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Let \(f(x)=|x|, g(x)=x^{2}-4\) \(F(x)=\sqrt{x},\) and \(G(x)=1 /(x-2) .\) Determine the following composite functions and give their domains. $$G \circ G$$

Short Answer

Expert verified
Answer: The composite function \(G \circ G\) is given by \((G \circ G)(x) = \frac{x - 2}{-2x + 5}\), and its domain is \(\{ x \in \mathbb{R} : x \neq \frac{5}{2} \}\).

Step by step solution

01

1. Compute the composite function G o G

The composite function \(G \circ G\) is defined as \((G \circ G)(x) = G(G(x))\). To compute this, we substitute the expression of \(G(x)\) into itself. Given \(G(x) = \frac{1}{x - 2}\), we will substitute this into the G function: \((G \circ G)(x) = G(G(x)) = G\left(\frac{1}{x - 2}\right)\) Now, replace the expression for \(G(x)\): \((G \circ G)(x) = \frac{1}{\left(\frac{1}{x - 2}\right) - 2}\)
02

2. Simplify the expression for G o G

We will simplify the expression to find the composite function \((G \circ G)(x)\): \((G \circ G)(x) = \frac{1}{\frac{1}{x - 2} - 2}\) To eliminate the complex fraction, we will multiply the numerator and denominator by \((x - 2)\): \((G \circ G)(x) = \frac{1(x - 2)}{\left(\frac{1}{x - 2}(x - 2)\right) - 2(x - 2)}\) Simplify the expression: \((G \circ G)(x) = \frac{x - 2}{1 - 2(x - 2)}\) Distribute the \(-2\) in the denominator: \((G \circ G)(x) = \frac{x - 2}{1 - 2x + 4}\) Combine the constants in the denominator: \((G \circ G)(x) = \frac{x - 2}{-2x + 5}\) Now we have found the composite function \((G \circ G)(x)\).
03

3. Find the domain of the function G o G

To find the domain of \((G \circ G)(x)\), we need to find the values of \(x\) for which the function is defined. For the given function, the denominator should not be equal to zero, otherwise, the function is undefined. Set the denominator to zero and solve for \(x\): \(-2x + 5 = 0\) Add \(2x\) to both sides: \(5 = 2x\) Divide both sides by \(2\): \(x = \frac{5}{2}\) So the function is undefined when \(x = \frac{5}{2}\). Therefore, to find the domain, we will exclude this value from the set of all real numbers: Domain of \((G \circ G)(x) = \{ x \in \mathbb{R} : x \neq \frac{5}{2} \}\) #Conclusion# The composite function \((G \circ G)(x)\) is given by: \((G \circ G)(x) = \frac{x - 2}{-2x + 5}\) The domain of \((G \circ G)(x)\) is: \(\{ x \in \mathbb{R} : x \neq \frac{5}{2} \}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Composition
Function composition is like a recipe where you create a new cuisine by combining and processing the ingredients in a specific order. Similarly, in mathematics, when you have two functions, say f(x) and g(x), composing them means applying one function to the result of the other function. The notation for this operation is \( (g \circ f)(x) \) which you should read as 'g of f of x'.

Imagine you have \( f(x) = x+1 \) and \( g(x) = x^2 \) and you wish to compose \( g \circ f \) which is \( g(f(x)) \) or \( (x+1)^2 \) through substitution. You apply \( f(x) \) first, adding 1 to your input, then apply \( g(x) \) to the result, squaring the new number you just obtained.

During the exercise, we performed a slightly more intricate dance of substitution with rational functions, ultimately simplifying it to yield the composite function \( (G \circ G)(x) = \frac{x - 2}{-2x + 5} \) without getting tripped up by complex fractions or undefined values - all in a day's work for function composition!
Domain of a Function
The domain of a function is the set of all possible inputs for which the function is defined. Mathematically, it's all the 'x' values you can plug into a function without causing mayhem like dividing by zero or taking a square root of a negative number in the real number system.

For the expression \( (G \circ G)(x) \) derived in the exercise, we ensure the values we use for 'x' don't cause the function to implode. Since dividing by zero is a no-no in math, we find the non-permissible value by setting the denominator equal to zero and solving for 'x'. Here, that naughty number is \( \frac{5}{2} \). The rest of the real numbers are like the cool kids at the party - they get in with no issues. Hence, the domain is all real numbers except \( \frac{5}{2} \), noted as \( \{ x \in \mathbb{R} : x eq \frac{5}{2} \} \).

Understanding the domain is crucial because it tells us the range of values we can explore without running into a mathematical faux pas, ensuring the function's integrity is maintained.
Rational Functions
Ponder on a fraction with polynomials on the top and bottom, and you've got yourself a rational function. In the spotlight is the rational function \( G(x) = \frac{1}{x - 2} \) which also stars in the composed function \( (G \circ G)(x) \).

Rational functions often show up with all sorts of asymptotic behavior and interesting graphical features which arise from their denominators. When the denominator equals zero, you hit a vertical asymptote, a line the function will cozily approach but never actually touch or cross.

In the exercise, we tiptoed around the denominator zero issue by determining the domain. This tells us where the function exists and behaves properly and helps us avoid unseemly mathematical situations. Similarly, when dealing with rational functions, always keep an eye on the denominator - it holds the secret to understanding much of the function's nature.

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