Problem 88 - College Algebra

Question

Find the composition when \(f(x)=x^{2}+2\) for all \(x \geq 0\) and \(g(x)=\sqrt{x-2}\). $$ (g \circ f)(a) ;(f \circ g)(a) $$

Step-by-Step Solution

Answer

Both \((g \circ f)(a) = a\) for \(a \geq 0\) and \((f \circ g)(a) = a\) for \(a \geq 2\).

Step 1: Understand the Composition Concept

Compositions of functions involve applying one function to the results of another. For \((g \circ f)(a)\), we first apply \(f\) to \(a\), then apply \(g\) to the result. For \((f \circ g)(a)\), we first apply \(g\) to \(a\), then apply \(f\) to the result.

Step 2: Find \((g \circ f)(a)\)

First, find \(f(a)\) using the function \(f(x) = x^2 + 2\). Substitute \(a\) to get \(f(a) = a^2 + 2\). Next, use \(g(x) = \sqrt{x-2}\) to find \(g(f(a))\). Substitute \(f(a)\) to get \(g(a^2 + 2) = \sqrt{(a^2 + 2) - 2} = \sqrt{a^2}\). Therefore, \((g \circ f)(a) = a\) for \(a \geq 0\).

Step 3: Find \((f \circ g)(a)\)

First, find \(g(a)\) using the function \(g(x) = \sqrt{x-2}\). For \(g(a)\) to be defined, \(a - 2 \geq 0\) or \(a \geq 2\). Then, \(g(a) = \sqrt{a - 2}\). Next, apply \(f\) on the result: \(f(g(a)) = f(\sqrt{a-2}) = (\sqrt{a-2})^2 + 2 = a - 2 + 2 = a\). Therefore, \((f \circ g)(a) = a\) for \(a \geq 2\).

Key Concepts

Functions

In mathematics, functions are like machines that take an input, do something with it, and then output a result. Every function is defined by a specific rule or set of instructions. For example, the function \( f(x) = x^2 + 2 \) takes any input \( x \) that is greater than or equal to zero, squares it, and then adds 2 to the result.
Another function, like \( g(x) = \sqrt{x - 2} \), waits for a number that is at least 2, subtracts 2 from it, and finds the square root. Functions often have names like \( f \), \( g \), or sometimes just letters like \( y \) to make them easy to refer to. Functions have inputs, known as "domains," and outputs, called "ranges." Understanding these will help you determine the values you put into the function and the values you get out.

Domain and Range

The domain of a function is all the possible inputs that can go into it. For example, the function \( f(x) = x^2 + 2 \) has a domain of all non-negative numbers \( x \geq 0 \). That's because you can square any non-negative number and get a valid result.
For the function \( g(x) = \sqrt{x-2} \), the domain consists only of numbers 2 and greater \( x \geq 2 \). You can't take the square root of a negative number without stepping into complex numbers, so \( g(x) \) can't accept numbers less than 2.

• **Domain**: The set of all possible inputs.
• **Range**: The set of all possible outputs.

This is important to understand because it tells you which values you can safely plug into a function without getting errors or undefined results.

Square Roots

Square roots are the opposite operation of squaring a number. For example, since \( 4^2 = 16 \), the square root of 16 is 4. The symbol for the square root is \( \sqrt{} \). Square roots are defined only for non-negative numbers when dealing with real numbers because the square of a real number is always non-negative.
For \( g(x) = \sqrt{x-2} \), the value inside the square root must be non-negative. This means you must choose \( x \) such that \( x-2 \geq 0 \), which simplifies to \( x \geq 2 \). This restriction ensures that \( g(x) \) outputs real numbers.

Composition of Functions

Composition of functions is like feeding the output of one function directly into another one. It's a way of combining functions to create a new function. When you see \((g \circ f)(a) \), it means you first apply \( f \) to \( a \) and then apply \( g \) to the result.
Let's break it down with an example:

• First, find \( f(a) = a^2 + 2 \).
• Take this result and input it into \( g \), giving \( g(f(a)) = \sqrt{a^2} = a \) for \( a \geq 0 \).

Likewise, for \((f \circ g)(a)\):

• First, find \( g(a) = \sqrt{a - 2} \).
• Apply \( f \) to this result: \( f(g(a)) = a \) for \( a \geq 2 \).

Composition allows us to see how functions can interact and transform inputs through a sequence of operations. This concept is powerful in mathematics as it helps build complex relationships between variables through simpler, connected steps.