}$ by substituting the expression for { g(x) $}$ into the function { f(x) $}$.

A: To simplify a composite function, you can use algebraic manipulations such as multiplying both sides of the equation by a common denominator or factoring out common terms.

  1. Evaluate the composite function { f \circ g $}$ where { f(x) = \frac{1}{x-2} $}$ and { g(x) = \frac{1}{x+3} $}$.
  2. Find the domain of the composite function { f \circ g $}$ where { f(x) = \frac{1}{x-1} $}$ and { g(x) = \frac{1}{x+2} $}$.
  1. To evaluate the composite function { f \circ g $}$, we need to substitute the expression for { g(x) $}$ into the function { f(x) $}$.

{ f(g(x)) = f\left(\frac{1}{x+3}\right) = \frac{1}{\frac{1}{x+3}-2} $}$

To simplify the expression, we can multiply the numerator and denominator by { x+3 $}$ to get rid of the fraction in the denominator.

{ f(g(x)) = \frac{1}{\frac{1}{x+3}-2} \cdot \frac{x+3}{x+3} = \frac{x+3}{1-2(x+3)} $}$

Therefore, the composite function { f \circ g $}$ is given by { f(g(x)) = \frac{x+3}{1-2(x+3)} $}$.

  1. To find the domain of the composite function { f \circ g $}$, we need to consider the restrictions on the domain of both functions { f(x) $}$ and { g(x) $}$. The function { f(x) = \frac{1}{x-1} $}$ is defined for all real numbers except { x = 1 $}$, since the denominator cannot be zero. The function { g(x) = \frac{1}{x+2} $}$ is defined for all real numbers except { x = -2 $}$, since the denominator cannot be zero.

However, when we substitute the expression for { g(x) $}$ into the function { f(x) $}$, we get a new function { f(g(x)) = \frac{x+3}{1-2(x+3)} $}$. This function is defined for all real numbers except { x = -\frac{1}{2} $}$, since the denominator cannot be zero.

Therefore, the domain of the composite function { f \circ g $}$ is all real numbers except { x = -\frac{1}{2} $}$.

In conclusion, the composition of functions is a powerful tool in mathematics that allows us to combine two or more functions to create a new function. By evaluating the composite function and finding its domain, we can gain a deeper understanding of the behavior of the function and its restrictions. In this article, we have answered some frequently asked questions about the composition of functions and provided additional examples to help illustrate the concept.