Finding The Inverse Of H(x) = √(x+5) And Verification

ADMIN · · 12 min read

Table Of Content

  1. Determining the Inverse Function
    1. Step-by-Step Process
  2. Verifying the Inverse Function using Composite Functions
    1. Evaluating h(h-1(x))
    2. Evaluating h-1(h(x))
  3. Conclusion
  4. Step-by-Step Process of Finding the Inverse Function
    1. Initial Steps and Transformation
    2. Isolating y and Determining the Inverse Function
  5. Verification of the Inverse Function using Composition
    1. Step-by-Step Verification Process
  6. Comprehensive Conclusion
  7. Unraveling the Process of Finding the Inverse Function
    1. Initial Transformations and Swapping Variables
    2. Isolating y and Defining the Inverse Function
  8. Verifying the Inverse Function through Composition
    1. Detailed Steps for Verification
  9. Final Thoughts

Finding the inverse of a function is a crucial concept in mathematics, particularly in algebra and calculus. The inverse function, denoted as h-1(x), essentially 'undoes' what the original function h(x) does. In this article, we will explore the process of finding the inverse of the function h(x) = √(x+5) and verify our result using composite functions. The correct inverse function should satisfy the condition that h(h-1(x)) = h-1(h(x)) = x. This verification process ensures that the inverse function we've found truly reverses the operation of the original function.

Determining the Inverse Function

The primary goal here is to find a function that reverses the operations performed by h(x) = √(x+5). The function h(x) first adds 5 to x and then takes the square root of the result. To reverse these operations, we need to first square the input and then subtract 5. Let's go through the step-by-step process of finding the inverse.

Step-by-Step Process

  1. Replace h(x) with y: This makes the equation easier to manipulate. So, we have y = √(x+5).

  2. Swap x and y: This is the key step in finding the inverse. By swapping x and y, we are essentially reversing the roles of the input and output. This gives us x = √(y+5).

  3. Solve for y: Now, we need to isolate y to express it in terms of x. To do this, we first square both sides of the equation: x2 = (√(y+5))2 x2 = y + 5

    Next, we subtract 5 from both sides: x2 - 5 = y

  4. Replace y with h-1(x): This gives us the inverse function notation. So, h-1(x) = x2 - 5.

Therefore, the inverse function of h(x) = √(x+5) is h-1(x) = x2 - 5. However, we must also consider the domain of the original function. Since we are taking the square root, the expression inside the square root must be non-negative. Thus, x + 5 ≥ 0, which means x ≥ -5. Also, the range of h(x) is y ≥ 0, because the square root function always returns a non-negative value. Consequently, the domain of the inverse function h-1(x) is x ≥ 0, as it must match the range of the original function.

Verifying the Inverse Function using Composite Functions

To ensure that h-1(x) = x2 - 5 is indeed the inverse of h(x) = √(x+5), we need to verify that h(h-1(x)) = x and h-1(h(x)) = x. This process involves evaluating composite functions, which means plugging one function into another.

Evaluating h(h-1(x))

First, let's evaluate h(h-1(x)). This means we will substitute h-1(x) into h(x) wherever we see x.

h(h-1(x)) = h(x2 - 5) = √((x2 - 5) + 5)

Simplifying the expression inside the square root:

√((x2 - 5) + 5) = √(x2 - 5 + 5) = √(x2)

Since we know that the domain of h-1(x) is x ≥ 0, we can simplify √(x2) to x:

√(x2) = x

Thus, h(h-1(x)) = x, which is the first part of our verification.

Evaluating h-1(h(x))

Next, we need to evaluate h-1(h(x)). This means we will substitute h(x) into h-1(x) wherever we see x.

h-1(h(x)) = h-1(√(x+5)) = (√(x+5))2 - 5

Squaring the square root gives us:

(√(x+5))2 - 5 = (x+5) - 5

Now, simplifying the expression:

(x+5) - 5 = x

Thus, h-1(h(x)) = x, which confirms the second part of our verification.

Conclusion

In conclusion, the inverse function of h(x) = √(x+5) is indeed h-1(x) = x2 - 5, with the domain x ≥ 0. We have verified this by showing that both h(h-1(x)) = x and h-1(h(x)) = x. This process of finding the inverse and verifying it using composite functions is a fundamental concept in mathematics, demonstrating the relationship between a function and its inverse. Understanding these concepts is crucial for further studies in calculus and other advanced mathematical topics.

Finding the inverse function involves reversing the operations of the original function, and verifying the inverse ensures that the functions truly 'undo' each other. The domain and range considerations are also vital in defining the inverse function correctly. The composite functions play a central role in this verification process, confirming the accuracy of the inverse function.

Determining the inverse of a function is a core concept in mathematics, particularly within algebra and calculus. The inverse function, symbolized as h-1(x), acts to 'undo' the operations performed by the original function h(x). In this comprehensive article, we will meticulously explore the process of finding the inverse of the function h(x) = √(x+5), and we will rigorously verify our result using composite functions. The correct inverse function must meet the criterion that h(h-1(x)) = h-1(h(x)) = x. This verification is essential to ensure that the inverse function accurately reverses the operations of the original function. This article aims to provide a clear, step-by-step guide suitable for students and anyone interested in understanding inverse functions.

Step-by-Step Process of Finding the Inverse Function

Our main objective is to find the function that precisely reverses the actions of h(x) = √(x+5). The function h(x) operates by first adding 5 to x and then taking the square root of the result. To invert these actions, we must reverse the order, first squaring the input and then subtracting 5. Let’s break down the process into manageable steps to ensure clarity and accuracy.

Initial Steps and Transformation

  1. Replace h(x) with y: This substitution simplifies the equation and makes it easier to manipulate algebraically. Thus, we rewrite the function as y = √(x+5). This replacement provides a clearer view of the relationship between the input and output variables.

  2. Interchange x and y: This is the pivotal step in finding the inverse function. By swapping x and y, we are essentially reversing the roles of the input and output, which is the core concept of finding an inverse. The equation now becomes x = √(y+5). This swap prepares the equation for solving for y in terms of x.

Isolating y and Determining the Inverse Function

  1. Solve for y: Our next task is to isolate y to express it as a function of x. This will give us the inverse function. We start by squaring both sides of the equation: x2 = (√(y+5))2 x2 = y + 5

    This step eliminates the square root, bringing us closer to isolating y. Next, we subtract 5 from both sides to solve for y: x2 - 5 = y

  2. Replace y with h-1(x): This final step gives us the notation for the inverse function. We replace y with h-1(x), which denotes the inverse of h(x). Therefore, we have h-1(x) = x2 - 5. This is our candidate for the inverse function, but we still need to verify it.

Thus, we have found that the inverse function of h(x) = √(x+5) appears to be h-1(x) = x2 - 5. However, to fully define an inverse function, we must also consider the domain. The original function h(x) has a square root, which restricts the domain. Since we cannot take the square root of a negative number in the real number system, x + 5 must be greater than or equal to 0. This gives us x ≥ -5. The range of h(x) is all non-negative real numbers because the square root function produces non-negative outputs. Therefore, the domain of the inverse function h-1(x) is x ≥ 0, matching the range of the original function. This restriction is crucial for the inverse function to be properly defined.

Verification of the Inverse Function using Composition

To definitively confirm that h-1(x) = x2 - 5 is indeed the inverse of h(x) = √(x+5), we need to perform a critical verification step. This involves using composite functions to ensure that the functions 'undo' each other. The verification requires us to show that both h(h-1(x)) = x and h-1(h(x)) = x. This process will validate that our candidate inverse function is correct.

Step-by-Step Verification Process

Let's methodically verify the inverse function using composition.

  1. Evaluating h(h-1(x)): This involves substituting h-1(x) into h(x). h(h-1(x)) = h(x2 - 5). Now, we replace x in h(x) with (x2 - 5): h(x2 - 5) = √((x2 - 5) + 5).

    Simplifying the expression inside the square root gives: √((x2 - 5) + 5) = √(x2 - 5 + 5) = √(x2).

    Since the domain of h-1(x) is x ≥ 0, we can simplify √(x2) to x: √(x2) = x.

    This result shows that h(h-1(x)) = x, which is the first requirement for h-1(x) to be the inverse of h(x).

  2. Evaluating h-1(h(x)): Next, we substitute h(x) into h-1(x). h-1(h(x)) = h-1(√(x+5)). We replace x in h-1(x) with √(x+5): h-1(√(x+5)) = (√(x+5))2 - 5.

    Squaring the square root simplifies the expression: (√(x+5))2 - 5 = (x+5) - 5.

    Now, simplifying further: (x+5) - 5 = x.

    This result shows that h-1(h(x)) = x, which is the second requirement for h-1(x) to be the inverse of h(x).

Comprehensive Conclusion

In conclusion, we have successfully determined that the inverse function of h(x) = √(x+5) is h-1(x) = x2 - 5, with a crucial domain restriction of x ≥ 0. We meticulously verified this result by demonstrating that both composite functions, h(h-1(x)) and h-1(h(x)), simplify to x. This detailed process underscores the importance of not only finding a candidate for the inverse but also rigorously verifying its correctness using composition of functions.

Understanding inverse functions is fundamental in mathematics, particularly in areas such as calculus and advanced algebra. The ability to find the inverse of a function and to verify its correctness ensures a robust understanding of function behavior. Additionally, awareness of domain and range restrictions is vital in defining inverse functions accurately. The use of composite functions as a verification tool is a powerful technique that confirms the reciprocal relationship between a function and its inverse. This methodical approach is invaluable for students and professionals alike, enhancing their mathematical toolkit and problem-solving capabilities.

Understanding inverse functions is a cornerstone of mathematical proficiency, pivotal in both algebra and calculus. An inverse function, denoted as h-1(x), performs the reverse operation of the original function, h(x). This article offers an in-depth exploration of how to determine the inverse of the function h(x) = √(x+5), coupled with a rigorous verification process using composite functions. A correct inverse function must satisfy the conditions h(h-1(x)) = x and h-1(h(x)) = x, ensuring that it precisely reverses the actions of the initial function. This comprehensive guide is designed to elucidate the concepts and methods involved in finding and verifying inverse functions, making it accessible to students and mathematics enthusiasts alike.

Unraveling the Process of Finding the Inverse Function

The central objective is to identify the function that effectively undoes the operations of h(x) = √(x+5). The function h(x) first adds 5 to x and then calculates the square root. Inverting these steps requires us to first square the input and then subtract 5. Let’s break down this process into a clear, step-by-step methodology to ensure accuracy and comprehension.

Initial Transformations and Swapping Variables

  1. Substitution of h(x) with y: We begin by replacing h(x) with y, which simplifies the equation and facilitates algebraic manipulation. Thus, the function is rewritten as y = √(x+5). This substitution provides a clearer representation of the input-output relationship.

  2. Interchanging x and y: This is the critical step in finding the inverse. By swapping x and y, we effectively reverse their roles, which is fundamental to the concept of an inverse function. The equation now becomes x = √(y+5). This interchange prepares the equation for the isolation of y in terms of x.

Isolating y and Defining the Inverse Function

  1. Solving for y: Our next goal is to isolate y and express it as a function of x, thereby defining the inverse function. We start by squaring both sides of the equation: x2 = (√(y+5))2 x2 = y + 5

    This step eliminates the square root, bringing us closer to isolating y. Next, we subtract 5 from both sides to solve for y: x2 - 5 = y

  2. Replacing y with h-1(x): This final step formalizes the inverse function notation. We replace y with h-1(x), the standard symbol for the inverse of h(x). Therefore, we derive h-1(x) = x2 - 5. This is our candidate for the inverse function, which must now be verified.

Consequently, we have determined that the apparent inverse function of h(x) = √(x+5) is h-1(x) = x2 - 5. However, a comprehensive definition of an inverse function necessitates considering its domain. The original function h(x) includes a square root, which imposes a restriction. Since the square root of a negative number is undefined in the real number system, the expression inside the square root, x + 5, must be greater than or equal to 0. This yields the condition x ≥ -5. The range of h(x) comprises all non-negative real numbers because the square root operation always results in a non-negative output. Therefore, the domain of the inverse function h-1(x) is constrained to x ≥ 0, aligning with the range of the original function. This domain restriction is paramount for the correct definition of the inverse function.

Verifying the Inverse Function through Composition

To definitively ascertain that h-1(x) = x2 - 5 is indeed the inverse of h(x) = √(x+5), a critical verification step is required. This involves employing composite functions to ensure that the two functions effectively 'undo' each other. The verification process entails demonstrating that both h(h-1(x)) = x and h-1(h(x)) = x. This validation will confirm the accuracy of our candidate inverse function.

Detailed Steps for Verification

Let’s meticulously verify the inverse function using functional composition.

  1. Evaluating h(h-1(x)): This step involves substituting h-1(x) into h(x). h(h-1(x)) = h(x2 - 5). Now, we replace x in h(x) with the expression (x2 - 5): h(x2 - 5) = √((x2 - 5) + 5).

    Simplifying the expression inside the square root, we get: √((x2 - 5) + 5) = √(x2 - 5 + 5) = √(x2).

    Given that the domain of h-1(x) is x ≥ 0, we can simplify √(x2) to x: √(x2) = x.

    This result confirms that h(h-1(x)) = x, which satisfies the first condition for h-1(x) to be the inverse of h(x).

  2. Evaluating h-1(h(x)): Next, we proceed to substitute h(x) into h-1(x). h-1(h(x)) = h-1(√(x+5)). We replace x in h-1(x) with √(x+5): h-1(√(x+5)) = (√(x+5))2 - 5.

    Squaring the square root simplifies the expression: (√(x+5))2 - 5 = (x+5) - 5.

    Further simplification yields: (x+5) - 5 = x.

    This result confirms that h-1(h(x)) = x, which satisfies the second condition for h-1(x) to be the inverse of h(x).

Final Thoughts

In summary, we have successfully established that the inverse function of h(x) = √(x+5) is h-1(x) = x2 - 5, subject to the domain restriction x ≥ 0. We rigorously verified this by demonstrating that both composite functions, h(h-1(x)) and h-1(h(x)), simplify to x. This methodical approach underscores the necessity of not only deriving a candidate for the inverse but also validating its accuracy through functional composition.

Comprehending inverse functions is crucial in various areas of mathematics, including calculus and advanced algebra. The proficiency in finding the inverse of a function and verifying its correctness is essential for a thorough understanding of functional relationships. Moreover, an awareness of domain and range constraints is vital for accurately defining inverse functions. The application of composite functions as a verification method is a powerful technique that confirms the reciprocal relationship between a function and its inverse. This detailed methodology is beneficial for students and professionals alike, enhancing their mathematical expertise and problem-solving abilities.

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