Problem 193 - Calculus Volume 1

Question

For the following exercises, a. find the inverse function, and b. find the domain and range of the inverse function. $$ f(x)=\sqrt{x-1} $$

Step-by-Step Solution

Answer

The inverse is \( f^{-1}(x) = x^2 + 1 \) with domain: all real \( x \), range: \( y \geq 1 \).

Step 1: Understand the Function

The given function is \( f(x) = \sqrt{x-1} \). This means that \( f(x) \) is defined only for values of \( x \) where \( x - 1 \geq 0 \), or \( x \geq 1 \).

Step 2: Swap and Solve for Inverse

To find the inverse function, swap \( x \) and \( y \) in the equation \( y = \sqrt{x-1} \) and solve for \( y \):\[ x = \sqrt{y-1} \]Square both sides of the equation to eliminate the square root:\[ x^2 = y - 1 \]Solve for \( y \):\[ y = x^2 + 1 \]Thus, the inverse function is \( f^{-1}(x) = x^2 + 1 \).

Step 3: Determine the Domain of the Inverse

The original function \( f(x) \) has an output of non-negative real numbers because it is a square root function. Therefore, the domain of the inverse function \( f^{-1}(x) = x^2 + 1 \) will be based on the range of the original function, which is \( y \geq 0 \). Thus, the inverse function is valid and defined for all real \( x \).

Step 4: Determine the Range of the Inverse

Since \( f^{-1}(x) = x^2 + 1 \), the smallest value it can take is when \( x = 0 \), which gives \( f^{-1}(0) = 0^2 + 1 = 1 \). As \( x \) increases or decreases, \( f^{-1}(x) \) also increases without bound. Therefore, the range of \( f^{-1}(x) \) is \( y \geq 1 \).

Key Concepts

Understanding Domain and Range

In mathematical functions, understanding domain and range is crucial. The **domain** refers to the set of all possible input values (or 'x' values) that a function can accept. For a square root function like \( f(x) = \sqrt{x-1} \), we need to ensure the expression inside the square root is non-negative to keep results real. Hence, we set the condition \( x - 1 \geq 0 \), leading to \( x \geq 1 \). This makes the domain of the function all real numbers greater than or equal to 1.

The **range** is the set of all possible output values (or 'y' values). For \( f(x) = \sqrt{x-1} \), the smallest output is 0 (when \( x = 1 \)), and as x increases, so does the output. Therefore, the range of the original function is non-negative real numbers (i.e., \( y \geq 0 \)).

When we find the inverse function, the domain and range switch roles. The domain of \( f^{-1}(x) = x^2 + 1 \) is the range of the original function, and the range becomes all numbers \( y \geq 1 \). This interplay between domain and range is vital for understanding how inverse functions work.

Exploring the Square Root Function

The square root function, symbolized as \( \sqrt{} \), is a fundamental mathematical operation. It finds the non-negative number which, when multiplied by itself, gives the original number. For example, \( \sqrt{4} = 2 \) because \( 2 \times 2 = 4 \).

In our function \( f(x) = \sqrt{x-1} \), the expression inside the square root dictates the domain restrictions. We cannot take the square root of a negative number in basic real functions, as it results in a complex number. This is why we require \( x - 1 \geq 0 \) to ensure real outputs.

Moreover, the behavior of the square root function ensures that outputs are always non-negative. This property directly influences the range of the function to only include non-negative numbers. The square root function grows slowly, implying that small changes in \( x \) can result in even smaller changes in \( f(x) \), especially as \( x \) becomes very large.

Decoding Function Notation

Function notation is a mathematical shorthand to represent functions compactly and systematically. When you encounter \( f(x) \), it signifies a function named 'f' with 'x' as the input variable. This notation provides clarity and makes manipulating functions more straightforward as you explore calculus and algebra.

Inverse functions use a similar notation; \( f^{-1}(x) \) indicates the inverse of the function \( f \). This doesn't mean reciprocal, rather it tells you how to reverse the process of \( f(x) \).

To find an inverse, you often swap 'x' and 'y' (like \( x = \sqrt{y-1} \)) and solve for the new 'y'. The result is the inverse function, here \( f^{-1}(x) = x^2 + 1 \). Understanding this notation helps you track how inputs and outputs transform between a function and its inverse, ensuring accurate mathematical communication.