Question

Show that composing the functions in either order gets us back to where we started. $$ y=\frac{10+x}{3} \text { and } x=3 y-10 $$

Short answer

Question: Show that composing the given functions in either order returns to the starting point. Functions: $$ y=\frac{10+x}{3} \quad \quad (1) $$ $$ x=3y-10 \quad \quad (2) $$ Answer: After composing the functions in both orders and simplifying the expressions, we found the identity functions \(y = y\) and \(x = x\), proving that composing the given functions in either order returns to the starting point.

Step-by-step solution

Substitute Function (2) into Function (1)

First, let's find \(y\) in terms of \(x\) by substituting function (2) into function (1): $$ y = \frac{10+(3y-10)}{3} $$ Now solve for \(y\) to see if it gets us back to where we started.

Solve for y in the First Composition

To solve for \(y\), first simplify the expression: $$ y = \frac{3y}{3} $$ Now, we can see that \(y = y\), which is the identity function for \(y\). Canceling the 3 in the numerator and the denominator and rearranging gives us the original function for y.

Substitute Function (1) into Function (2)

Now let's find \(x\) in terms of \(y\) by substituting function (1) into function (2): $$ x=3\left(\frac{10+x}{3}\right)-10 $$ Now solve for \(x\) to see if it gets us back to where we started.

Solve for x in the Second Composition

To solve for \(x\), first simplify the expression: $$ x=10+x-10 $$ Now we can see that \(x = x\), which is the identity function for \(x\). After canceled out 10, we get the original function for x.

Conclusion

In this exercise, we showed that by composing the functions in either order, we get back to where we started. By substituting function (2) into function (1), we got the identity function \(y = y\). Similarly, by substituting function (1) into function (2), we found the identity function \(x = x\). This proves that composing the functions gets us back to the starting point.

Key concepts

Inverses of Functions

In mathematics, the inverse of a function reverses the operation done by the original function. Imagine it as a process of retracing steps to return to the initial point. For a function to have an inverse, its output must map back uniquely to its input.

• If we have a function, say, \(f\), its inverse is often noted as \(f^{-1}\).
• To confirm two functions are inverses of each other, composing them in both orders should yield the identity function (more on this later).

This means that the composition of a function and its inverse satisfies two properties:

• \(f(f^{-1}(x)) = x\)
• \(f^{-1}(f(y)) = y\)

In our example, showing these steps confirms that both functions effectively undo each other, leading us back to our original input.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions using various algebraic rules and techniques. In this exercise, we perform algebraic manipulation to verify function inverses.

• First, we substitute and simplify expressions. This is essential in transforming either function into its partner's form.
• Next, careful canceling of terms is necessary, such as when simplifications led from \(y = \frac{3y}{3}\) to \(y = y\).

A keen understanding of arithmetic operations is crucial. Without simplifying convoluted expressions, we can't verify the identity functions, which shows successful inverse composition.

Identity Function

The identity function serves as a neutral element in function compositions. Its primary characteristic is leaving its input unchanged:

• The identity function for real numbers is \(f(x) = x\).
• In function composition, achieving the identity function proves that the operations effectively cancel each other out.

In our context, when we composed the two provided functions, they returned their inputs:

• \(f(f^{-1}(x)) = x\) resulted in the identity function \(x = x\).
• \(f^{-1}(f(y)) = y\) likewise returned \(y = y\).

Establishing the identity function confirms that each function precisely undoes the operation of the other. This forms the core of proving our original functions are indeed inverses.