Problem 73 - Calculus for Biology and Medicine

Question

(a) Show that \(f(x)=1 / x^{3}, x>0\), is one to one, and find its inverse together with its domain. (b) Graph \(f(x)\) and \(f^{-1}(x)\) in one coordinate system, together with the line \(y=x\), and convince yourself that the graph of \(f^{-1}(x)\) can be obtained by reflecting the graph of \(f(x)\) about the line \(y=x\)

Step-by-Step Solution

Answer

(a) \(f(x)\) is one-to-one; inverse is \(f^{-1}(x) = x^{-1/3}\) with domain \(x > 0\). (b) Graphs are reflections across \(y=x\).

Step 1: Show that \(f(x)\) is One-to-One

To determine whether \(f(x) = \frac{1}{x^{3}}\) is one-to-one on \(x > 0\), we check if different inputs produce different outputs. Assume \(f(a) = f(b)\) for some \(a, b > 0\). Then:\[\frac{1}{a^3} = \frac{1}{b^3}\]Taking reciprocals and using the laws of exponents, this simplifies to:\[a^3 = b^3\]Taking the cube root of both sides gives us \(a = b\). Since \(a\) and \(b\) must be equal, \(f(x)\) is one-to-one.

Step 2: Find the Inverse \(f^{-1}(x)\)

To find the inverse of \(f(x) = \frac{1}{x^3}\), replace \(f(x)\) with \(y\), resulting in:\[y = \frac{1}{x^3}\]Swap \(x\) and \(y\) to solve for \(y\):\[x = \frac{1}{y^3}\]Taking the reciprocal of both sides and solving for \(y\) gives:\[y^3 = \frac{1}{x}\]\[y = \sqrt[3]{\frac{1}{x}}\]So the inverse function is \(f^{-1}(x) = x^{-1/3}\), defined for \(x > 0\).

Step 3: Determine the Domain of the Inverse Function

Since the inverse function is \(f^{-1}(x) = x^{-1/3}\), and it involves taking the cube root, it's defined for all real numbers where \(x > 0\). Hence, the domain of \(f^{-1}(x)\) is \(x > 0\).

Step 4: Graph \(f(x)\), \(f^{-1}(x)\), and \(y = x\)

By graphing \(f(x)=\frac{1}{x^3}\) and its inverse \(f^{-1}(x)=x^{-1/3}\), we place them on the same coordinate plane. The line \(y = x\) is the axis of symmetry. The graphs of \(f(x)\) and \(f^{-1}(x)\) will be reflections across the line \(y = x\). Check this by observing that for any point \((a, f(a))\) on \(f(x)\), the point \((f(a), a)\) will lie on \(f^{-1}(x)\).

Key Concepts

One-to-One Function

A one-to-one function has a unique characteristic. Every input corresponds to one and only one output. Conversely, each output is associated with a single input.
For instance, consider the function given by the formula \(f(x) = \frac{1}{x^3}\) for \(x > 0\). To check its one-to-one nature, assume that two different inputs might result in the same output: \(f(a) = f(b)\). This implies:

• \(\frac{1}{a^3} = \frac{1}{b^3}\)
• Which simplifies upon manipulation to \(a^3 = b^3\).
• Taking cube roots of both sides gives \(a = b\).

So indeed, different inputs cannot yield the same output, confirming that it is a one-to-one function. This property is crucial as it allows us to define an inverse function, providing a reversible map between inputs and outputs.

Domain of a Function

The domain of a function refers to the complete set of possible input values. For any function \(f(x)\), the domain includes all the x-values for which \(f(x)\) produces real and defined outputs.
In our example, \(f(x) = \frac{1}{x^3}\), the domain is restricted to \(x > 0\). This restriction arises because the function involves division by \(x^3\), which is undefined for \(x = 0\). Additionally, since only positive real numbers are considered, \(x\) must be greater than zero to maintain the function's definition.
When considering the inverse function, \(f^{-1}(x) = x^{-1/3}\), the domain also remains for \(x > 0\). This is because taking the cube root of a positive number results in a real number, ensuring the inverse function's validity for the same positive domain. Understanding domains helps in grasping where functions are applicable and reversible.

Function Reflection

Reflection involves transforming a function's graph across a specific line. When dealing with inverses of functions, this line of symmetry is usually \(y = x\).
When graphing \(f(x) = \frac{1}{x^3}\) together with its inverse \(f^{-1}(x) = x^{-1/3}\), and the line \(y = x\), you can visualize their relationship. Each point on the graph of \(f(x)\) reflects over the line \(y = x\) to a corresponding point on \(f^{-1}(x)\).

• This means if there is a point \((a, f(a))\) on \(f(x)\), it reflects as the point \((f(a), a)\) on \(f^{-1}(x)\).
• The reflection process preserves the one-to-one nature of the function in both directions: from function to inverse and vice versa.

Function reflection not only enhances understanding of inverse relationships but also provides a visual proof of these mathematical concepts.