Problem 76 - Algebra and Trigonometry

Question

The relative value of currencies fluctuates every day. When this problem was written, one Canadian dollar was worth 0.8159 U.S. dollar. (a) Find a function \(f\) that gives the U.S. dollar value \(f(x)\) of \(x\) Canadian dollars. (b) Find \(f^{-1} .\) What does \(f^{-1}\) represent? (c) How much Canadian money would \(\$ 12,250\) in U.S. currency be worth?

Step-by-Step Solution

Answer

(a) \( f(x) = 0.8159x \); (b) \( f^{-1}(y) = \frac{y}{0.8159} \), it converts U.S. dollars to Canadian dollars; (c) \$12,250 U.S. is approximately \$15,012.25 Canadian.

Step 1: Defining the Function for Currency Conversion

Given the exchange rate, one Canadian dollar is worth 0.8159 U.S. dollars. The function \( f(x) \) will convert Canadian dollars to U.S. dollars, so it can be expressed as \( f(x) = 0.8159x \). Here, \( x \) is the amount in Canadian dollars, and \( f(x) \) is the equivalent amount in U.S. dollars.

Step 2: Finding the Inverse Function

To find the inverse function \( f^{-1} \), we start with the equation \( y = 0.8159x \), and solve for \( x \) in terms of \( y \). Rearranging gives \( x = \frac{y}{0.8159} \). Thus, the inverse function is \( f^{-1}(y) = \frac{y}{0.8159} \). This function \( f^{-1} \) represents the conversion from U.S. dollars back to Canadian dollars.

Step 3: Computing the Canadian Equivalent of $12,250 U.S.

Using the inverse function \( f^{-1}(y) = \frac{y}{0.8159} \), we substitute \( y = 12,250 \) to find the equivalent in Canadian dollars. This results in \( f^{-1}(12,250) = \frac{12,250}{0.8159} \). Calculating this gives approximately \( 15,012.25 \) Canadian dollars.

Key Concepts

Inverse Function

An inverse function essentially reverses the effect of the original function. If a function, like our currency conversion, takes an input and produces an output, then the inverse function does the opposite. For example, if the function \( f(x) \) converts from Canadian dollars to U.S. dollars, the inverse \( f^{-1}(x) \) will convert from U.S. dollars back to Canadian dollars.

• Purpose: The primary purpose is to undo the conversion process and retrieve the original value.
• Mathematical Representation: If \( y = f(x) \), then \( x = f^{-1}(y) \).

Understanding inverse functions is crucial in currency conversion, since it allows us to switch currencies back and forth.

Exchange Rate

The exchange rate is a key concept in currency conversion, referring to the value of one currency in relation to another. It indicates how much of one currency you can exchange for another.

• Example in Context: In our exercise, 1 Canadian dollar equals 0.8159 U.S. dollars. This rate helps determine the function \( f(x) = 0.8159x \).
• Fluctuation: Exchange rates can fluctuate due to various market factors, influencing international trade and travel.

Being aware of the current exchange rate is essential when dealing with currency exchanges, as it affects cost calculations and financial decisions.

Function Notation

Function notation is a mathematical way to describe the relationship between quantities. It's a shorthand that makes understanding and manipulating functions more manageable. In our problem, \( f(x) \) is used to represent the conversion from Canadian to U.S. dollars.

• Expression: The notation \( f(x) = 0.8159x \) specifies that applying \( f \) to \( x \) gives the amount in U.S. dollars.
• Clarity: Function notation provides a clear, systematic way of describing how different variables are linked.

Using function notation simplifies complex relationships by distilling them into understandable formulas.

Algebraic Manipulation

Algebraic manipulation involves using algebra to reorganize equations and solve for variables. It's crucial for finding inverse functions or solving equations.

• Process: To find the inverse function in our exercise, we started with \( y = 0.8159x \) and rearranged this to solve for \( x \), leading to \( x = \frac{y}{0.8159} \).
• Skills Needed: Understanding multiplication, division, and rearranging equations is fundamental for effective algebraic manipulation.

This skill enables one to tackle a variety of mathematical tasks, including those involving currency conversion, by transforming and solving the equations involved.