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Chapter 13: Problem 34

Find the inverse of each one-to-one function. $$f(x)=\frac{2}{5} x+1$$

Short Answer

Expert verified
The inverse of the given one-to-one function \(f(x) = \frac{2}{5}x + 1\) is \(f^{-1}(x) = \frac{5}{2}(x - 1)\).

Step by step solution

01

Replace f(x) with y

We replace the function notation f(x) with y: $$y = \frac{2}{5}x + 1$$
02

Swap x and y variables

Now, we swap the x and y variables: $$x = \frac{2}{5}y + 1$$
03

Solve for y

To solve for y, we need to isolate y on one side of the equation. Here is the process: 1. Subtract 1 from both sides of the equation: $$x - 1 = \frac{2}{5}y$$ 2. Multiply both sides of the equation by \(\frac{5}{2}\) to get y by itself: $$\frac{5}{2}(x - 1) = y$$
04

Replace y with the inverse function notation

Finally, we replace y with the inverse function notation \(f^{-1}(x)\): $$f^{-1}(x) = \frac{5}{2}(x - 1)$$ And that's our inverse function!

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

One-to-One Functions
A one-to-one function is a special kind of function that never assigns the same output to two different inputs. This means that each input in the function's domain corresponds to a unique output in its range.
  • If a function is one-to-one, it passes both the horizontal line test and the vertical line test.
  • The horizontal line test ensures that no horizontal line crosses the graph of the function more than once.
  • Passing these tests guarantees that the function has an inverse.
Understanding one-to-one functions is crucial because only these functions have inverses that are also functions. For example, in our exercise, the given function \(f(x) = \frac{2}{5}x + 1\) is a linear function, which is typically one-to-one. This property ensures that there's a valid inverse for this function.
Algebraic Manipulation
Algebraic manipulation is the process of rearranging equations to solve for unknown variables. It's a vital skill in solving for inverse functions. In our exercise, we used algebraic manipulation to find the inverse of \(f(x) = \frac{2}{5}x + 1\).
Let's break down the steps:
  • We first replace \(f(x)\) with \(y\) for simplicity, getting \(y = \frac{2}{5}x + 1\).
  • Next, we swap \(x\) and \(y\) to form the equation \(x = \frac{2}{5}y + 1\). This swapping step is essential for finding the inverse.
  • We solve for \(y\) by isolating it. This involves subtracting 1 from both sides, then multiplying by \(\frac{5}{2}\) to clear the fraction and isolate \(y\).
  • Finally, we represent \(y\) as \(f^{-1}(x)\), confirming the inverse is \(f^{-1}(x) = \frac{5}{2}(x - 1)\).
Each step in algebraic manipulation should be carried out with care to ensure accuracy and clarity in solving inverses.
Linear Functions
Linear functions are a type of function characterized by a straight line graph. Their general form is \(f(x) = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
Key properties of linear functions include:
  • They have a constant rate of change, represented by the slope \(m\).
  • They produce a straight line when graphed.
  • Linear functions such as \(f(x) = \frac{2}{5}x + 1\) are one-to-one, making it possible to find inverses.
Linear functions are straightforward to work with because their inverses can be found using simple algebraic steps. In our problem, we see that the function's inverse is also a linear function, \(f^{-1}(x) = \frac{5}{2}(x - 1)\), which reinforces the simplicity and symmetry characteristic of linear equations.

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Most popular questions from this chapter

Given that \(\log 5=0.6990\) and \(\log 9=0.9542,\) use the propertics of logarithms to approximate the following $$\log \frac{9}{5}$$

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