/*! 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 149 Find the inverse of \(f(x)=x^{2}... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the inverse of \(f(x)=x^{2}+4, x \geq 0\) (Section \(1.8, \text { Example } 7)\).

Short Answer

Expert verified
The inverse of the function \(f(x) = x^{2} + 4, x \geq 0\) is \(f^{-1}(x) = \sqrt{x - 4}, x \geq 4\).

Step by step solution

01

Interchanging \(f(x)\) and \(x\)

To find the inverse of a function, we switch the roles of \(f(x)\) and \(x\). In this case, we convert the equation \(f(x) = x^{2} + 4\) into \(x = f^{-1}(x)^{2} + 4\)
02

Solving for \(f^{-1}(x)\)

Then, solve for \(f^{-1}(x)\) by isolating \(f^{-1}(x)\) on one side of the equation. This would lead to \(f^{-1}(x) = \sqrt{x - 4}\). As we are given that \(x \geq 0\) originally, it's clear that the square root is real, and since \(f^{-1}(x) = \sqrt{x - 4}\) is only defined for \(x \geq 4\), we will only take the positive square root as the inverse.
03

Check if the inverse function is correct

Lastly, we check to see if \(f(f^{-1}(x)) = x\) is true using the new function \(f^{-1}(x)\). If we let \(f(x) = f^{-1}(x)^{2} + 4\) and \(f^{-1}(x) = \sqrt{x - 4}\), substituting \(f^{-1}(x)\) into \(f(x)\) gives \(f(f^{-1}(x)) = (\sqrt{x - 4})^2 + 4 = x\), which is the same as the \(x\) started with, thus the inverse function is correct.

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Key Concepts

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

Function Composition
Function composition involves combining two functions to create a new function. This process can be thought of as plugging one function into another. It's an essential tool when checking if two functions are inverses.
For our exercise, we found the inverse function of a given function and check its accuracy using composition. The check involves plugging the inverse function back into the original function and verifying the result is equal to the input.
In mathematical terms, if we have two functions, say \( f(x) \) and \( g(x) \), then the composition of \( f \) and \( g \) is written as \( f(g(x)) \). To confirm if \( f(x) \) and \( f^{-1}(x) \) are inverses, we must ensure:
  • \( f(f^{-1}(x)) = x \)
  • \( f^{-1}(f(x)) = x \)
In our problem, this was demonstrated by showing \((\sqrt{x-4})^2 + 4 = x\), confirming the inverse calculation.
Domain and Range
Understanding the domain and range of functions and their inverses is vital. The domain of a function is all the possible input values (typically \( x \)), while the range is all possible outputs (\( f(x) \)).
For our function \( f(x) = x^2 + 4 \), given \( x \geq 0 \), this constraint defines its domain. Consequently, the range is all real numbers greater than or equal to 4 because squaring non-negative numbers starts from 0, and adding 4 shifts this upward.
When finding an inverse function, the domain and range swap roles. Thus, the inverse \( f^{-1}(x) = \sqrt{x - 4} \) has a domain where \( x \geq 4 \) and a range consisting of all non-negative real numbers \( y \geq 0 \). This switch ensures the function and its inverse satisfy all values originally covered or produced.
Solving Equations
Solving the equation to find the inverse function is a systematic process. After performing the switch between \( f(x) \) and \( x \), we solve algebraically to isolate the inverse term.
Starting with \( x = f^{-1}(x)^2 + 4 \), rearrange to isolate \( f^{-1}(x) \) on one side. We subtract 4 from both sides to find \( x - 4 = f^{-1}(x)^2 \).
Next, applying the square root to both sides, we solve for \( f^{-1}(x) \):
  • Take the square root: \( f^{-1}(x) = \sqrt{x - 4} \).
  • Ensure the values fulfill the original domain condition \( x \geq 0 \).
This process highlights key solving techniques, emphasizing careful consideration of domains and ensuring results are feasible given initial function constraints.

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