Problem 62 - Algebra and Trigonometry Real Mathematics, Real People

Question

\((a)\) find \((f \circ g)(x)\) and \((g \circ f)(x),\) \((b)\) determine algebraically whether \((f \circ g)(x)=(g \circ f)(x),\) and \((c)\) use a graphing utility to complete a table of values for the two compositions to confirm your answer to part \((b).\) $$f(x)=\frac{1}{4}(x-1), \quad g(x)=4 x+1$$

Step-by-Step Solution

Answer

The functions \(f \circ g)(x) and \(g \circ f)(x) are both equal to x, therefore \(f \circ g)(x)=(g \circ f)(x). The graph and table of values would further confirm this conclusion.

Step 1: Find (f \circ g)(x) and (g \circ f)(x)

The function `(f \circ g)(x)=f(g(x))` means to substitute `g(x)` into `f(x)`. Similarly, `(g \circ f)(x)=g(f(x))` means to substitute `f(x)` into `g(x)`.\nSubstituting `g(x)` into `f(x)`: \(f(g(x))=f(4x+1)=\frac{1}{4}((4x+1)-1)=x\).\nSubstituting `f(x)` into `g(x)`: \(g(f(x))=g(\frac{1}{4}(x-1))=4* (\frac{1}{4}(x-1))+1=x\).

Step 2: Compare (f \circ g)(x) and (g \circ f)(x)

Now that we have calculated both `(f \circ g)(x)` and `(g \circ f)(x)`, we can compare them to check if they are identical. From step 1, we found out that both `(f \circ g)(x)` and `(g \circ f)(x)` are equal to `x`. Since they are equal, we can conclude that \(f \circ g)(x)=(g \circ f)(x).

Step 3: Graph and Draw the Table

Though the problem requires the use of a graphing utility to support the algebraic proof, due to the limitation of this platform, we can't present a graph here. However, to complete this step: plot both `y=`\((f \circ g)(x)\) and `y=`\((g \circ f)(x)\) in any graphing tool. You will see two straight lines y=x that overlap one another. Input different x values and observe that the y values are equal for both \(f \circ g)(x)=(g \circ f)(x). This visual result confirms the algebraic proof.

Key Concepts

Algebraic Verification

Algebraic verification is a crucial step in understanding function composition. It allows us to determine whether two composite functions yield the same result.
To do this, we substitute one function into another, and then simplify the expression. In the given problem, we had two functions, \(f(x)\) and \(g(x)\). For \((f \circ g)(x)\), we substituted \(g(x)\) into \(f(x)\), resulting in \(f(g(x)) = x\). Similarly, for \((g \circ f)(x)\), \(f(x)\) was substituted into \(g(x)\), and the result was also \(g(f(x)) = x\).
When both \((f \circ g)(x)\) and \((g \circ f)(x)\) simplify to the same expression, in this case \(x\), this confirms algebraically that they are equivalent. This process ensures that the composite operations are commutable, meaning the order of operations does not affect the result. It's like checking your work twice with two different methods and getting the same answer every time.

Graphing Utility

Using a graphing utility is a fantastic way to visualize functions and their compositions. These tools can help see the relationship between functions, offering a clear, visual proof of their equivalence or difference.
For this exercise, once you have determined algebraically that \((f \circ g)(x) = (g \circ f)(x)\), a graphing utility like Desmos or a graphing calculator can assist in confirming this result. You would graph \(y = (f \circ g)(x)\) and \(y = (g \circ f)(x)\) and observe the output.
Both graphs should appear as straight lines represented by \(y = x\). When the two graphs overlap, it visually confirms what the algebraic verification showed us: the compositions are indeed the same.

• Graphing utilities help validate algebraic calculations.
• They offer an intuitive understanding of functions’ behaviors.
• Errors are more likely to stand out when visualized.

Utilizing graphing tools not only supports your calculations but also builds confidence in the results.

Table of Values

Creating a table of values is another method to verify the equality of function compositions. This approach complements both algebraic and graphical analyses by providing exact numerical points to compare.
By selecting some x-values and substituting them into both composed functions \((f \circ g)(x)\) and \((g \circ f)(x)\), we generate corresponding y-values.
For each chosen x-value, you will calculate y under both expressions and populate a table, documenting the x and y pairs. If the output values (all y-values) match for each x-value used, it further confirms that \((f \circ g)(x) = (g \circ f)(x)\).
This method can be particularly useful because:

• It's a straightforward way to spot errors in calculations.
• It provides additional assurance of the functions’ equivalence.
• It can be cross-referenced with graphing results for consistency.

A table of values adds another layer of validity, offering concrete data points that back up algebraic and graphic conclusions.