Problem 54 - College Algebra with Corequisite Support

Question

For the following exercises, find the inverse of the functions with \(a, b, c\) positive real numbers. $$ f(x)=\sqrt[3]{a x+b} $$

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(x) = \frac{x^3 - b}{a} \).

Step 1: Understand the Given Function

The given function is \( f(x) = \sqrt[3]{ax + b} \). We need to find the inverse of this function, meaning we need to find a function \( f^{-1}(x) \) such that \( f(f^{-1}(x)) = x \).

Step 2: Replace f(x) with y

To find the inverse, start by setting \( f(x) = y \). This gives us the equation \( y = \sqrt[3]{ax + b} \).

Step 3: Solve for x in terms of y

To find \( x \) in terms of \( y \), first, eliminate the cube root by cubing both sides of the equation: \( y^3 = ax + b \).

Step 4: Isolate x

Subtract \( b \) from both sides to get \( y^3 - b = ax \). Then, divide by \( a \) to solve for \( x \): \( x = \frac{y^3 - b}{a} \).

Step 5: Write the Inverse Function

Now that we have \( x = \frac{y^3 - b}{a} \), replace \( y \) with \( x \) to express the inverse function in terms of \( x \): \( f^{-1}(x) = \frac{x^3 - b}{a} \).

Key Concepts

Cube Roots

The cube root of a number is a value that, when multiplied by itself three times, equals the original number. It is the opposite operation of cubing a number. In mathematical terms, the cube root of a number \( x \) is written as \( \sqrt[3]{x} \). This notation can sometimes be confusing because it looks similar to the square root, but remember, it involves three multiplications instead of two.Cube roots are useful in various mathematical scenarios, including solving equations involving cubic terms. When dealing with a cube root, consider that:

• The cube root of a positive number is positive.
• The cube root of a negative number is negative.
• Cube roots apply to real numbers since any real number has a real cube root.

Practically, if you want to eliminate a cube root in an equation, taking the cube of both sides will help to resolve the function to its basic form. For example, if you have \( \sqrt[3]{x} = y \), cubing both sides gives \( x = y^3 \).

Solving Equations

Solving equations is a fundamental aspect of algebra that involves finding unknown values that satisfy a given mathematical statement. In our case, finding the inverse of the function \( f(x) = \sqrt[3]{ax + b} \) required solving the equation in terms of \( x \).To solve an equation successfully:

• Identify what needs to be isolated. In inverse function problems, you usually solve for \( x \) in terms of \( y \).
• Perform operations that will help isolate the variable. Start with basic operations like addition, subtraction, multiplication, and division.
• After isolating the term of interest, make sure to test your solution by substituting back to ensure it is correct.

For example, when dealing with \( y = \sqrt[3]{ax + b} \), you aim to get rid of the cube root by cubing both sides, which leads you to another more straightforward equation, \( y^3 = ax + b \). From here, you can solve for \( x \) progressively.

Algebraic Manipulation

Algebraic manipulation refers to rearranging and simplifying expressions in algebra to find solutions. It is crucial in solving equations, especially when finding inverse functions.In the process of finding inverses, algebraic manipulation can involve:

• Rearranging terms to isolate variables. This means moving terms from one side of an equation to the other by reverse operations.
• Simplifying complex fractions or equations by breaking them down into more manageable parts.
• Using basic algebraic operations to maintain equality while making the expression simpler.

For instance, when working with \( y^3 = ax + b \), you subtract \( b \) and divide by \( a \) to solve for \( x \), thus obtaining \( x = \frac{y^3 - b}{a} \). This demonstrates a step-by-step approach involving algebraic manipulation to derive the inverse function \( f^{-1}(x) = \frac{x^3 - b}{a} \). These manipulations help in transforming the equation into a form where calculating the inverse becomes a direct task.