Finding F Inverse Of -3 Given F Of X Equals X Plus 2 Over X Plus 9

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In this article, we will explore the process of finding the inverse of a function and then evaluating it at a specific point. We will focus on the function f(x)=x+2x+9f(x) = \frac{x+2}{x+9} and determine the value of its inverse, f−1(−3)f^{-1}(-3). This involves several key steps, including understanding the concept of an inverse function, finding the algebraic expression for the inverse, and finally, substituting the given value into the inverse function.

Understanding Inverse Functions

Before we dive into the specifics of our function, let's first establish a solid understanding of what an inverse function is. In mathematical terms, an inverse function essentially "undoes" the operation of the original function. If we apply a function ff to a value xx to get yy (i.e., f(x)=yf(x) = y), then applying the inverse function f−1f^{-1} to yy should give us back xx (i.e., f−1(y)=xf^{-1}(y) = x). This fundamental property is crucial for understanding and working with inverse functions. Graphically, the inverse function is a reflection of the original function across the line y=xy = x. This reflection highlights the swapping of the input and output roles between the function and its inverse. Not all functions have inverses; a function must be bijective (both injective, meaning one-to-one, and surjective, meaning onto) to have an inverse. In simpler terms, this means that each input must map to a unique output (injective) and every possible output must be mapped to by some input (surjective).

To determine if a function has an inverse, we often use the horizontal line test. If any horizontal line intersects the graph of the function at most once, then the function is injective and may have an inverse. The concept of inverse functions is not just a theoretical exercise; it has significant practical applications across various fields. For example, in cryptography, inverse functions are used to decrypt messages. In computer graphics, they are used for transformations and projections. In calculus, understanding inverse functions is essential for implicit differentiation and related rates problems. The process of finding an inverse function involves a few key steps, which we will delve into in the next section. These steps generally include swapping the roles of xx and yy in the function's equation and then solving for yy. The resulting expression represents the inverse function. However, it's important to remember that the domain and range of the original function and its inverse are swapped. This means that the domain of ff becomes the range of f−1f^{-1}, and the range of ff becomes the domain of f−1f^{-1}. This reciprocal relationship is a fundamental aspect of inverse functions and must be considered when working with them.

Finding the Inverse of f(x)=x+2x+9f(x) = \frac{x+2}{x+9}

Now, let's apply the concept of inverse functions to our specific function, f(x)=x+2x+9f(x) = \frac{x+2}{x+9}. The process of finding the inverse involves a systematic approach. First, we replace f(x)f(x) with yy to make the equation easier to manipulate. So, we have y=x+2x+9y = \frac{x+2}{x+9}. The next crucial step is to swap the variables xx and yy. This reflects the fundamental property of inverse functions, where the input and output roles are reversed. After swapping, we get x=y+2y+9x = \frac{y+2}{y+9}. The goal now is to isolate yy on one side of the equation. This will give us the expression for the inverse function, f−1(x)f^{-1}(x). To isolate yy, we first multiply both sides of the equation by (y+9)(y+9). This eliminates the fraction and gives us x(y+9)=y+2x(y+9) = y+2. Next, we distribute xx on the left side, resulting in xy+9x=y+2xy + 9x = y + 2. Now, we need to gather all terms containing yy on one side and all other terms on the other side. Subtracting yy from both sides gives us xy−y+9x=2xy - y + 9x = 2. Then, subtracting 9x9x from both sides gives us xy−y=2−9xxy - y = 2 - 9x. To further isolate yy, we factor it out from the left side of the equation: y(x−1)=2−9xy(x - 1) = 2 - 9x. Finally, we divide both sides by (x−1)(x - 1) to solve for yy: y=2−9xx−1y = \frac{2 - 9x}{x - 1}. This expression represents the inverse function, f−1(x)f^{-1}(x). Therefore, we can write f−1(x)=2−9xx−1f^{-1}(x) = \frac{2 - 9x}{x - 1}. This algebraic manipulation is a standard technique for finding inverse functions. It involves carefully applying algebraic operations to isolate the desired variable. The resulting expression for f−1(x)f^{-1}(x) is a rational function, similar in form to the original function f(x)f(x). However, it's important to note that the domain of f−1(x)f^{-1}(x) is restricted by the denominator, x−1x - 1. Specifically, xx cannot be equal to 1, as this would result in division by zero. This restriction is a consequence of the fact that the range of f(x)f(x) becomes the domain of f−1(x)f^{-1}(x).

Evaluating f−1(−3)f^{-1}(-3)

With the inverse function f−1(x)=2−9xx−1f^{-1}(x) = \frac{2 - 9x}{x - 1} in hand, our next step is to evaluate it at x=−3x = -3. This means we substitute −3-3 for xx in the expression for f−1(x)f^{-1}(x). So, we have f−1(−3)=2−9(−3)−3−1f^{-1}(-3) = \frac{2 - 9(-3)}{-3 - 1}. Now, we simplify the expression. First, we calculate the numerator: 2−9(−3)=2+27=292 - 9(-3) = 2 + 27 = 29. Next, we calculate the denominator: −3−1=−4-3 - 1 = -4. Therefore, f−1(−3)=29−4f^{-1}(-3) = \frac{29}{-4}, which can also be written as f−1(−3)=−294f^{-1}(-3) = -\frac{29}{4}. This is the value of the inverse function at x=−3x = -3. In this evaluation, we followed the standard order of operations, ensuring that multiplication is performed before addition and subtraction. The resulting fraction, −294-\frac{29}{4}, is an improper fraction, which is perfectly acceptable as a final answer. We could also express it as a mixed number, −714-7\frac{1}{4}, or as a decimal, −7.25-7.25. However, the fraction form is often preferred in mathematical contexts as it is exact and avoids any potential rounding errors. The process of evaluating an inverse function at a specific point is a straightforward application of the algebraic expression we derived for the inverse. It involves simply substituting the given value and simplifying the resulting expression. This skill is essential for working with inverse functions in various contexts, including solving equations, graphing, and analyzing function behavior. The value of f−1(−3)f^{-1}(-3) tells us that if we input −294-\frac{29}{4} into the original function f(x)f(x), the output will be −3-3. This confirms the fundamental property of inverse functions, where they "undo" the operation of the original function. The ability to evaluate inverse functions is a valuable tool in mathematics and its applications.

Conclusion

In summary, we have successfully found the inverse of the function f(x)=x+2x+9f(x) = \frac{x+2}{x+9} and evaluated it at x=−3x = -3. The inverse function was determined to be f−1(x)=2−9xx−1f^{-1}(x) = \frac{2 - 9x}{x - 1}, and the value of f−1(−3)f^{-1}(-3) was found to be −294-\frac{29}{4}. This process involved understanding the concept of inverse functions, applying algebraic techniques to find the inverse function, and then substituting the given value to evaluate the inverse. The ability to find and evaluate inverse functions is a crucial skill in mathematics, with applications in various fields. This article provides a clear and detailed explanation of the steps involved, making it a valuable resource for anyone learning about inverse functions. The key steps include swapping the variables xx and yy, solving for yy, and then substituting the given value into the resulting expression. The concept of inverse functions is not just a mathematical abstraction; it has real-world applications in areas such as cryptography, computer graphics, and calculus. Understanding inverse functions allows us to solve problems and analyze situations in a more comprehensive way. The reciprocal relationship between a function and its inverse is a fundamental concept in mathematics, and mastering it is essential for success in higher-level mathematics courses and applications. This article has provided a solid foundation for understanding and working with inverse functions, and it serves as a valuable reference for future learning and problem-solving.