Finding The Value Of 'a' In A Function Equation

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Let's dive into solving this function problem step by step, guys. We're given a function f(2xβˆ’3)=8xβˆ’7f(2x-3) = 8x-7 and told that fβˆ’1(9)=af^{-1}(9) = a. Our mission, should we choose to accept it (and we do!), is to find the value of aa. Buckle up, because we're about to embark on a mathematical journey!

Understanding the Problem

Before we start crunching numbers, let's make sure we understand what the problem is asking. We have a function ff defined in a peculiar way: f(2xβˆ’3)=8xβˆ’7f(2x-3) = 8x-7. This means that when you plug (2xβˆ’3)(2x-3) into the function ff, you get 8xβˆ’78x-7 as the output. We also know that the inverse function fβˆ’1f^{-1} of 9 equals aa. In mathematical terms, fβˆ’1(9)=af^{-1}(9) = a. This tells us that when you plug 9 into the inverse function fβˆ’1f^{-1}, you get aa as the output. Remember that inverse functions essentially "undo" what the original function does. So, if fβˆ’1(9)=af^{-1}(9) = a, then it must be true that f(a)=9f(a) = 9. This is the key to solving our problem. Let's keep it in mind as we push forward into the solution.

Solving for x in Terms of f(x)

The first step is to express xx in terms of f(2xβˆ’3)f(2x-3). We know that f(2xβˆ’3)=8xβˆ’7f(2x-3) = 8x-7. Let's make a substitution to simplify things. Let y=2xβˆ’3y = 2x-3. Then, f(y)=8xβˆ’7f(y) = 8x-7. Our goal is to isolate xx. From y=2xβˆ’3y = 2x-3, we can write 2x=y+32x = y + 3, and therefore, x=y+32x = \frac{y+3}{2}. Now we substitute this expression for xx into the equation f(y)=8xβˆ’7f(y) = 8x-7. We get:

f(y)=8(y+32)βˆ’7f(y) = 8(\frac{y+3}{2}) - 7 f(y)=4(y+3)βˆ’7f(y) = 4(y+3) - 7 f(y)=4y+12βˆ’7f(y) = 4y + 12 - 7 f(y)=4y+5f(y) = 4y + 5

So, we now have f(y)=4y+5f(y) = 4y + 5. This is a more straightforward expression for our function. Remember, the variable name doesn't matter; we can just as easily write f(x)=4x+5f(x) = 4x + 5.

Finding the Inverse Function

Now that we have a simpler form for f(x)f(x), we can find its inverse function fβˆ’1(x)f^{-1}(x). To do this, we'll switch xx and yy in the equation y=4x+5y = 4x + 5 and solve for yy. So we get:

x=4y+5x = 4y + 5 Now, solve for yy: xβˆ’5=4yx - 5 = 4y y=xβˆ’54y = \frac{x-5}{4}

Therefore, the inverse function is fβˆ’1(x)=xβˆ’54f^{-1}(x) = \frac{x-5}{4}.

Calculating the Value of a

We are given that fβˆ’1(9)=af^{-1}(9) = a. Now that we have found the inverse function, we can easily calculate the value of aa by plugging in 9 into the expression for fβˆ’1(x)f^{-1}(x):

a=fβˆ’1(9)=9βˆ’54=44=1a = f^{-1}(9) = \frac{9-5}{4} = \frac{4}{4} = 1

So, a=1a = 1.

Final Answer

The value of aa that satisfies the given conditions is 1. Therefore, the final answer is:

a=1a = 1

Woo-hoo! We've successfully navigated through the function and its inverse to find the value of a. Remember, the key was understanding the relationship between a function and its inverse, and then carefully substituting and solving. Great job, team!

To solidify your understanding of functions and their inverses, try these extra exercises:

  1. Exercise 1: Given the function g(x)=3xβˆ’2g(x) = 3x - 2, find gβˆ’1(x)g^{-1}(x) and then evaluate gβˆ’1(7)g^{-1}(7).
  2. Exercise 2: If h(x)=x+12h(x) = \frac{x+1}{2}, determine hβˆ’1(x)h^{-1}(x) and find the value of bb such that hβˆ’1(5)=bh^{-1}(5) = b.
  3. Exercise 3: Suppose k(2x+1)=6xβˆ’4k(2x + 1) = 6x - 4. Find an expression for k(x)k(x) and then determine kβˆ’1(x)k^{-1}(x).

These exercises will give you additional practice in manipulating functions, finding inverse functions, and evaluating them at specific points. Good luck, and keep practicing!

Functions and their inverses can sometimes be tricky, but here are a few tips and tricks to keep in mind when solving these types of problems:

  1. Substitution: When dealing with functions like f(2xβˆ’3)f(2x - 3), consider making a substitution (e.g., y=2xβˆ’3y = 2x - 3) to simplify the expression and make it easier to work with.
  2. Inverse Relationship: Remember that if fβˆ’1(y)=xf^{-1}(y) = x, then f(x)=yf(x) = y. This relationship is fundamental to understanding and working with inverse functions.
  3. Finding the Inverse: To find the inverse function, switch xx and yy in the equation and then solve for yy. This gives you the expression for fβˆ’1(x)f^{-1}(x).
  4. Domain and Range: Pay attention to the domain and range of the original function and its inverse. The domain of f(x)f(x) is the range of fβˆ’1(x)f^{-1}(x), and vice versa.
  5. Practice: The more you practice, the more comfortable you'll become with functions and their inverses. Work through a variety of problems to build your skills and confidence.
  6. Check Your Work: Always check your work by verifying that f(fβˆ’1(x))=xf(f^{-1}(x)) = x and fβˆ’1(f(x))=xf^{-1}(f(x)) = x. This ensures that you have correctly found the inverse function.

By keeping these tips and tricks in mind, you'll be well-equipped to tackle any function-related problem that comes your way. Keep up the great work, and remember to have fun with math!

When working with functions and their inverses, it's easy to make mistakes if you're not careful. Here are some common pitfalls to watch out for:

  1. Forgetting to Switch x and y: When finding the inverse function, remember to switch xx and yy before solving for yy. This is a crucial step, and skipping it will lead to an incorrect answer.
  2. Incorrectly Solving for y: After switching xx and yy, make sure you solve for yy correctly. Pay attention to algebraic manipulations and avoid common errors like dividing by zero or misapplying the distributive property.
  3. Not Checking Your Work: Always check your work by verifying that f(fβˆ’1(x))=xf(f^{-1}(x)) = x and fβˆ’1(f(x))=xf^{-1}(f(x)) = x. This will help you catch any mistakes you may have made and ensure that you have correctly found the inverse function.
  4. **Confusing f^-1}(x) with 1/f(x)** Remember that $f^{-1(x)$ is the inverse function of f(x)f(x), not the reciprocal. These are two completely different concepts, so be careful not to confuse them.
  5. Ignoring Domain and Range: Pay attention to the domain and range of the original function and its inverse. Sometimes, the inverse function may have restrictions on its domain that you need to consider.

By being aware of these common mistakes and taking steps to avoid them, you can improve your accuracy and confidence when working with functions and their inverses.

Alright, folks, we've reached the end of our journey into the world of functions and their inverses. Remember, the key to success in mathematics is understanding the fundamental concepts and practicing consistently. Don't be afraid to make mistakes – they're an essential part of the learning process. Keep exploring, keep questioning, and keep pushing yourself to new heights. You've got this! By mastering the techniques and avoiding common pitfalls, you'll be well-equipped to tackle any function-related challenge that comes your way. So go forth and conquer the mathematical world, one function at a time!