Problem 14 - Algebra A Combined Function

Question

Given the one-to-one function \(f(x)=x^{3}+2,\) find the following. (Hint: You do not need to find the equation for \(f^{-1}\).) a. \(f(-2)\) b. \(f^{-1}(-6)\)

Step-by-Step Solution

Answer

a. \(-6\); b. \(-2\)

Step 1: Evaluate f(-2)

To find \(f(-2)\), substitute \(-2\) for \(x\) in the function \(f(x) = x^3 + 2\). This gives \(f(-2) = (-2)^3 + 2 = -8 + 2 = -6\). Therefore, \(f(-2) = -6\).

Step 2: Determine f^{-1}(-6)

Given that \(f(x) = x^3 + 2\), to find \(f^{-1}(-6)\), set \(f(x) = -6\) and solve for \(x\). So, \(x^3 + 2 = -6\). Simplify to \(x^3 = -6 - 2\), thus, \(x^3 = -8\). Solve for \(x\) by taking the cube root of \(-8\), which is \(-2\). Therefore, \(f^{-1}(-6) = -2\).

Key Concepts

One-to-One Functions

A one-to-one function is a special type of function where each output is associated with exactly one input. This characteristic is crucial because it ensures that the function has an inverse, meaning we can work our way back from output to a specific input. You can check if a function is one-to-one by the horizontal line test: if no horizontal line intersects the graph of the function more than once, then the function is one-to-one. This quality helps greatly when dealing with inverse functions, as it guarantees their existence. In the exercise, the function \( f(x) = x^3 + 2 \) is one-to-one because each output from the cubic expression plus 2 is unique.

Function Evaluation

Function evaluation involves substituting a given input into the function to find the corresponding output. For instance, in our exercise, \( f(-2) \) requires substituting \(-2\) into the function \( f(x) = x^3 + 2 \). Breaking it down, you would calculate the cube of \(-2\), which is \(-8\), and then add 2, resulting in \(-6\). Therefore, \( f(-2) \) is evaluated to be \(-6\). Using these straightforward operations and substitution steps, you quickly find the value of the function at a specific point.

Cubic Functions

Cubic functions are polynomial functions of degree three, and they take the form \( f(x) = ax^3 + bx^2 + cx + d \). These functions have a curve with at least one turning point and can potentially have one or all roots real or complex.
Cubic functions are known for their distinct, S-like curve.

• The permanent feature is that they extend to infinity both upwards and downwards.
• They can intersect the x-axis at most three times.

In the exercise, \( f(x) = x^3 + 2 \) is a cubic function without any squared or linear terms, and the graph is a simple curve that passes vertically through the y-intercept at 2.

Solving Equations

Solving equations involves finding the value(s) of the variable that make the equation true. In our exercise, to determine \( f^{-1}(-6) \), we set \( f(x) = -6 \). This means solving \( x^3 + 2 = -6 \). First, subtract 2 from both sides to isolate the cubic term, resulting in \( x^3 = -8 \). Then, find the cube root of \(-8\), which is \(-2\). By systematically manipulating the equation, you uncover that \( f^{-1}(-6) = -2 \).
This example highlights that solving equations often involves reversing operations, carefully undoing addition, subtraction, and rooting, among others, to find the solution.