Function Composition: Videos & Practice Problems

Function composition is a fundamental concept in mathematics that involves combining two functions to create a new function. When composing functions, instead of substituting a number into a function, you substitute one function into another. This process can initially seem challenging, but with practice, it becomes clearer.

To evaluate a function at a specific value, you replace the variable with that value. For example, if you have the function \( f(x) = x^2 + 3x - 10 \) and you want to evaluate it at \( f(7) \), you would calculate:

\( f(7) = 7^2 + 3(7) - 10 = 49 + 21 - 10 = 60 \).

In contrast, when composing functions, such as finding \( f(g(x)) \), you replace the variable in \( f(x) \) with the entire function \( g(x) \). For instance, if \( g(x) = x - 2 \), then:

\( f(g(x)) = f(x - 2) = (x - 2)^2 + 3(x - 2) - 10 \).

Upon simplifying, this results in:

\( f(g(x)) = x^2 - x - 12 \).

Function composition can be denoted as \( f \circ g \), which indicates that \( g \) is applied first, followed by \( f \). The notation emphasizes that \( f \) is the outer function and \( g \) is the inner function.

To illustrate further, consider the functions \( f(x) = x + 4 \) and \( g(x) = x^2 - 3 \). To find \( f(g(x)) \), substitute \( g(x) \) into \( f(x) \):

\( f(g(x)) = f(x^2 - 3) = (x^2 - 3) + 4 = x^2 + 1 \).

For the reverse composition \( g(f(x)) \), substitute \( f(x) \) into \( g(x) \):

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

Expanding this using the FOIL method gives:

\( g(f(x)) = x^2 + 8x + 16 - 3 = x^2 + 8x + 13 \).

In summary, function composition allows you to create new functions by substituting one function into another, and understanding this process is crucial for solving more complex mathematical problems.

Evaluating composed functions involves understanding how to combine two functions and then determine their output based on a specific input. When given two functions, such as f(x) = x² and g(x) = x - 1, the composition f(g(x)) can be found by substituting g(x) into f(x). This process begins by calculating f(g(x)) = f(x - 1), which simplifies to (x - 1)². By applying the FOIL method, we expand this to x² - 2x + 1, providing the composed function.

To evaluate this composition at a specific value, such as 3, we replace x in the composed function with 3. This results in f(g(3)) = 3² - 2(3) + 1, which simplifies to 9 - 6 + 1 = 4. Thus, f(g(3)) = 4.

Alternatively, a shortcut method can be employed for evaluating composed functions. This involves first calculating the inner function at the desired input. For instance, finding g(3) gives us 3 - 1 = 2. We then substitute this result into the outer function: f(g(3)) = f(2), which simplifies to 2² = 4. This method can be quicker but is not always applicable, especially when the problem requires finding the composition before evaluating.

In summary, while both methods yield the same result, the choice of method may depend on the specific requirements of the problem. Understanding how to compose functions and evaluate them effectively is crucial in mastering function operations.

Understanding the domain of composed functions is essential for evaluating them correctly. When dealing with composed functions, such as \( f(g(x)) \), it is crucial to identify the restrictions imposed by both the inner function \( g(x) \) and the outer function \( f(x) \).

Consider the functions \( f(x) = \frac{1}{x - 2} \) and \( g(x) = \sqrt{x} \). To find the domain of the composition \( f(g(x)) \), we follow a systematic approach. First, we determine the restrictions for the inner function \( g(x) \). Since \( g(x) = \sqrt{x} \), it is clear that \( x \) must be greater than or equal to 0, as the square root of a negative number is undefined. Thus, the restriction for \( g(x) \) is:

\( x \geq 0 \)

Next, we substitute \( g(x) \) into \( f(x) \) to form the composed function:

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

Now, we need to identify any additional restrictions that arise from this composition. Specifically, the denominator of the fraction cannot equal zero. Therefore, we set up the equation:

\( \sqrt{x} - 2 \neq 0 \)

Solving this gives:

\( \sqrt{x} \neq 2 \)

Squaring both sides results in:

\( x \neq 4 \)

At this point, we have two restrictions: \( x \geq 0 \) from \( g(x) \) and \( x \neq 4 \) from the composition. To express the domain of the composed function \( f(g(x)) \), we combine these restrictions. The domain can be described as all values from 0 to 4, excluding 4, and all values greater than 4:

Domain: \( [0, 4) \cup (4, \infty) \)

This comprehensive approach ensures that we account for all restrictions when determining the domain of composed functions, allowing for accurate evaluations and further mathematical operations.

Function decomposition is the process of breaking down a composite function into its individual components, essentially reversing the function composition. This can be a challenging task, as it requires identifying the original functions from a single composed function. However, with practice and the right strategies, it can become an engaging and creative exercise.

To illustrate function decomposition, consider the function \( h(x) = \sqrt{x - 2} \). The goal is to express this function as a composition of two functions, \( h(x) = f(g(x)) \). A common approach is to identify the inner function first. In this case, we can set \( g(x) = x - 2 \), which is the expression inside the square root. Consequently, the outer function can be defined as \( f(x) = \sqrt{x} \). Thus, we have:

\( h(x) = f(g(x)) = f(x - 2) = \sqrt{x - 2} \)

This decomposition is valid because substituting \( g(x) \) back into \( f(x) \) yields the original function \( h(x) \).

Interestingly, there are multiple correct ways to decompose a function. For instance, one could also define \( g(x) \) as the entire function \( \sqrt{x - 2} \) and set \( f(x) = x \). This would also satisfy the composition:

\( h(x) = f(g(x)) = f(\sqrt{x - 2}) = \sqrt{x - 2} \)

Moreover, creativity can play a role in decomposition. For example, one could define \( g(x) = \sqrt{x - 2} - 1000 \) and \( f(x) = x + 1000 \). This unconventional approach still results in the original function when composed:

\( h(x) = f(g(x)) = f(\sqrt{x - 2} - 1000) = (\sqrt{x - 2} - 1000) + 1000 = \sqrt{x - 2} \)

In summary, function decomposition allows for various interpretations and methods, making it a flexible and creative mathematical process. Understanding how to manipulate and identify the components of composite functions is essential for mastering this concept.

To solve the problem of decomposing the function \( h(x) = \frac{1}{x^2 + 3x - 10} \) into a composition of two functions \( f \) and \( g \), we need to identify suitable forms for these functions such that \( h(x) = f(g(x)) \).

In this case, we can define the inner function \( g(x) \) as the expression in the denominator of \( h(x) \). Therefore, we set:

\( g(x) = x^2 + 3x - 10 \)

Next, the outer function \( f(x) \) will be the operation applied to \( g(x) \). Since \( h(x) \) is the reciprocal of \( g(x) \), we can express \( f(x) \) as:

\( f(x) = \frac{1}{x} \)

Thus, the decomposition of the function can be summarized as:

\( h(x) = f(g(x)) = f(x^2 + 3x - 10) = \frac{1}{x^2 + 3x - 10} \)

This method of decomposition is intuitive, as it clearly separates the components of the function into an inner and outer function. While there are other possible decompositions, this approach effectively captures the essence of the original function.