Problem 53 - College Algebra

Question

For the following exercises, find the inverse of the functions with \(a, b, c\) positive real numbers. \(f(x)=\sqrt{a x^{2}+b}\)

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(x) = \sqrt{\frac{x^2 - b}{a}} \).

Step 1: Understand the Function

The given function is \( f(x) = \sqrt{a x^2 + b} \). This function takes an input \( x \), applies the operations inside the square root, and then returns the square root of the result. Our goal is to find \( f^{-1}(x) \), which is the inverse function.

Step 2: Replace \(f(x)\) with \(y\)

To find the inverse, start by replacing \( f(x) \) with \( y \). Now the equation looks like \( y = \sqrt{a x^2 + b} \).

Step 3: Solve for \(x\) in terms of \(y\)

Square both sides to eliminate the square root: \( y^2 = a x^2 + b \). Next, solve for \( x^2 \) by subtracting \( b \) from both sides: \( y^2 - b = a x^2 \). Divide both sides by \( a \) to isolate \( x^2 \): \( x^2 = \frac{y^2 - b}{a} \).

Step 4: Solve for \(x\)

Take the square root of both sides to solve for \( x \): \( x = \pm \sqrt{\frac{y^2 - b}{a}} \). Since \( a, b, c \) are positive real numbers, the range of \( f(x) \) is \([\sqrt{b}, \infty)\). Therefore, consider \( x = \sqrt{\frac{y^2 - b}{a}} \) to keep the function one-to-one.

Step 5: Express the Inverse Function

Since we solved for \( x \) in terms of \( y \) and need to express the inverse function, interchange the roles of \( x \) and \( y \). Thus the inverse function is \( f^{-1}(x) = \sqrt{\frac{x^2 - b}{a}} \).

Key Concepts

Function Operations

Function operations involve performing mathematical processes on functions, just like you would with numbers. When you have a function, such as \( f(x) = \sqrt{a x^2 + b} \), several operations are outlined:

• Taking an input \( x \) and adjusting it with the operations inside the function.
• Calculating the expression under the square root.
• Finding the square root of the resulting value.

These steps illustrate how a function modifies an input to produce an output. When finding an inverse function, the goal is to reverse these operations to solve for the original input. This involves swapping the input and output, and working through the operations in the reverse order to arrive back at the original input from an output.

Square Roots

Square roots are fundamental in mathematics and have key properties that are exploited in function operations and solving equations. The function \( f(x) = \sqrt{a x^2 + b} \) involves using the square root operation as its primary mathematical process. Here’s what you need to know:

• Squaring a number \( x \) (i.e., \( x^2 \)) is finding the area of a square with side length \( x \).
• Taking the square root \( (\sqrt{ }) \) reverses this process and finds the side length of that square.

When finding an inverse that includes a square root, you'll typically square both sides of an equation to remove the root. In our equation, \( y = \sqrt{a x^2 + b} \), squaring both sides yields \( y^2 = a x^2 + b \). This is a crucial step because it transforms the expression into a polynomial equation, making it easier to isolate \( x \). Remember, solving equations with square roots often means carefully considering positive and negative roots.

Solving Equations

Solving equations is about isolating the variable you’re interested in. When given an equation like \( y = \sqrt{a x^2 + b} \), the goal is to find the expression for \( x \) relative to \( y \). The process is generally as follows:

• Rearrange the equation to isolate terms involving \( x \).
• Apply inverse operations, such as squaring both sides to eliminate square roots: \( y^2 = a x^2 + b \).
• Solve for \( x^2 \) by isolating it: \( x^2 = \frac{y^2 - b}{a} \).
• Finally, calculate \( x \) using the square root: \( x = \sqrt{\frac{y^2 - b}{a}} \).

It’s important to keep track of operations to ensure no step violates mathematical rules. Moreover, the solution’s validity must be evaluated within the function’s context. Since \( x \) must remain a valid number under the function’s domain, solutions might restrict x to positive or negative values depending on context.

Domain and Range

Domain and range are vital concepts in understanding functions and their inverses. A function’s domain is the set of possible inputs, whereas its range is the set of possible outputs. For the function \( f(x) = \sqrt{a x^2 + b} \), these concepts can be described as follows:

• Domain: It's determined by the values of \( x \) that make the expression under the square root non-negative since you cannot take the square root of a negative number in real numbers. For example, \( a x^2 + b \geq 0 \).
• Range: The output of \( f(x) \) is the result of the square root operation. As \( x \) varies over the domain, the range is \([\sqrt{b}, \infty)\), starting from the smallest value the square root can achieve and extending indefinitely.

When dealing with inverse functions, the domain and range interchange. Hence understanding these concepts ensures that the inverse function remains valid and one-to-one, crucial for accurately reversing the function operations.