Problem 72 - Contemporary Precalculus

Question

If \(n\) is an odd positive integer, show that \(f(x)=x^{n}\) has an inverse function and find the rule of the inverse function. [Hint: Exercise \(71 \text { is the case when } n=5 .]\)

Step-by-Step Solution

Answer

Question: Show that the given function \(f(x) = x^n\) has an inverse function if n is an odd positive integer. If it does, provide the rule for the inverse function. Answer: The function \(f(x) = x^n\) has an inverse function if n is an odd positive integer. The rule for the inverse function is \(f^{-1}(x) = \sqrt[n]{x}\).

Step 1: Proving Injectivity (One-to-One)

To show that the function is injective, we must prove that for any \(x_1, x_2 \in \mathbb{R}\), if \(f(x_1) = f(x_2)\), then \(x_1 = x_2\). So, we assume that \(f(x_1) = f(x_2)\) and prove \(x_1 = x_2\). \(f(x_1) = f(x_2)\) implies \(x_1^n = x_2^n\). Since n is odd, we can write n = 2k+1, where k is a non-negative integer. Therefore, \((x_1^{n}) = x_1^{(2k+1)}\) and \((x_2^{n}) = x_2^{(2k+1)}\). As a result, \(x_1^{(2k+1)} = x_2^{(2k+1)}\), and the only solution to this equation is that \(x_1 = x_2\). Hence, the function is injective.

Step 2: Proving Surjectivity (Onto)

To show that the function is surjective, we must prove that for any \(y \in \mathbb{R}\), there exists an \(x \in \mathbb{R}\) such that \(f(x) = y\). Let \(y = x^n\) be an arbitrary real number. Now, we must find an \(x\) for which this equation holds true. To do so, we can solve for x: \(y = x^n \Rightarrow x = \sqrt[n]{y}\). Since n is odd, the nth root of y exists, and we always have only one real solution for \(x\) for any real number \(y\). Therefore, the function is surjective.

Step 3: Finding the Inverse Function Rule

Now that we have proven that f is both injective and surjective, we know that f has an inverse function, denoted as \(f^{-1}(x)\). We already solved for \(x\) in the surjective proof: \(x = \sqrt[n]{y}\). Thus, the rule for the inverse function is \(f^{-1}(y) = \sqrt[n]{y}\) or equivalently \(f^{-1}(x) = \sqrt[n]{x}\).

Key Concepts

Injectivity

When we talk about an injective function, we are describing a one-to-one relationship between input and output values. This means that each element of the domain is paired with a distinct and unique element of the range. To establish that a function is injective, we must demonstrate that if f(x₁) = f(x₂), it logically follows that x₁ = x₂.

In our exercise, the function f(x) = x^n, where n is an odd positive integer, is injective because raising distinct real numbers to an odd power yields distinct results. This characteristic is crucial because it is the first step in proving that the function has an inverse, a reverse operation that 'undoes' the effect of the function on its inputs. Without injectivity, an inverse cannot be properly defined, as there would be ambiguity in reversing the function's output back to a singular input.

Surjectivity

Conversely, a surjective function, or an 'onto' function, ensures that every element in the range is an output of the function for some input from the domain. In layman's terms, this means that the function 'covers' the entire range – there's no value left out that can't be reached by the function. To prove surjectivity, for any given y in the range, there must be an x in the domain such that f(x) = y.

For the power function in our exercise, every real number can be expressed as an odd power of another real number due to the existence of roots for all real numbers when the exponent is odd. Thus, with no value of y left without a corresponding x, the function meets the criteria for being surjective. Proving surjectivity is essential because it reassures us that an inverse function can map back to every possible input.

Odd Positive Integer

An odd positive integer is a whole number greater than zero that cannot be evenly divided by two. In the context of our function f(x) = x^n, the 'n' signifies the exponent, and the function's behavior is decisively influenced by whether 'n' is odd or even. For odd exponents, the function's outputs preserve the sign of the input — meaning negative inputs yield negative outputs and positive inputs yield positive outputs.

This property is central to the proof of injectivity and surjectivity for the function. Moreover, odd exponents guarantee that every output has exactly one corresponding input. This underpins the concept that for each result of an odd-powered function, there's only one specific root from which it came, which directly ties into the next critical concept, the 'nth root'.

nth Root

The term nth root specifically pertains to the inverse operation of raising a number to the power of 'n'. When n is an odd positive integer, each real number has precisely one real nth root. For instance, the cube root (√[3]{x}) of a number gives the value that, when cubed, results in 'x'. This single unique root exists for every real number when dealing with odd roots.

Hence, the inverse function for our exercise is defined using the concept of the nth root to 'undo' the original powering function. The proof of surjectivity earlier leads to the definition of the inverse function as f^-1(x) = √[n]{x}. This mathematical representation shows that for our function, each output can be traced back to one particular input via the nth root, reinforcing the function's reversibility, which is essential for an inverse function to exist.