/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Find the inverse function of \(f... [FREE SOLUTION] | 91影视

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Find the inverse function of \(f\). $$f(x)=4 x+7$$

Short Answer

Expert verified
The inverse function is \( f^{-1}(x) = \frac{x - 7}{4} \).

Step by step solution

01

Understand the Problem

We need to find the inverse of the given function \( f(x) = 4x + 7 \). The inverse function, \( f^{-1}(x) \), will 鈥渦ndo鈥 the action of \( f(x) \).
02

Write \( y = f(x) \)

Start by writing the function in terms of \( y \): \( y = 4x + 7 \). This helps us to treat it like an equation to solve for \( x \).
03

Solve for \( x \)

To find the inverse function, solve the equation \( y = 4x + 7 \) for \( x \). First, subtract 7 from both sides to get \( y - 7 = 4x \).
04

Isolate \( x \)

Divide both sides by 4 to solve for \( x \): \( x = \frac{y - 7}{4} \).
05

Write the Inverse Function \( f^{-1}(x) \)

Replace \( y \) with \( x \) in the equation from Step 4 to express the inverse function: \( f^{-1}(x) = \frac{x - 7}{4} \).

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Key Concepts

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

Function Composition
When we explore functions and their inverses, one important concept is function composition. Function composition involves applying one function to the result of another. In the context of inverse functions, this allows us to confirm whether two functions, such as \( f(x) \) and its inverse \( f^{-1}(x) \), truly "undo" each other.

To understand this, think of function \( f(x) = 4x + 7 \) and its inverse. The question is: if you take a number, apply \( f(x) \), and then apply \( f^{-1}(x) \), will you get back the original number? In mathematical terms, if \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \), then they are true inverses.

Here's how you check it:
  • Start with \( f(f^{-1}(x)) = f\left(\frac{x - 7}{4}\right) = 4\left(\frac{x - 7}{4}\right) + 7 = x - 7 + 7 = x \).
  • Similarly, verify that \( f^{-1}(f(x)) = f^{-1}(4x + 7) = \frac{4x + 7 - 7}{4} = \frac{4x}{4} = x \).
This confirms that applying \( f \) followed by \( f^{-1} \), or vice versa, returns the initial input, verifying the inverse relationship.
Algebraic Manipulation
Algebraic manipulation is a critical skill to understand inverse functions. It involves rearranging and simplifying equations to make them easier to solve, just as we do when finding inverses.

Let's recall the function \( y = 4x + 7 \). To find the inverse, we rearranged this as a solvable equation by:
  • Subtracting 7 from both sides: \( y - 7 = 4x \).
  • Then dividing by 4 to isolate \( x \): \( x = \frac{y - 7}{4} \).
Once we isolate \( x \), we redefine it as a function of \( y \) or vice-versa, which is the essence of finding an inverse function. This step converts the original function into its inverse, \( f^{-1}(x) = \frac{x - 7}{4} \).

Practicing these manipulations strengthens understanding and lays the groundwork for more complex operations in algebra.
Linear Functions
Linear functions, such as \( f(x) = 4x + 7 \), are characterized by their simplicity and straightforwardness. They graph as straight lines and are often expressed as \( y = mx + c \), where \( m \) is the slope and \( c \) is the y-intercept.

In our linear function \( f(x) = 4x + 7 \), the slope \( m \) is 4, indicating the function increases by 4 units in \( y \) for every 1 unit increase in \( x \). The y-intercept \( c \) is 7, meaning the line crosses the y-axis at point (0, 7).

Understanding these parameters helps when finding inverses. The inverse \( f^{-1}(x) = \frac{x - 7}{4} \) mirrors the original relationship with a change in slope to \( \frac{1}{m} \) for the inverse, reflecting how the input and output are switched. This swap is core to understanding inverses and is especially clear with linear functions due to their constant rate of change.

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