Problem 45 - Elementary and Intermediate Algebra

Question

Each of the following functions is one-to-one. Find the inverse of each function and express it using \(f^{-1}(x)\) notation. \(f(x)=x^{3}+8\)

Step-by-Step Solution

Answer

The inverse of the function is \(f^{-1}(x) = \sqrt[3]{x - 8}\).

Step 1: Understand the Function

First, identify the given function, which is \(f(x) = x^3 + 8\). This is a cubic function shifted up by 8 units.

Step 2: Set Function Equal to y

Replace \(f(x)\) with \(y\) to set the equation as \(y = x^3 + 8\).

Step 3: Isolate the Cubic Term

Subtract 8 from both sides to isolate the cubic term: \(y - 8 = x^3\).

Step 4: Solve for x

Take the cube root of both sides to solve for \(x\): \(x = \sqrt[3]{y - 8}\).

Step 5: Write the Inverse Function

Replace \(x\) with \(f^{-1}(x)\) and \(y\) with \(x\) to express the inverse function: \(f^{-1}(x) = \sqrt[3]{x - 8}\).

Step 6: Verify the Inverse Function

Check that \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\) to verify our inverse. For \(f(f^{-1}(x))\): \(f(\sqrt[3]{x - 8}) = (\sqrt[3]{x - 8})^3 + 8 = x - 8 + 8 = x\). Similarly, for \(f^{-1}(f(x))\): \(f^{-1}(x^3 + 8) = \sqrt[3]{(x^3 + 8) - 8} = \sqrt[3]{x^3} = x\). Both conditions are satisfied.

Key Concepts

Cubic Functions

Cubic functions are a type of polynomial function where the highest power of the variable is three. These functions have the general form \(f(x) = ax^3 + bx^2 + cx + d\), where \(a\), \(b\), \(c\), and \(d\) are constants and \(a eq 0\). Such functions can create a variety of curves on a graph, typically producing an 'S' shaped curve with one inflection point.
However, our function \(f(x) = x^3 + 8\) is a simpler case, as it only consists of a cubic term \(x^3\) and a constant \(+8\), meaning it is a vertically shifted version of the basic cubic graph \(x^3\).
The graph of a cubic function like this one can pass through negative, zero, and positive regions due to its curvature, leading to different behaviors depending on the input value \(x\). This variation in shape and orientation helps them model phenomena in physics or economics like volume or long-term growth rates. They are also very useful mathematical tools in calculus.

One-to-One Functions

One-to-one functions are crucial when finding inverses. A function is one-to-one if every value of the output corresponds to one unique input. This means if \(f(a) = f(b)\), then \(a = b\).
They can be identified on a graph using the horizontal line test. If any horizontal line intersects the graph more than once, the function is not one-to-one. Why is this important? Because only one-to-one functions have inverses that are also functions.
For a cubic function like \(f(x) = x^3 + 8\), you can show it is one-to-one by checking its derivative. The derivative \(f'(x) = 3x^2\) is always positive (except at \(x = 0\), but it doesn’t change from positive), indicating that the function is strictly increasing. Hence, each output value is generated by exactly one input value, confirming it is one-to-one and thus has an inverse.

Function Notation

Function notation is a convenient way to represent and work with functions and their inverses. The standard notation \(f(x)\) denotes a function \(f\) applied to an input \(x\), yielding the output. For example, in \(f(x) = x^3 + 8\), if \(x = 2\), then \(f(2) = 2^3 + 8 = 16\).
When expressing an inverse function, we use the notation \(f^{-1}(x)\). This tells us we are talking about a new function that undoes the work of \(f(x)\). For any \(x\) in the domain of \(f^{-1}\), \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\).
Using these notations clarifies the relationship between a function and its inverse, which is essential in solving equations involving these transformations. Understanding and correctly employing function and inverse function notation is key in algebra and calculus, enabling clear communication of complex mathematical ideas.