Problem 41 - Calculus: One and Several Variables

Question

Find \(g\) such that \(f \circ g=F\) given that $$f(x)=x^{2}+1 , F(x)=\left(2 x^{3}-1\right)^{2}+1$$

Step-by-Step Solution

Answer

The function g(x) that makes \(f \circ g = F\) is \(g(x) = 2x^3 - 1\)

Step 1: Write F(x) in terms of f

To express F(x) in terms of f, we have to identify the expression that replaces x in the definition of f which is \(f(x) = x^2 + 1\) to form F(x). Looking at \( F(x) = (2x^3 - 1)^2 + 1 \), we see that \(f(x)\) is in there but instead of \(x\), there is \(2x^3 - 1\). Hence, we write it as \(F(x) = f(2x^3 - 1)\).

Step 2: Equate F(x) to f(g(x)) to solve for g(x)

Knowing that the composition of \(f\) and \(g\) should equal to \(F\), we can equate \(f(g(x))\) to \(F(x)\) to solve for \(g(x)\). As \(F(x) = f(2x^3 - 1)\) we can set them equal to each other to solve for \(g\). Hence, \(f(g(x)) = f(2x^3 - 1)\).

Step 3: Identify the function g(x)

From fugacity in step 2, we have \(f(g(x)) = f(2x^3 - 1)\). Because \(f\) is a function, then \(g(x)\) is equal to \(2x^3 - 1\). This is because for \(f(g(x))\) to be equal to \(f(2x^3 - 1)\) then \(g(x)\) must be the equivalent of \(2x^3 - 1\).

Key Concepts

Inverse Functions

Inverse functions are an essential concept in algebra that helps us understand function compositions and how to reverse the effect of functions. For a function to have an inverse, it must be bijective. This means it should be both one-to-one (each input has a unique output) and onto (every possible output is covered).

When dealing with function composition, the role of inverse functions becomes apparent. If you have a function \( f \) and its inverse \( f^{-1} \), composing them will bring you back to your original input:

• This is seen as \( f(f^{-1}(x)) = x \).
• Similarly, \( f^{-1}(f(x)) = x \).

In our problem, we were given that \( f(g(x)) = F(x) \), and we had to find \( g(x) \). Recognizing how inverse functions work helps us see that \( g(x) \, (\text{in some contexts}) \) could be viewed as a reverse operation related to \( f(x) \) that gets composed to give \( F(x) \). While \( g(x) \) isn't \( f^{-1}(x) \) in this problem, knowing about inverses helps with understanding the function pairing involved.

Polynomial Functions

Polynomial functions are one of the most studied types of functions in mathematics. They are made up of terms which are variables raised to positive integer powers, combined using addition, subtraction, and multiplication.

When considering polynomial functions such as \( f(x) = x^2 + 1 \) from the exercise, we recognize:

• The term \( x^2 \) is a polynomial of degree 2.
• It is quadratic since its highest exponent is 2.

In the exercise, we also work with another polynomial hidden inside \( F(x) = (2x^3 - 1)^2 + 1 \). According to polynomial function rules, the expression \( 2x^3 - 1 \) is of degree 3, often called cubic due to its highest variable power being 3.

These polynomial functions play a crucial role in composing new function expressions and transformations. Understanding the degree of a function helps predict its behavior, such as how it will expand or contract on a graph. Identifying polynomials was key to expressionally organizing \( F(x) \) in the given problem.

Algebraic Manipulation

Algebraic manipulation is all about rearranging expressions to simplify problems or express them in a desired form. This skill is critical for solving equations, particularly in the context of function composition.

In the problem, algebraic manipulation allowed us to express \( F(x) \) in terms of \( f \) by identifying the substitution inside the function:

• We recognized that \( F(x) = (2x^3 - 1)^2 + 1 \) can be written as \( f(2x^3 - 1) \).
• This led to finding that \( g(x) = 2x^3 - 1 \) through manipulation.

Such manipulation involves recognizing which terms or expressions can be replaced according to given functions. By substituting accurately, complex expressions simplify, making it possible to determine unknown pieces in the equation just like we did with finding \( g(x) \) in the exercise.

Developing a strong grasp of algebraic manipulation simplifies the path towards understanding more complex algebraic relationships and and solves equations efficiently.