Composite Functions: Find (f∘g)(x) And (g∘f)(x)

Alright, guys, let's dive into the fascinating world of composite functions! Today, we're tackling the challenge of finding $(f \circ g)(x)$ and $(g \circ f)(x)$ when given different pairs of functions $f(x)$ and $g(x)$. Composite functions might sound intimidating, but trust me, once you get the hang of it, it's like riding a bike... a mathematical bike, that is!

Understanding Composite Functions

Before we jump into the examples, let's make sure we're all on the same page about what composite functions actually are. Think of it like this: instead of just plugging a number into a function, we're plugging an entire function into another function. So, $(f \circ g)(x)$ means we first apply the function $g$ to $x$, and then we take the result and plug it into the function $f$. In mathematical notation:

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

Similarly, $(g \circ f)(x)$ means we first apply the function $f$ to $x$, and then plug the result into the function $g$:

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

The order matters! $(f \circ g)(x)$ is generally not the same as $(g \circ f)(x)$. It's like putting on your socks and then your shoes versus putting on your shoes and then your socks – you'll get a different (and probably uncomfortable) result!

Now, let's get our hands dirty with some examples. We'll break down each step to make sure you're following along.

Example a: $f(x) = 2x^2 + x$ and $g(x) = 3 - 2x$

In this first example, we are given two functions: a quadratic function $f(x) = 2x^2 + x$ and a linear function $g(x) = 3 - 2x$. We will explore how these functions interact when composed together, calculating both $(f \circ g)(x)$ and $(g \circ f)(x)$ to illustrate that the order of composition dramatically affects the resulting function. Understanding this principle is crucial for manipulating and solving more complex mathematical problems involving function composition. So, let's proceed step by step to unravel this example.

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

First, we need to find $(f \circ g)(x) = f(g(x))$. This means we'll take the function $g(x)$ and substitute it into $f(x)$ wherever we see an $x$.

$f(g(x)) = 2(g(x))^2 + g(x)$

Now, substitute $g(x) = 3 - 2x$:

$f(g(x)) = 2(3 - 2x)^2 + (3 - 2x)$

Next, expand and simplify. Remember that $(3 - 2x)^2 = (3 - 2x)(3 - 2x) = 9 - 12x + 4x^2$:

$f(g(x)) = 2(9 - 12x + 4x^2) + (3 - 2x)$

Distribute the 2:

$f(g(x)) = 18 - 24x + 8x^2 + 3 - 2x$

Finally, combine like terms:

$f(g(x)) = 8x^2 - 26x + 21$

So, $(f \circ g)(x) = 8x^2 - 26x + 21$.

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

Now, let's find $(g \circ f)(x) = g(f(x))$. This time, we'll take the function $f(x)$ and substitute it into $g(x)$ wherever we see an $x$.

$g(f(x)) = 3 - 2(f(x))$

Substitute $f(x) = 2x^2 + x$:

$g(f(x)) = 3 - 2(2x^2 + x)$

Distribute the -2:

$g(f(x)) = 3 - 4x^2 - 2x$

Rearrange the terms (optional, but often preferred for clarity):

$g(f(x)) = -4x^2 - 2x + 3$

So, $(g \circ f)(x) = -4x^2 - 2x + 3$.

Notice that $(f \circ g)(x)$ and $(g \circ f)(x)$ are completely different functions! This highlights the importance of the order of composition.

Example b: $f(x) = \frac{1}{2x - 3}$ and $g(x) = 2x + 5$

Here, we have a rational function $f(x) = \frac{1}{2x - 3}$ and a linear function $g(x) = 2x + 5$. This example will show us how to manage composite functions when one of the functions involves fractions. The key here is careful substitution and simplification. This kind of practice is essential for anyone looking to master the manipulation of algebraic expressions and understand function behaviors more deeply. Therefore, we will proceed meticulously to ensure every step is clear.

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

We want to find $(f \circ g)(x) = f(g(x))$. Substitute $g(x)$ into $f(x)$:

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

Now, substitute $g(x) = 2x + 5$:

$f(g(x)) = \frac{1}{2(2x + 5) - 3}$

Distribute the 2:

$f(g(x)) = \frac{1}{4x + 10 - 3}$

Simplify:

$f(g(x)) = \frac{1}{4x + 7}$

So, $(f \circ g)(x) = \frac{1}{4x + 7}$.

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

Now, let's find $(g \circ f)(x) = g(f(x))$. Substitute $f(x)$ into $g(x)$:

$g(f(x)) = 2(f(x)) + 5$

Substitute $f(x) = \frac{1}{2x - 3}$:

$g(f(x)) = 2(\frac{1}{2x - 3}) + 5$

Simplify. To add the terms, we need a common denominator:

$g(f(x)) = \frac{2}{2x - 3} + \frac{5(2x - 3)}{2x - 3}$

$g(f(x)) = \frac{2 + 10x - 15}{2x - 3}$

$g(f(x)) = \frac{10x - 13}{2x - 3}$

So, $(g \circ f)(x) = \frac{10x - 13}{2x - 3}$.

Again, $(f \circ g)(x)$ and $(g \circ f)(x)$ are different.

Example c: $f(x) = \sqrt{x + 2}$ and $g(x) = x^2$

In our final example, we delve into functions involving square roots and squaring, specifically $f(x) = \sqrt{x + 2}$ and $g(x) = x^2$. This scenario introduces the added complexity of domain restrictions, as the square root function is only defined for non-negative values. Therefore, we must consider these restrictions when determining the composite functions $(f \circ g)(x)$ and $(g \circ f)(x)$. This example not only reinforces the process of function composition but also highlights the importance of domain awareness in mathematical analysis. Let's carefully examine how these functions interact, keeping in mind any limitations imposed by the square root.

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

We want to find $(f \circ g)(x) = f(g(x))$. Substitute $g(x)$ into $f(x)$:

$f(g(x)) = \sqrt{g(x) + 2}$

Substitute $g(x) = x^2$:

$f(g(x)) = \sqrt{x^2 + 2}$

So, $(f \circ g)(x) = \sqrt{x^2 + 2}$. Notice that $x^2 + 2$ is always positive (or zero at $x=0$), so there are no domain restrictions here.

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

Now, let's find $(g \circ f)(x) = g(f(x))$. Substitute $f(x)$ into $g(x)$:

$g(f(x)) = (f(x))^2$

Substitute $f(x) = \sqrt{x + 2}$:

$g(f(x)) = (\sqrt{x + 2})^2$

Simplify:

$g(f(x)) = x + 2$

So, $(g \circ f)(x) = x + 2$.

However, we need to consider the domain of the original $f(x) = \sqrt{x + 2}$. The expression inside the square root must be non-negative, so $x + 2 \ge 0$, which means $x \ge -2$. Therefore, even though $(g \circ f)(x) = x + 2$ is defined for all real numbers, the composite function $(g \circ f)(x)$ is only defined for $x \ge -2$.

Key Takeaways

• Order matters: $(f \circ g)(x)$ is generally not the same as $(g \circ f)(x)$.
• Substitution is key: Carefully substitute the inner function into the outer function.
• Simplify: Expand and combine like terms to get the final expression.
• Domain awareness: Always consider the domain of the original functions, especially when dealing with square roots or rational functions. The domain of the composite function is restricted by the domains of both the inner and outer functions.

Practice makes perfect! The more you work with composite functions, the easier they'll become. So, grab some more examples and start composing! You'll be a composite function pro in no time!