Solving Composite Function Inverse: A Step-by-Step Guide

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Hey guys! Let's dive into a cool math problem today that involves composite functions and their inverses. We're going to tackle a problem where we need to find the value of (f∘g)βˆ’1(6)(f \circ g)^{-1}(6), given the inverse functions fβˆ’1(x)=3x+2f^{-1}(x) = 3x + 2 and gβˆ’1(x)=3x+42xβˆ’8g^{-1}(x) = \frac{3x+4}{2x-8}. It might sound a bit intimidating at first, but don't worry, we'll break it down step-by-step so it's super easy to understand.

Understanding the Problem

Before we jump into the solution, let's make sure we understand what the problem is asking. We have two functions, ff and gg, and we know their inverses, fβˆ’1f^{-1} and gβˆ’1g^{-1}. Remember, the inverse of a function basically "undoes" what the original function does. The notation (f∘g)(x)(f \circ g)(x) means the composite function, where we first apply the function gg to xx, and then apply the function ff to the result. So, (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x)). We're asked to find the value of the inverse of this composite function, (f∘g)βˆ’1(f \circ g)^{-1}, when it's evaluated at x=6x = 6. In other words, we want to find the value yy such that (f∘g)(y)=6(f \circ g)(y) = 6. This means f(g(y))=6f(g(y)) = 6. To solve this, we will need to use the properties of inverse functions and work our way inside out. This requires a solid understanding of function composition and inverse functions. Remember that the inverse of a composite function has a specific property: (f∘g)βˆ’1(x)=(gβˆ’1∘fβˆ’1)(x)(f \circ g)^{-1}(x) = (g^{-1} \circ f^{-1})(x). This property is crucial for solving this problem efficiently. We can use this to our advantage by finding fβˆ’1f^{-1} and gβˆ’1g^{-1} and then composing them in the reverse order. This is often a more straightforward approach than trying to find the composite function f∘gf \circ g first and then finding its inverse. Understanding this property is key to navigating this type of problem successfully. Additionally, it's important to remember the domain restrictions, especially for gβˆ’1(x)g^{-1}(x), where xβ‰ 4x \neq 4. These restrictions can affect the possible solutions and must be considered when interpreting the final result. Ignoring these restrictions can lead to incorrect answers. Therefore, always pay close attention to the given conditions and domain limitations when solving function problems.

Key Concepts and Formulas

Okay, before we start crunching numbers, let's quickly review some key concepts and formulas that will be super helpful:

  • Inverse Function: If f(a)=bf(a) = b, then fβˆ’1(b)=af^{-1}(b) = a. Think of it like a reverse operation. If ff turns aa into bb, then fβˆ’1f^{-1} turns bb back into aa. This is a fundamental concept for this problem.
  • Composite Function: (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x)). This means we apply gg first, and then apply ff to the result. Got it?
  • Inverse of a Composite Function: (f∘g)βˆ’1(x)=(gβˆ’1∘fβˆ’1)(x)(f \circ g)^{-1}(x) = (g^{-1} \circ f^{-1})(x). This is a super important property that we'll use to solve the problem. Remember this, guys! It basically says that the inverse of a composite function is the composition of the inverses in the reverse order. This formula is derived from the fundamental properties of inverse functions and function composition. To fully appreciate its significance, consider what it implies: finding the inverse of a composite function directly can be complex, but by using this property, we can break down the problem into simpler steps involving individual inverse functions. This is a powerful tool for simplifying complex problems in mathematics. Understanding the derivation of this formula can also provide deeper insights into the nature of function inverses and their interactions. For instance, it highlights the importance of order when dealing with composite functions and their inverses. The reverse order in the formula is not arbitrary; it is a direct consequence of how functions and their inverses operate in sequence. By mastering this concept, you'll be well-equipped to tackle a wide range of problems involving composite functions and their inverses.

Solving the Problem Step-by-Step

Alright, let's get down to business! We need to find (f∘g)βˆ’1(6)(f \circ g)^{-1}(6). Remember that cool formula we just talked about? (f∘g)βˆ’1(x)=(gβˆ’1∘fβˆ’1)(x)(f \circ g)^{-1}(x) = (g^{-1} \circ f^{-1})(x). So, we can rewrite our problem as finding (gβˆ’1∘fβˆ’1)(6)(g^{-1} \circ f^{-1})(6). This is a major simplification! It allows us to work with the functions we already know, fβˆ’1f^{-1} and gβˆ’1g^{-1}. The beauty of this approach is that it transforms a complex problem into a series of manageable steps. Instead of grappling with the composite function and its inverse directly, we can focus on the individual inverse functions and their composition. This strategy is not only effective but also highlights the elegance of mathematical problem-solving – breaking down a complex problem into simpler, solvable parts. This step-by-step approach is a common theme in mathematics and is a valuable skill to develop. It encourages a systematic way of thinking and can be applied to a variety of problems beyond just function inverses. Moreover, it reduces the likelihood of errors, as each step can be carefully checked and verified. By mastering this technique, you will gain confidence in your ability to tackle challenging mathematical problems. So, let's embrace this approach and move forward with the solution, knowing that we have a clear and methodical plan.

  1. Find f⁻¹(6):

    We know that fβˆ’1(x)=3x+2f^{-1}(x) = 3x + 2. So, to find fβˆ’1(6)f^{-1}(6), we simply substitute x=6x = 6 into the equation:
    fβˆ’1(6)=3(6)+2=18+2=20f^{-1}(6) = 3(6) + 2 = 18 + 2 = 20. Easy peasy! This is a straightforward application of the given formula for fβˆ’1(x)f^{-1}(x). Substituting the value of xx and performing the arithmetic gives us the value of the inverse function at that point. This step underscores the importance of understanding function notation and how to evaluate functions at specific values. It's a foundational skill in mathematics and is crucial for more advanced topics such as calculus and differential equations. The clarity and simplicity of this step are a testament to the power of using the correct formula and applying it methodically. By carefully substituting the value and performing the calculations, we ensure accuracy and avoid potential errors. This meticulous approach is essential in mathematics, where a small mistake can lead to a completely different answer. So, let's celebrate this simple yet crucial step and move forward with confidence.

  2. Find g⁻¹(f⁻¹(6)):

    Now we know that fβˆ’1(6)=20f^{-1}(6) = 20. So, we need to find gβˆ’1(20)g^{-1}(20). We're given that gβˆ’1(x)=3x+42xβˆ’8g^{-1}(x) = \frac{3x+4}{2x-8}. Let's plug in x=20x = 20: gβˆ’1(20)=3(20)+42(20)βˆ’8=60+440βˆ’8=6432=2g^{-1}(20) = \frac{3(20)+4}{2(20)-8} = \frac{60+4}{40-8} = \frac{64}{32} = 2. Awesome! This step builds upon the previous one, utilizing the result we obtained for fβˆ’1(6)f^{-1}(6) to evaluate gβˆ’1(x)g^{-1}(x). The process involves substituting the value into the given formula for gβˆ’1(x)g^{-1}(x) and simplifying the expression. This highlights the interconnectedness of the steps in solving the problem. Each step relies on the results of the previous ones, emphasizing the importance of accuracy and attention to detail throughout the solution. The calculation itself involves basic arithmetic operations, but the potential for errors exists if the substitution or simplification is not performed carefully. Therefore, it's crucial to double-check each step to ensure correctness. This practice not only leads to the correct answer but also reinforces good mathematical habits. By breaking down the problem into smaller, manageable steps, we reduce the complexity and increase the likelihood of success. So, let's proceed with confidence, knowing that we have a solid foundation built on accurate calculations and a clear understanding of the process.

  3. Therefore, (f ∘ g)⁻¹(6) = 2:

    Since (f∘g)βˆ’1(6)=(gβˆ’1∘fβˆ’1)(6)(f \circ g)^{-1}(6) = (g^{-1} \circ f^{-1})(6) and we found that (gβˆ’1∘fβˆ’1)(6)=2(g^{-1} \circ f^{-1})(6) = 2, we can conclude that (f∘g)βˆ’1(6)=2(f \circ g)^{-1}(6) = 2. We did it! This final step is the culmination of all our previous efforts. It brings together the results we obtained in the earlier steps and provides the answer to the original problem. The logical flow from the initial problem statement to the final answer is clear and concise, demonstrating the power of a systematic approach to problem-solving. By using the property of inverse functions and breaking down the problem into smaller, manageable steps, we were able to arrive at the solution with confidence. This process not only provides the answer but also reinforces the underlying mathematical concepts and techniques. The satisfaction of reaching the final answer is a reward for our perseverance and attention to detail. It's a testament to the fact that even complex problems can be solved with a clear understanding of the principles and a methodical approach. So, let's celebrate our success and recognize the value of the problem-solving skills we have honed along the way.

Conclusion

So, there you have it! We successfully found that (f∘g)βˆ’1(6)=2(f \circ g)^{-1}(6) = 2. Remember, the key was understanding the properties of inverse functions and composite functions, and then breaking the problem down into smaller, manageable steps. You guys got this! Function problems, especially those involving inverses and compositions, can seem daunting at first glance. However, by mastering the fundamental concepts and applying them systematically, you can tackle even the most challenging problems with confidence. This problem-solving process is not just about finding the correct answer; it's about developing critical thinking skills that are applicable in various areas of life. The ability to break down complex problems into simpler components, identify relevant information, and apply appropriate strategies is a valuable asset in both academic and professional settings. So, keep practicing, keep exploring, and keep challenging yourselves. The world of mathematics is full of fascinating concepts and intriguing problems waiting to be discovered. Embrace the journey, and you'll be amazed at what you can achieve.