Composite Function And Domain: A Step-by-Step Guide

Hey guys! Today, we're diving into the fascinating world of composite functions and how to determine their domains. We'll be working with two functions, $f(x)=\frac{1}{x+4}$ and $g(x)=x^2+4x-16$, and we'll find the composite function $(f \circ g)(x)$ and its domain. Let's get started!

Finding the Composite Function $(f \circ g)(x)$

So, what exactly is a composite function? Well, it's basically plugging one function into another. In our case, $(f \circ g)(x)$ means we're taking the function $g(x)$ and plugging it into the function $f(x)$.

Here's how we do it step-by-step:

1. Write down the functions:

   • $f(x) = \frac{1}{x+4}$
   • $g(x) = x^2 + 4x - 16$

2. Substitute $g(x)$ into $f(x)$: This means replacing the 'x' in $f(x)$ with the entire expression for $g(x)$.

   $(f \circ g)(x) = f(g(x)) = \frac{1}{g(x) + 4} = \frac{1}{(x^2 + 4x - 16) + 4}$

3. Simplify the expression: Now, let's simplify the denominator.

   $(f \circ g)(x) = \frac{1}{x^2 + 4x - 16 + 4} = \frac{1}{x^2 + 4x - 12}$

4. Factor the denominator (if possible): Factoring the quadratic expression in the denominator can help us identify any values of x that would make the denominator zero, which are critical for determining the domain.

   $x^2 + 4x - 12 = (x+6)(x-2)$

   So, $(f \circ g)(x) = \frac{1}{(x+6)(x-2)}$

Therefore, the simplified form of the composite function is $(f \circ g)(x) = \frac{1}{(x+6)(x-2)}$. This is a crucial step, and making sure you get the algebra right here is super important. Double-check your work! A small error here can throw off the rest of the problem. Remember, composite functions are all about carefully substituting and simplifying. Once you understand this process, you can tackle more complex functions with ease. Practice is key, so try out a few more examples to solidify your understanding. Also, keep in mind that the order matters. The composite function $(g \circ f)(x)$ would be calculated differently, with $f(x)$ being plugged into $g(x)$. Understanding the concept of function composition opens doors to more advanced topics in calculus and analysis, so mastering it now will definitely pay off in the long run. Don't be afraid to ask for help or look up more examples if you are struggling. Keep practicing, and you will master this skill in no time!

Determining the Domain of $(f \circ g)(x)$

Alright, now that we've found the composite function, let's figure out its domain. Remember, the domain of a function is the set of all possible input values (x-values) for which the function is defined.

For rational functions like $(f \circ g)(x) = \frac{1}{(x+6)(x-2)}$, we need to watch out for values of x that make the denominator equal to zero, because division by zero is a big no-no!

Here's how to find the domain:

1. Identify values that make the denominator zero: We already factored the denominator: $(x+6)(x-2)$. So, the denominator is zero when $x+6 = 0$ or $x-2 = 0$.

   Solving these equations gives us $x = -6$ and $x = 2$.

2. Exclude these values from the set of all real numbers: This means our domain includes all real numbers except -6 and 2.

3. Express the domain in interval notation: In interval notation, we write the domain as a union of intervals that include all allowable values. Since -6 and 2 are excluded, we use parentheses.

   Domain: $(-\infty, -6) \cup (-6, 2) \cup (2, \infty)$

So, the domain of $(f \circ g)(x)$ is $(-\infty, -6) \cup (-6, 2) \cup (2, \infty)$. Understanding the domain is crucial because it tells us where our function is well-behaved. In this case, the function is defined for all real numbers except -6 and 2. At these points, the function is undefined. When defining the domain, make sure you consider any restrictions imposed by the original functions, $f(x)$ and $g(x)$, as well. In our example, $f(x)$ has a restriction at $x=-4$, and we'll consider how this might affect the overall domain. Always remember that the domain represents the set of all allowable inputs that will produce a real number output. When in doubt, you can graph the function and visually confirm the domain. Careful consideration of all these factors will help you accurately determine the domain of the composite function.

Considering the Domain of Original Functions

Now, let's circle back to the original functions, $f(x)$ and $g(x)$, and see if their domains impose any further restrictions on the domain of the composite function $(f \circ g)(x)$.

1. Domain of $f(x)$: $f(x) = \frac{1}{x+4}$ is defined for all real numbers except $x = -4$. So, the domain of $f(x)$ is $(-\infty, -4) \cup (-4, \infty)$.

2. Domain of $g(x)$: $g(x) = x^2 + 4x - 16$ is a polynomial, and polynomials are defined for all real numbers. So, the domain of $g(x)$ is $(-\infty, \infty)$.

3. Impact on $(f \circ g)(x)$: Since $g(x)$ is plugged into $f(x)$, we need to make sure that $g(x)$ does not equal -4 for any value of x, because that would make $f(g(x))$ undefined. In other words, we need to solve the equation $g(x) = -4$.

   $x^2 + 4x - 16 = -4$ $x^2 + 4x - 12 = 0$ $(x+6)(x-2) = 0$ $x = -6$ or $x = 2$

   We already found that $x = -6$ and $x = 2$ make the denominator of $(f \circ g)(x)$ equal to zero. This confirms that these values must be excluded from the domain.

4. Final Domain of $(f \circ g)(x)$: Considering both the restriction from the denominator of $(f \circ g)(x)$ and the domain of $f(x)$, we find that the domain of $(f \circ g)(x)$ is indeed $(-\infty, -6) \cup (-6, 2) \cup (2, \infty)$. It's very important to check the range of the inner function to determine if it contains elements outside of the domain of the outer function. This ensures a comprehensive understanding of the domain of composite functions. Ignoring this step can lead to incorrect conclusions about the set of all permissible input values. When determining the domain of a composite function, it's crucial to ensure that all domain restrictions from both the inner and outer functions are accounted for.

Conclusion

Alright, guys, we've successfully found the composite function $(f \circ g)(x) = \frac{1}{(x+6)(x-2)}$ and determined its domain to be $(-\infty, -6) \cup (-6, 2) \cup (2, \infty)$. Remember, finding the composite function involves substituting one function into another and simplifying. Determining the domain involves identifying values that make the denominator zero and excluding them, while also considering any restrictions imposed by the original functions. Practice makes perfect, so keep working on these types of problems, and you'll become a pro in no time!