Problem 24 - Algebra 2

Question

For each function \(f,\) find \(f^{-1}\) and the domain and range of \(f\) and \(f^{-1} .\) Determine whether \(f^{-1}\) is a function. $$ f(x)=\sqrt{x-5} $$

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}(x) = x^2 + 5\), with domain (-\(\infty\), \(\infty\)) and range [5, \(\infty\)]. The domain of \(f\) is [5, \(\infty\)], and the range of \(f\) is [0, \(\infty\)]. \(f^{-1}\) is a function.

Step 1: Find the inverse function

Start by replacing \(f(x)\) with \(y\), to get \(y = \sqrt{x-5}\). Next, swap \(x\) and \(y\) to get \(x = \sqrt{y-5}\). Then, square both sides and solve for \(y\) to get the inverse function. This gives \(y = x^2 + 5\), or \(f^{-1}(x) = x^2 + 5\).

Step 2: Find the domain and range of \(f\)

The function \(f\) is defined for all \(x\) such that \(x-5 \geq 0\), or \(x \geq 5\). So, the domain of \(f\) is [5, \(\infty\)]. Since \(f(x) = \sqrt{x-5}\), \(f(x)\) can produce any nonnegative real number, so the range of \(f\) is [0, \(\infty\)].

Step 3: Find the domain and range of \(f^{-1}\)

The inverse function \(f^{-1}(x) = x^2 + 5\) is defined and real for all \(x\), so its domain is (-\(\infty\), \(\infty\)). Since \(f^{-1}(x) = x^2 + 5\), its lowest possible value is 5, and it increases from there. Thus, the range of \(f^{-1}\) is [5, \(\infty\)].

Step 4: Determine if \(f^{-1}\) is a function

Since every input \(x\) in the domain of \(f^{-1}\) gives exactly one output, \(f^{-1}\) is a function.

Key Concepts

Domain and Range

Understanding the domain and range of a function is crucial when working with inverse functions. The domain of a function consists of all the input values (or x-values) that the function accepts. Meanwhile, the range is all the possible output values (or y-values) produced by the function.

In our exercise, the given function is \( f(x) = \sqrt{x-5} \). Here, the expression under the square root, \(x-5\), must be non-negative for \(f\) to be defined. This means \(x-5 \geq 0\), leading to a domain of \([5, \infty)\).

For the range, since \(f(x)\) gives us the square root values, and square roots are always non-negative, the range is \([0, \infty)\).

For the inverse function \( f^{-1}(x) = x^2 + 5 \), the situation is different. It is defined for all real numbers, so its domain is \((-\infty, \infty)\). However, since \(x^2 + 5\) is always 5 or greater (since squared numbers cannot be negative), the range is \([5, \infty)\).

Knowing these allows us to see where we can apply each function and understand their limitations.

Function Notation

Function notation is a way to name and describe functions in a clear and concise manner. It helps us distinguish between different functions and clarify what operation a function performs.

Using \(f(x)\), \(g(x)\), or \(h(x)\) is common practice, where the letter typically represents the function's name, and \(x\) is the variable representing the input. For example, in our case, \( f(x) = \sqrt{x-5} \), \(f\) is the name of the function, and \(x\) is its input.

When dealing with inverse functions, this notation becomes even more helpful. The inverse of \(f(x)\) is represented as \(f^{-1}(x)\), which is not to be confused with \(1/f(x)\). Here, \(f^{-1}(x)\) stands for a function that reverses the operation of \(f\). So while \(f(x) = \sqrt{x-5}\) turns \(x\) into \(\sqrt{x-5}\), the inverse \(f^{-1}(x) = x^2 + 5\) would turn it back.

Function notation, therefore, simplifies understanding and manipulating multiple functions and their inverses, making it an essential tool in mathematics.

Solving Equations

Solving equations is a fundamental skill in mathematics, necessary for finding inverse functions. To solve an equation means to find the value of the variable that makes the equation true.

For finding an inverse like in our problem, we began with \( y = \sqrt{x-5} \). We then swapped \(x\) and \(y\) which is the step required to reflect across the line \(y=x\) in theory, giving us \(x = \sqrt{y-5}\).

Next, you manipulate the equation to solve for \(y\). Here, we squared both sides to eliminate the square root, transforming it to \( x^2 = y - 5 \), followed by adding 5 to both sides to isolate \(y\). This results in \( y = x^2 + 5 \), our inverse function.

Understanding each step when solving these equations ensures that you can consistently find inverse functions and verify if they are truly inverses by checking if their compositions return the identity functions \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\). Thus, solving equations allows us to delve deeper into the world of functions and their inverses.