Inverse Function And Domain Of F(x)=8+√(4-x)

Jul 14, 2025 by ADMIN 45 views

$Let $f(x)=8+\sqrt{4-x}$. Determine the inverse function and its domain.$

Finding the Inverse Function

To determine the inverse function, denoted as $f^{-1}(x)$ , for the given function $f(x) = 8 + \sqrt{4-x}$ , we need to follow a systematic approach. The concept of an inverse function is fundamental in mathematics, as it essentially "undoes" the operation performed by the original function. This process involves swapping the roles of the independent variable $x$ and the dependent variable $y$ , and then solving for $y$ in terms of $x$ . This new expression will represent the inverse function. Before we dive into the algebraic manipulation, it's crucial to understand the underlying principles. The inverse function exists if and only if the original function is one-to-one, meaning that each output value corresponds to a unique input value. Graphically, this can be verified using the horizontal line test, where a horizontal line intersects the graph of the function at most once. Furthermore, the domain of the inverse function is the range of the original function, and vice versa. This relationship highlights the symmetrical nature of functions and their inverses, which are reflections of each other across the line $y=x$ . Understanding these concepts will help us navigate the process of finding the inverse function more effectively and ensure that our solution is mathematically sound.

Replace $f(x)$ with $y$ : $y = 8 + \sqrt{4 - x}$
Swap $x$ and $y$ : $x = 8 + \sqrt{4 - y}$
Isolate the square root term: $x - 8 = \sqrt{4 - y}$
Square both sides to eliminate the square root: $(x - 8)^2 = 4 - y$
Isolate $y$ : $y = 4 - (x - 8)^2$
Express the inverse function: $f^{-1}(x) = 4 - (x - 8)^2$

Thus, the inverse function is found to be $f^{-1}(x) = 4 - (x - 8)^2$ . This quadratic form might seem straightforward, but it's crucial to consider the domain and range implications, which we will delve into in the subsequent sections. The algebraic steps we've taken are crucial, but they must be contextualized within the broader framework of inverse functions. For instance, squaring both sides of the equation can introduce extraneous solutions if we're not careful about the domain restrictions. Similarly, the range of the original function will directly impact the domain of the inverse function, and vice versa. Therefore, it's not just about manipulating equations; it's about understanding the behavior of functions and their inverses, and how they interact with each other. The next step involves determining the domain of this inverse function, which requires us to consider the range of the original function. This interconnectedness between a function and its inverse is a recurring theme in mathematics, and it highlights the importance of a holistic approach when solving problems.

Determining the Domain of the Inverse Function

To determine the domain of the inverse function $f^{-1}(x) = 4 - (x - 8)^2$ , we must consider the range of the original function $f(x) = 8 + \sqrt{4 - x}$ . The domain of the original function is determined by the expression inside the square root, which must be non-negative: $4 - x \geq 0$ . This inequality implies that $x \leq 4$ . Now, let's find the range of $f(x)$ . Since the square root term $\sqrt{4 - x}$ is always non-negative, its minimum value is 0. This occurs when $x = 4$ , giving $f(4) = 8 + \sqrt{4 - 4} = 8$ . As $x$ decreases, the value of $4 - x$ increases, and so does the square root term. Therefore, the function $f(x)$ is decreasing on its domain. There is no upper bound on $f(x)$ as $x$ approaches negative infinity, since the square root term becomes arbitrarily large. Consequently, the range of $f(x)$ is $[8, \infty)$ .

Since the domain of the inverse function is the range of the original function, the domain of $f^{-1}(x)$ is $[8, \infty)$ . However, we must also consider the domain of the inverse function in its derived form, $f^{-1}(x) = 4 - (x - 8)^2$ . This is a quadratic function, which would typically have a domain of all real numbers. But, because it's an inverse function, we must restrict the domain based on the range of the original function. If we didn't apply this restriction, the inverse function would not pass the vertical line test, and it wouldn't truly be an inverse. Therefore, we explicitly state that the domain of $f^{-1}(x)$ is $[8, \infty)$ . This underscores a crucial aspect of finding inverse functions: algebraic manipulation is just one part of the process. The other part involves careful consideration of domain and range restrictions to ensure that the resulting function is a true inverse. Furthermore, understanding the graphical representation of the function and its inverse can provide valuable insights into these restrictions. The inverse function is a reflection of the original function across the line $y=x$ , and this graphical interpretation can help us visualize the domain and range relationships more clearly. In summary, determining the domain of the inverse function involves a combination of algebraic analysis and understanding the fundamental properties of functions and their inverses.

Restriction on the Inverse Function

Due to the restriction on the domain of the original function $f(x)$ , we must consider the impact on the range and, consequently, on the domain of the inverse function. The original function $f(x) = 8 + \sqrt{4 - x}$ is only defined for $x \leq 4$ . The square root function always yields a non-negative value, so $\sqrt{4 - x} \geq 0$ . Therefore, $f(x) = 8 + \sqrt{4 - x} \geq 8$ . This means the range of $f(x)$ is $[8, \infty)$ . When we found the inverse function algebraically as $f^{-1}(x) = 4 - (x - 8)^2$ , we need to consider this range restriction. The domain of $f^{-1}(x)$ is the range of $f(x)$ , so $x$ in $f^{-1}(x)$ must be greater than or equal to 8.

This restriction is vital because the quadratic function $4 - (x - 8)^2$ without a domain restriction would be a parabola opening downward, and it would not be a one-to-one function. However, by restricting the domain to $x \geq 8$ , we are considering only one half of the parabola, which does pass the horizontal line test and is indeed the inverse of the original function. This illustrates a crucial point about inverse functions: sometimes, we need to restrict the domain of a function to make it one-to-one and ensure that its inverse is also a function. The interplay between the algebraic form of the function and its graphical representation becomes particularly important in these situations. Visualizing the graph of the function and its potential inverse can provide valuable insights into the necessary domain restrictions. For instance, in this case, the original function is a decreasing function, and its inverse is a portion of a parabola. Without the domain restriction, the inverse would fail the vertical line test, indicating that it's not a true function. Therefore, the restriction is not just a mathematical formality; it's an essential part of defining the inverse function correctly. In summary, determining the correct domain for the inverse function requires a careful consideration of the range of the original function and the properties of the inverse function itself.

Final Answer

Therefore, the inverse function is:

f^{-1}(x) = 4 - (x - 8)^2

and its domain is:

Domain of $f^{-1}(x) = [8, \infty)$

In conclusion, finding the inverse of a function and its domain involves a multi-step process that combines algebraic manipulation with a thorough understanding of function properties. The key steps involve swapping the variables, solving for the new dependent variable, and critically evaluating the domain and range restrictions. The domain of the inverse function is intrinsically linked to the range of the original function, and this relationship is crucial for correctly defining the inverse. Furthermore, it's essential to recognize that not all functions have inverses, and those that do must satisfy the one-to-one property. When dealing with functions like the one in this problem, which involves a square root and a quadratic form in the inverse, domain restrictions become particularly important. Squaring both sides of an equation can introduce extraneous solutions, and the domain of the inverse function must be chosen carefully to ensure that it truly undoes the operation of the original function. By paying close attention to these details, we can successfully navigate the process of finding inverse functions and accurately determine their domains.