Composite Functions: Find $f \circ G, G \circ F, F \circ F, G \circ G$

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Let's dive into the world of composite functions! Given two functions, $f(x)$ and $g(x)$, a composite function is essentially applying one function to the result of another. It's like a mathematical assembly line. We are given $f(x)=\frac{x-7}{x+2}$ and $g(x)=\frac{x+3}{x-4}$, and our mission is to find the composite functions $(f \circ g)(x)$, $(g \circ f)(x)$, $(f \circ f)(x)$, and $(g \circ g)(x)$. Buckle up; it's going to be a fun ride!

(a) Finding $(f \circ g)(x)$

First, let's tackle $(f \circ g)(x)$, which means $f(g(x))$. In simpler terms, we're going to plug the entire function $g(x)$ into the function $f(x)$ wherever we see an $x$.

So, we have:

$f(x) = \frac{x-7}{x+2}$ and $g(x) = \frac{x+3}{x-4}$

Therefore, $f(g(x)) = f(\frac{x+3}{x-4}) = \frac{\frac{x+3}{x-4} - 7}{\frac{x+3}{x-4} + 2}$

Now, let's simplify this expression. To do that, we need to get rid of the fractions within the fraction. We can multiply both the numerator and the denominator by $(x-4)$:

$\frac{(\frac{x+3}{x-4} - 7)(x-4)}{(\frac{x+3}{x-4} + 2)(x-4)} = \frac{(x+3) - 7(x-4)}{(x+3) + 2(x-4)}$

Expanding the terms, we get:

$\frac{x+3 - 7x + 28}{x+3 + 2x - 8} = \frac{-6x + 31}{3x - 5}$

Thus, $(f \circ g)(x) = \frac{-6x + 31}{3x - 5}$.

In summary, to find $f \circ g$, we substitute $g(x)$ into $f(x)$ and simplify the resulting expression. This involves algebraic manipulation to eliminate complex fractions and combine like terms. The final result, $(f \circ g)(x) = \frac{-6x + 31}{3x - 5}$, represents the new function formed by this composition. Understanding this process is crucial for working with more complex functions and solving related problems. Practice with different functions will solidify this concept. Keep in mind that the domain of the composite function is also important, and it depends on the domains of both $f(x)$ and $g(x)$.

(b) Finding $(g \circ f)(x)$

Next up, we want to find $(g \circ f)(x)$, which is $g(f(x))$. This time, we'll plug $f(x)$ into $g(x)$ wherever we see an $x$.

We have:

$f(x) = \frac{x-7}{x+2}$ and $g(x) = \frac{x+3}{x-4}$

Therefore, $g(f(x)) = g(\frac{x-7}{x+2}) = \frac{\frac{x-7}{x+2} + 3}{\frac{x-7}{x+2} - 4}$

Again, we need to simplify this. Multiply the numerator and the denominator by $(x+2)$:

$\frac{(\frac{x-7}{x+2} + 3)(x+2)}{(\frac{x-7}{x+2} - 4)(x+2)} = \frac{(x-7) + 3(x+2)}{(x-7) - 4(x+2)}$

Expanding the terms, we get:

$\frac{x-7 + 3x + 6}{x-7 - 4x - 8} = \frac{4x - 1}{-3x - 15}$

So, $(g \circ f)(x) = \frac{4x - 1}{-3x - 15}$. We can also write this as $(g \circ f)(x) = -\frac{4x - 1}{3x + 15}$.

In essence, the process of finding $g \circ f$ involves substituting the function $f(x)$ into $g(x)$ and simplifying the resulting expression. This requires algebraic techniques similar to those used in finding $f \circ g$, such as eliminating complex fractions and combining like terms. The result, $(g \circ f)(x) = -\frac{4x - 1}{3x + 15}$, is another new function derived from the composition of $g$ and $f$. It's important to recognize that the order of composition matters; $f \circ g$ is generally not the same as $g \circ f$. Furthermore, the domain of $g \circ f$ will depend on the domains of both $f(x)$ and $g(x)$, with particular attention needed to ensure that $f(x)$ is in the domain of $g(x)$.

(c) Finding $(f \circ f)(x)$

Now, let's find $(f \circ f)(x)$, which is $f(f(x))$. This means we're plugging $f(x)$ into itself.

We have:

$f(x) = \frac{x-7}{x+2}$

Therefore, $f(f(x)) = f(\frac{x-7}{x+2}) = \frac{\frac{x-7}{x+2} - 7}{\frac{x-7}{x+2} + 2}$

To simplify, we multiply the numerator and the denominator by $(x+2)$:

$\frac{(\frac{x-7}{x+2} - 7)(x+2)}{(\frac{x-7}{x+2} + 2)(x+2)} = \frac{(x-7) - 7(x+2)}{(x-7) + 2(x+2)}$

Expanding the terms, we get:

$\frac{x-7 - 7x - 14}{x-7 + 2x + 4} = \frac{-6x - 21}{3x - 3}$

We can simplify this further by factoring out a 3 from both the numerator and the denominator:

$\frac{3(-2x - 7)}{3(x - 1)} = \frac{-2x - 7}{x - 1}$

Thus, $(f \circ f)(x) = \frac{-2x - 7}{x - 1}$.

In the case of $f \circ f$, we are composing the function $f(x)$ with itself. This means substituting $f(x)$ into $f(x)$. The simplification process is similar to previous examples, involving the elimination of complex fractions and combining like terms. The result, $(f \circ f)(x) = \frac{-2x - 7}{x - 1}$, is a new function that represents the composition of $f(x)$ with itself. Understanding this type of composition is important in the study of dynamical systems, where repeated applications of a function are analyzed. The domain of $f \circ f$ will depend on the domain of $f(x)$, ensuring that the output of the inner $f(x)$ is within the domain of the outer $f(x)$. The composition of a function with itself can reveal interesting properties and behaviors of the function.

(d) Finding $(g \circ g)(x)$

Finally, let's find $(g \circ g)(x)$, which is $g(g(x))$. We're plugging $g(x)$ into itself this time.

We have:

$g(x) = \frac{x+3}{x-4}$

Therefore, $g(g(x)) = g(\frac{x+3}{x-4}) = \frac{\frac{x+3}{x-4} + 3}{\frac{x+3}{x-4} - 4}$

To simplify, we multiply the numerator and the denominator by $(x-4)$:

$\frac{(\frac{x+3}{x-4} + 3)(x-4)}{(\frac{x+3}{x-4} - 4)(x-4)} = \frac{(x+3) + 3(x-4)}{(x+3) - 4(x-4)}$

Expanding the terms, we get:

$\frac{x+3 + 3x - 12}{x+3 - 4x + 16} = \frac{4x - 9}{-3x + 19}$

Thus, $(g \circ g)(x) = \frac{4x - 9}{-3x + 19}$.

When considering the composition $g \circ g$, we substitute $g(x)$ into $g(x)$, similar to the process used for $f \circ f$. The algebraic simplification follows the same pattern of eliminating complex fractions and combining like terms. The result, $(g \circ g)(x) = \frac{4x - 9}{-3x + 19}$, is a new function representing the composition of $g(x)$ with itself. Understanding this type of composition is important for various applications, including iterative processes and mathematical modeling. As with other composite functions, the domain of $g \circ g$ depends on the domain of $g(x)$, ensuring that the output of the inner $g(x)$ is a valid input for the outer $g(x)$. The repeated application of a function to itself can lead to interesting and complex behaviors, making the study of such compositions a valuable area of mathematical exploration.

In conclusion, we have successfully found all the required composite functions:

(a) $(f \circ g)(x) = \frac{-6x + 31}{3x - 5}$ (b) $(g \circ f)(x) = -\frac{4x - 1}{3x + 15}$ (c) $(f \circ f)(x) = \frac{-2x - 7}{x - 1}$ (d) $(g \circ g)(x) = \frac{4x - 9}{-3x + 19}$

Remember, the key to composite functions is understanding the order of operations and carefully simplifying the resulting expressions. Keep practicing, and you'll become a composite function master in no time!

For further learning on composite functions, you can visit Khan Academy's page on composite functions.