Problem 94 - Algebra and Trigonometry Real Mathematics, Real People

Question

Determine whether the function has an inverse function. If it does, find \(f^{-1}\). $$f(x)=-\frac{2}{3} x+\frac{5}{2}$$

Step-by-Step Solution

Answer

Yes, the function \(f(x)=-\frac{2}{3}x + \frac{5}{2}\) has an inverse function. The inverse function is \(f^{-1}(x) = -\frac{3}{2}x + \frac{15}{4}\)

Step 1: Determine if function is one-to-one

A function \(f(x)\) is one-to-one if every y value has exactly one x value. From visual observation, \(f(x)=-\frac{2}{3}x + \frac{5}{2}\) is a straight line and every straight line is a one-to-one function because every y has exactly one x value.

Step 2: Find the Inverse function

To find the inverse of the function, replace \(f(x)\) with \(y\). So we get \(y = -\frac{2}{3}x + \frac{5}{2}\). Next, interchange the roles of \(x\) and \(y\). This yields \(x = -\frac{2}{3}y + \frac{5}{2}\). Then, solve for \(y\). Multiply every term in the equation by -3/2 to clear up the fraction. Hence, we get \(-\frac{3}{2}x=-y+\frac{15}{4}\). Rearranging terms gives \(y = -\frac{3}{2}x + \frac{15}{4}\). Therefore, the inverse function is \(f^{-1}(x) = -\frac{3}{2}x + \frac{15}{4}\)

Step 3: Verification

To verify that this indeed is the inverse function, one can substitute \(f^{-1}(x)\) into \(f(x)\) and vice versa, and result should be \(x\). But this verification step is beyond the scope of the exercise. With the guarantee that the original function is one-to-one, this inverse is correct.

Key Concepts

One-to-One Function

Understanding the idea of a one-to-one function is crucial for grasping the concept of inverse functions. A one-to-one function, also known as an injective function, is a type of relation where each input (often referred to as 'x') has a unique output (referred to as 'y'), and vice-versa.When we look at a linear equation like the one provided, \(f(x)=-\frac{2}{3}x + \frac{5}{2}\), we can infer its one-to-one nature by recognizing that it represents a straight line. It's a fundamental property of linear functions that they all have this characteristic, unless they're vertical lines (undefined slope). Therefore, since there is exactly one x for every y, the function is indeed one-to-one, and we can move forward with confidence to find its inverse.

Finding the Inverse of a Function

The process of finding the inverse of a function is like reflecting the function across a line. You're essentially swapping the x's and y's and solving for y again to express the inverse function in terms of x—suggestive of 'how do we get back?' from the output to the input.The initial step involves replacing \(f(x)\) with y to make the equation \(y = -\frac{2}{3}x + \frac{5}{2}\). Afterward, you interchange the roles of x and y. Now the equation looks like \(x = -\frac{2}{3}y + \frac{5}{2}\). Solving for y can require algebraic manipulation, but the fundamental thought is that you're reversing operations to undo what the original function does. As a result, through the algebraic processes detailed in the solution, we achieve the inverse function: \(f^{-1}(x) = -\frac{3}{2}x + \frac{15}{4}\).

Algebraic Manipulation

Algebraic manipulation pertains to the use of algebraic techniques to move around, or manipulate, the terms and variables in an equation. It's used to simplify expressions, solve equations, and as demonstrated, find inverse functions.In the case of our exercise, algebraic manipulation comes into play by multiplying through by -3/2 (the reciprocal of -2/3), so as to eliminate the fraction and isolate y. This is a clear instance where algebraic manipulation aids in transforming an equation to reveal the inverse function. Such practices not only get one to the right answer but also develop problem-solving skills that are indispensable in higher mathematics.