Problem 89 - College Algebra

Question

For the following exercises, find the composition when \(f(x)=x^{2}+2\) for all \(x \geq 0\) and \(g(x)=\sqrt{x-2}\). $$(f \circ g)(11) ;(g \circ f)(11)$$

Step-by-Step Solution

Answer

Both compositions, \\( (f \circ g)(11) \\) and \\( (g \circ f)(11) \\), equal 11.

Step 1: Understand Composition of Functions

The composition of two functions, denoted \(f \circ g\), means we apply function \(g\) first and then apply function \(f\) to the result. Similarly, \(g \circ f\) means we apply function \(f\) first and then apply function \(g\).

Step 2: Calculate \(g(11)\)

Since we need to find \(f(g(11))\), we first calculate the output of \(g(11)\). Based on the given function \(g(x) = \sqrt{x-2}\), plug in 11: \ g(11) = \sqrt{11-2} = \sqrt{9} = 3\.

Step 3: Evaluate \(f(g(11))\)

Now we substitute \(g(11)\) into \(f\) to find \(f(g(11))\): \(f(x) = x^2 + 2\). Input the result from the previous step, \(f(3) = 3^2 + 2 = 9 + 2 = 11\). Thus, \( (f \circ g)(11) = 11\).

Step 4: Calculate \(f(11)\)

Now find \(g(f(11))\) by first computing \(f(11)\). Using function \(f(x) = x^2 + 2\), substitute 11: \ f(11) = 11^2 + 2 = 121 + 2 = 123\.

Step 5: Evaluate \(g(f(11))\)

Substitute the value of \(f(11)\) into \(g\): \(g(x) = \sqrt{x-2}\). So, \ g(123) = \sqrt{123-2} = \sqrt{121} = 11\. Therefore, \( (g \circ f)(11) = 11\).

Key Concepts

Function Evaluation

Function evaluation is the process of substituting a specific value into a function's formula to determine its output. Imagine a function as a machine that takes an input and provides an output based on its rule.

For example, if we have a function defined as \( f(x) = x^2 + 2 \), and we want to evaluate it at \( x = 3 \), we replace \( x \) with 3 in the expression. This gives us \( f(3) = 3^2 + 2 = 9 + 2 = 11 \).

• Identify the function's formula.
• Substitute the specific input value into the formula.
• Simplify the expression to find the result.

Evaluating functions correctly is crucial for understanding more advanced concepts, such as composing functions, which involves plugging one function's output into another.

Domain of Functions

The domain of a function is the set of all possible input values (typically \( x \) values) that allow the function to produce a valid output. Not all mathematical expressions are defined for all inputs, which is why understanding the domain is important.

For the function \( g(x) = \sqrt{x-2} \), the value under the square root must be non-negative (i.e., greater than or equal to zero) for it to yield a real number output. This constraint gives us the domain \( x \geq 2 \).

• Linear functions: Typically, all real numbers are allowed.
• Square root functions: Only values where the expression inside the root is non-negative.
• Fractional functions: Values making the denominator zero are excluded.

For function compositions, the domain is determined by the innermost function's domain and the subsequent capability of the outer function to accept the innermost function's outputs.

Function Notation

Function notation is a way to succinctly express the operation of functions and clearly indicate inputs and outputs. It uses letters like \( f \) or \( g \) to name a function and associates it with an input variable like \( x \).

When we write \( f(x) \), it indicates that \( f \) is a function of \( x \), and the expression provides the rule for calculating the output based on \( x \). This notation makes it easy to change inputs and perform operations on functions, like compositions.

• \( f(x) = x^2 + 2 \) implies a function named \( f \) with input \( x \).
• \( f(g(x)) \) shows a composition where the output of \( g \) is used as the input to \( f \).
• The notation clarifies which function is applied first in compositions, improving understanding.

Understanding function notation is essential for both evaluating functions and exploring more complex operations.