Question

Finding the inverse of a cubic polynomial is equivalent to solving a cubic equation. A special case that is simpler than the general case is the cubic \(y=f(x)=x^{3}+\) ax. Find the inverse of the following cubics using the substitution (known as Vieta's substitution) \(x=z-a /(3 z) .\) Be sure to determine where the function is one-to-one. $$f(x)=x^{3}+2 x$$

Short answer

Question: Determine the inverse of the cubic function \(f(x) = x^3 + 2x\) and state the domain where the function is one-to-one. Answer: The inverse of the given cubic function is given by the equation: $$z^3 + \frac{4}{3z^2} - \frac{8}{27z^3} = 2z^2 - 8x + 4.$$ The function is one-to-one for all x.

Step-by-step solution

Identify the function and its variables

We are given the function: $$f(x) = x^3 + 2x$$ We need to find its inverse function, \(f^{-1}(x)\). Here, \(a=2\).

Apply Vieta's substitution

We will use Vieta's substitution: \(x = z - \frac{a}{3z}\), where \(a=2\): $$x = z - \frac{2}{3z}$$

Express the original function in terms of z

Now, we replace \(x\) with the expression in terms of z, in the original function: $$y = \left(z - \frac{2}{3z}\right)^3 + 2\left(z - \frac{2}{3z}\right)$$

Simplify the expression

Expand and simplify the expression: \begin{align*} y &= \left(z^3 - \frac{6z^2}{3z} + \frac{12z}{9z^2} - \frac{8}{27z^3}\right) + 2\left(z - \frac{2}{3z}\right) \\ y &= z^3 - 2z^2 + \frac{4}{3z^2} - \frac{8}{27z^3} + 2z - \frac{4}{3z} \end{align*}

Combine like terms and reorganize the expression

Bring terms with z to one side and terms with y to the other side: \begin{align*} z^3 - 2z^2 - 8y + 4 + \frac{4}{3z^2} - \frac{8}{27z^3} &= 0 \\ z^3 - 2z^2 + \frac{4}{3z^2} - \frac{8}{27z^3} - 8y + 4 &= 0 \end{align*}

Make z the subject of the expression

We can rewrite the equation in terms of \(z\): $$z^3 + \frac{4}{3z^2} - \frac{8}{27z^3} = 2z^2 - 8y + 4$$ We have now expressed the inverse function in terms of z. To get the inverse function, \(f^{-1}(x)\), we can replace \(y\) with \(x\).

Determine the domain of the inverse function

Consider the derivative of the original function to determine where it is one-to-one: $$f'(x) = 3x^2 + 2$$ The derivative is never zero or undefined, hence the original function is always increasing. Therefore, the inverse function exists for all x.

Final result

The inverse of the given cubic function \(f(x) = x^3 + 2x\) is: $$z^3 + \frac{4}{3z^2} - \frac{8}{27z^3} = 2z^2 - 8x + 4$$ The function is one-to-one for all x.

Key concepts

Vieta's Substitution

Vieta's substitution is a clever mathematical technique used to solve cubic equations. It entails transforming the variable into a form that simplifies the equation. Specifically for a cubic of the form f(x) = x^3 + ax, the substitution is x = z - a/(3z).

By using this substitution, the seemingly complicated cubic function can be broken down into simpler terms involving z. This transformation is particularly helpful in finding inverse functions of cubic polynomials, making the formidable task of solving for x much more manageable. It's named after the French mathematician François Viète, who contributed significantly to the theory of equations.

Cubic Equation

A cubic equation is a polynomial equation of degree three, typically written in the form ax^3 + bx^2 + cx + d = 0, where a, b, c, and d are constants. Cubic equations have at least one real root, and they can have three real roots or one real root and a pair of complex roots.

Solving these equations can be achieved using various methods, such as factoring, graphing, and substitution techniques, including Vieta's substitution. It's essential to understand the nature of cubic equations, as they appear frequently in various mathematical and real-world problems.

One-to-One Function

A one-to-one function, also known as an injective function, has a unique output for every input. This means that no two different values of x produce the same value of y.

For a function to have an inverse, it must be a one-to-one function. To verify if a cubic function is one-to-one, one can calculate its derivative to see if it's always increasing or always decreasing, indicating that it's a one-to-one function. For the cubic function in our exercise, f'(x) = 3x^2 + 2 is never zero, which confirms that the function is always increasing and hence, one-to-one.

Inverse Function

An inverse function, denoted as f^{-1}(x), essentially reverses the role of the input and output of the original function f(x). Finding the inverse obligates the original function to be a one-to-one function, so every input has a unique output, and vice-versa.

To compute the inverse function, we solve the equation y = f(x) for x. Vieta's substitution is part of this process for cubic equations. Once we've expressed x in terms of y, we can then replace the y with x to get the final expression for f^{-1}(x). In the exercise given, the resultant equation after substitution gives us the inverse.

Derivative

The derivative of a function is a fundamental concept in calculus that represents the rate of change of the function with respect to its variable. For any continuous and differentiable function f(x), the derivative, denoted as f'(x), helps determine the function's increasing or decreasing behavior and points of extrema.

In the context of finding the inverse of a cubic function, the derivative is critical in confirming whether the function is one-to-one. As demonstrated in the exercise, once we know that the derivative of the cubic function is always positive, we are assured that the function is strictly increasing and, therefore, has an inverse for all x values.