Problem 81 - Algebra and Trigonometry

Question

In the margin notes in this section we pointed out that the inverse of a function can be found by simply reversing the operations that make up the function. For instance, in Example 6 we saw that the inverse of $$f(x)=3 x-2 \quad \text { is } \quad f^{-1}(x)=\frac{x+2}{3}$$ because the "reverse" of "multiply by 3 and subtract 2 " is "add 2 and divide by 3 " Use the same procedure to find the inverse of the following functions. (a) \(f(x)=\frac{2 x+1}{5} \quad\) (b) \(f(x)=3-\frac{1}{x}\) (c) \(f(x)=\sqrt{x^{3}+2} \quad\) (d) \(f(x)=(2 x-5)^{3}\) Now consider another function: $$f(x)=x^{3}+2 x+6$$ Is it possible to use the same sort of simple reversal of operations to find the inverse of this function? If so, do it. If not, explain what is different about this function that makes this task difficult.

Step-by-Step Solution

Answer

It is not possible to easily reverse the function \(f(x)=x^3 + 2x + 6\) into an explicit inverse function.

Step 1: Inverse of (a)

Given the function \(f(x) = \frac{2x + 1}{5}\), we start by reversing the operations. First, multiply both sides by 5 to clear the fraction: \(y = \frac{2x + 1}{5} \to 5y = 2x + 1\). Then, subtract 1 from both sides: \(5y - 1 = 2x\). Finally, divide by 2 to isolate \(x\): \(x = \frac{5y - 1}{2}\). Therefore, the inverse function is \(f^{-1}(x) = \frac{5x - 1}{2}\).

Step 2: Inverse of (b)

For \(f(x) = 3 - \frac{1}{x}\), reverse the operations. Start by moving \(-\frac{1}{x}\) to the other side: \(y = 3 - \frac{1}{x} \to 3 - y = \frac{1}{x}\). Next, take the reciprocal to solve for \(x\): \(x = \frac{1}{3 - y}\). Thus, the inverse function is \(f^{-1}(x) = \frac{1}{3 - x}\).

Step 3: Inverse of (c)

Given \(f(x) = \sqrt{x^3 + 2}\), first isolate the square root by squaring both sides: \((y)^2 = x^3 + 2\). Next, subtract 2 from both sides to isolate the cube: \(y^2 - 2 = x^3\). Finally, take the cube root: \(x = \sqrt[3]{y^2 - 2}\). The inverse function is \(f^{-1}(x) = \sqrt[3]{x^2 - 2}\).

Step 4: Inverse of (d)

For \(f(x) = (2x - 5)^3\), take the cube root of both sides: \(\sqrt[3]{y} = 2x - 5\). Add 5 to both sides: \(\sqrt[3]{y} + 5 = 2x\). Divide by 2 to find \(x\): \(x = \frac{\sqrt[3]{y} + 5}{2}\). The inverse function is \(f^{-1}(x) = \frac{\sqrt[3]{x} + 5}{2}\).

Step 5: Analyzing the inverse of the new function

For \(f(x) = x^3 + 2x + 6\), notice that the function involves a polynomial of degree 3 plus a linear term. To find an inverse by reversing operations, the function needs to be easily invertible at each step, like simple operations (add, subtract, multiply, divide). Here, the addition of terms \(x^3\) and \(2x\) complicates reversing the process explicitly due to the mixture of nonlinear terms, which cannot be easily separated or individually reversed in simple algebraic steps.

Key Concepts

Function operations

Understanding function operations is crucial when working with inverse functions. These operations involve such actions as addition, subtraction, multiplication, and division within the context of a function. To find the inverse of a function, you effectively "undo" these operations in reverse order.
For example, if a function involves multiplying by a constant and then adding another constant, finding its inverse involves reversing these steps.

• First, undo the addition by subtracting the constant on both sides.
• Then, undo the multiplication by dividing by the constant on both sides.

This systematic approach ensures that we arrive at the original input value of the function, flipping the process performed initially.
Each operation has its opposite: addition versus subtraction and multiplication versus division. Keeping this in mind helps when solving for a function's inverse.

Polynomial functions

Polynomial functions are a staple in algebra, characterized by expressions with variables raised to whole number powers. A polynomial might be simple like \(x^2\) or more complex like \(x^3 + 2x + 6\).
Polynomials generally have one or more terms, and each of these includes a variable raised to a power and a coefficient (which is a constant). Understanding the degree of a polynomial, which is the highest power of the variable present, is essential as it dictates the function's behavior.

• The higher the degree, the more complex the function.
• An important property is that linear polynomials (degree 1) are usually straightforward to invert.
• However, as the degree increases, finding the inverse can become more complicated.

When dealing with inverse functions, polynomials with mixed terms such as \(x^3 + 2x\) add complexity, often making it difficult to isolate and reverse using simple algebra.
This complexity arises because polynomial functions of higher degrees can have intricate graphs, lacking the one-to-one correspondence necessary for inversion.

Algebraic manipulation

Algebraic manipulation involves the skillful rearrangement of equations to solve for a desired variable. It's akin to playing a strategy game, where the goal is to isolate the variable of interest by performing permissible mathematical operations.
Mastering algebraic manipulation is essential for solving inverse functions, as it allows you to derive the formula for the inverse from the original function. The process often includes:

• Isolating terms systematically, usually by first reversing any addition or subtraction involved.
• Next, tackling any multiplication or division.
• Finally, dealing with higher-level operations like powers or roots if present.

This manipulation helps transform an equation into a form that reveals the inverse.
As tasks get more complex, involving nonlinear terms or powers, the algebraic manipulation might require creative problem-solving and a deep understanding of algebraic principles.