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Chapter 3: Problem 40

Find the inverse function of \(f\) $$ f(x)=3-5 x $$

Short Answer

Expert verified
The inverse function is \( f^{-1}(x) = \frac{x-3}{-5} \).

Step by step solution

01

Understand the Function

The given function is \( f(x) = 3 - 5x \). We need to find its inverse function, denoted as \( f^{-1}(x) \). Generally, the inverse function reverses the action of the original function.
02

Replace f(x) with y

To find the inverse, start by replacing \( f(x) \) with \( y \). This gives us the equation \( y = 3 - 5x \).
03

Solve for x

Rearrange the equation \( y = 3 - 5x \) to find \( x \) in terms of \( y \). To do this, subtract 3 from both sides: \( y - 3 = -5x \).
04

Isolate x

Now divide both sides by \(-5\) to isolate \( x \):\[ x = \frac{y - 3}{-5} \]
05

Replace y with x to Express the Inverse Function

The last step is to replace \( y \) with \( x \). Therefore, the inverse function is:\[ f^{-1}(x) = \frac{x - 3}{-5} \]

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Key Concepts

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

Linear Equations
Linear equations are equations of the first degree, meaning they involve no powers higher than one of the variables. A simple form is \( y = mx + b \), representing a straight line when graphed. Here, \( m \) is the slope, indicating the steepness of the line, and \( b \) is the y-intercept, where the line crosses the y-axis.
  • Slope (m): Describes the change in \( y \) for a unit change in \( x \). In the function \( f(x) = 3 - 5x \), the slope is -5.
  • Y-intercept (b): The value of \( y \) when \( x = 0 \). For \( f(x) = 3 - 5x \), the y-intercept is 3.
Linear equations are fundamental because they represent the simplest function form and are used extensively in various problems, like finding inverse functions as shown in the original problem.
Function Notation
Function notation is a way of representing functions in mathematics by using symbols and is helpful for clarity and simplicity. For example, \( f(x) \) is the traditional notation for a function named \( f \), with \( x \) being its variable or input.
In the equation \( f(x) = 3 - 5x \), \( f \) denotes the function, and \( 3 - 5x \) represents the operation performed on \( x \). This notation helps in distinguishing between different functions and understanding their behavior.
  • Input (x): The variable that you substitute into the function.
  • Output (f(x)): The result you get after evaluating the function with the input.
  • Inverse (f^{-1}(x)): A function that "undoes" the original function, reversing inputs and outputs.
Utilizing function notation, particularly with inverse functions, helps in easily representing and manipulating mathematical relationships.
Solving Equations
Solving equations involves finding the value of variables that make the equation true. It is a critical aspect of algebra that requires several steps and techniques based on the structure of the equation.
  • Identifying the Equation: Start by identifying the form of the equation and the unknown variable. For example, when given \( y = 3 - 5x \), \( x \) is the variable you're solving for.
  • Rearranging Terms: Use algebraic operations to isolate the variable of interest. Subtraction, addition, multiplication, or division can help move terms across the equation.
  • Isolate the Variable: Follow each manipulation step-by-step to solve for the variable. In our example, you subtract 3 and then divide by -5 to extract \( x \).
  • Check Your Solution: Substitute the solved values back into the original equation to verify correctness.
Understanding the method of solving equations is essential in deriving the inverse function, a practical application seen in algebra regarding function transformations.

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Most popular questions from this chapter

Solving an Equation for an Unknown Function In Exercise 69 of Section 3.6 you were asked to solve equations in which the unknowns were functions. Now that we know about inverses and the identity function (see Exercise 92 ) we can use algebra to solve such equations. For instance, to solve \(f \circ g=h\) for the unknown function \(f,\) we perform the following steps: $$ \begin{aligned} f \circ g &=h \\ f \circ g \circ g^{-1} &=h \circ g^{-1} \\\ f \circ I &=h \circ g^{-1} \\ f &=h \circ g^{-1} \end{aligned} $$ So the solution is \(f=h \circ g^{-1} .\) Use this technique to solve the equation \(f \circ g=h\) for the indicated unknown function. (a) Solve for \(f,\) where \(g(x)=2 x+1\) and $$ h(x)=4 x^{2}+4 x+7 $$ (b) Solve for \(g,\) where \(f(x)=3 x+5\) and $$ h(x)=3 x^{2}+3 x+2 $$

A family of functions is given. In parts (a) and (b) graph all the given members of the family in the viewing rectangle indicated. In part (c) state the conclusions that you can make from your graphs. \(f(x)=(x-c)^{2}\) (a) \(c=0,1,2,3 ; \quad[-5,5]\) by \([-10,10]\) (b) \(c=0,-1,-2,-3 ; \quad[-5,5]\) by \([-10,10]\) (c) How does the value of \(c\) affect the graph?

Multiple Discounts You have a S50 coupon from the manufacturer good for the purchase of a cell phone. The store where you are purchasing your cell phone is offering a 20\(\%\) discount on all cell phones. Let \(x\) represent the regular price of the cell phone. (a) Suppose only the 20\(\%\) discount applies. Find a function \(f\) that models the purchase price of the cell phone as a function of the regular price \(x .\) (b) Suppose only the \(\$ 50\) coupon applies. Find a function \(g\) that models the purchase price of the cell phone as a function of the sticker price \(x\) (c) If you can use the coupon and the discount, then the purchase price is either \(f \circ g(x)\) or \(g \circ f(x),\) depending on the order in which they are applied to the price. Find both \(f \circ g(x)\) and \(g \circ f(x) .\) Which composition gives the lower price?

Find the functions \(f \circ g, g \circ f, f \circ f,\) and \(g \circ g\) and their domains. $$ f(x)=x-4, \quad g(x)=|x+4| $$

Find \(f+g, f-g \cdot f g,\) and \(f / g\) and their domains. $$ f(x)=\frac{2}{x}, g(x)=\frac{4}{x+4} $$
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