Problem 35 - Intermediate Algebra

Question

Determine the indicated functional values. (Objective 2 ) If \(f(x)=\sqrt{3 x-2}\) and \(g(x)=-x+4\), find \((f \circ g)(1)\) and \((g \circ f)(6)\).

Step-by-Step Solution

Answer

\( (f \circ g)(1) = \sqrt{7} \) and \( (g \circ f)(6) = 0 \).

Step 1: Understand the Composition of Functions

When asked to find \((f \circ g)(x)\), this means we need to determine \(f(g(x))\). Similarly, \((g \circ f)(x)\) would require finding \(g(f(x))\).

Step 2: Calculate \(f(g(1))\)

First, evaluate \(g(1)\). Using the function \(g(x) = -x + 4\), substitute \(x = 1\). Hence, \(g(1) = -1 + 4 = 3\). Next, use \(f(x) = \sqrt{3x - 2}\) and substitute \(x = 3\) into \(f(x)\). Therefore, \(f(g(1)) = f(3) = \sqrt{3 \times 3 - 2} = \sqrt{9 - 2} = \sqrt{7}\).

Step 3: Calculate \(g(f(6))\)

First, evaluate \(f(6)\). Using the function \(f(x) = \sqrt{3x - 2}\), substitute \(x = 6\). Hence, \(f(6) = \sqrt{3 \times 6 - 2} = \sqrt{18 - 2} = \sqrt{16} = 4\). Next, use \(g(x) = -x + 4\) and substitute \(x = 4\) into \(g(x)\). Therefore, \(g(f(6)) = g(4) = -4 + 4 = 0\).

Key Concepts

Functional Values

Functional values are the outputs we get when we input particular values into a function. When we see a function like \(f(x)\) or \(g(x)\), what we're really looking at is a special relationship where each input \(x\) gives exactly one output. For example, in our exercise, we have functions \(f(x) = \sqrt{3x - 2}\) and \(g(x) = -x + 4\). Here, if we put 1 into \(g(x)\), the functional value we get is 3, because when we calculate \(g(1)\), it becomes \(-1 + 4 = 3\).
To solve the problem, we first found the functional value of \(g(1)\) and used it as an input for the next function, \(f(x)\). It shows how interconnected these functions can be as part of a chain of calculations.

Composition of Functions

The composition of functions involves putting one function inside another, creating a new function. It’s like a two-step process where the output of one function becomes the input of another. In math terms, this is denoted as \((f \circ g)(x)\), which means \(f(g(x))\). In our example, \(f(g(1))\) means you first calculate \(g(1)\) and then put its result into \(f(x)\).

• First, determine \(g(1)\), which is 3 using \(g(x) = -x + 4\).
• Then, find \(f(3)\) using \(f(x) = \sqrt{3x - 2}\), giving \(\sqrt{7}\).

Another example is \((g \circ f)(x)\), where you would calculate \(f(x)\) and use its result in \(g(x)\). This approach highlights the procedural nature of composite functions, revealing how sequences of operations can be layered.

Inverse Functions

Inverse functions reverse the effects of the original function, essentially doing the opposite work. If you have a function \(f\) that takes an input \(x\) and outputs \(y\), its inverse, denoted \(f^{-1}\), will take \(y\) back to \(x\). This characteristic is crucial in many applied scenarios where you need to backtrack or reverse a calculation.
In the context of composition, if you have two functions such that \(f(g(x)) = x\), \(g\) would be considered the inverse of \(f\), provided their compositions yield the original input. In simpler terms, applying \(g\) after \(f\) (or vice versa) returns you to your starting point. While our exercise does not explicitly find inverse functions, understanding them can amplify your grasp on how composite functions can "unweave" or "undo" each other's effects.