Problem 14 - Calculus for Biology and Medicine

Question

Suppose that $$ f(x)=x^{3}, \quad x \in \mathbf{R} $$ and $$ g(x)=1-x, \quad x \in \mathbf{R} $$ (a) Show that $$ (f \circ g)(x)=(1-x)^{3}, \quad x \in \mathbf{R} $$ (b) Show that $$ (g \circ f)(x)=1-x^{3}, \quad x \in \mathbf{R} $$

Step-by-Step Solution

Answer

(f \circ g)(x)=(1-x)^3 and (g \circ f)(x)=1-x^3.

Step 1: Understanding Function Composition

Function composition means applying one function to the result of another. For example, if we have two functions \( f \) and \( g \), the composition \((f \circ g)(x)\) means that we apply \(g\) first to \(x\), then apply \(f\) to the result of \(g(x)\). Similarly, \((g \circ f)(x)\) means applying \(f\) first, then \(g\).

Step 2: Compute \((f \circ g)(x)\)

To find \((f \circ g)(x)\), first compute \(g(x)\), which is \(1-x\). Then substitute \(1-x\) into \(f(x)\):\[(f \circ g)(x) = f(g(x)) = f(1-x) = (1-x)^3\]

Step 3: Verify \((f \circ g)(x) = (1-x)^3\)

From Step 2, we see that \((f \circ g)(x) = (1-x)^3\). As \(f(x) = x^3\), replacing \(x\) with \(1-x\) results in \((1-x)^3\). This matches the expression given in part (a).

Step 4: Compute \((g \circ f)(x)\)

To find \((g \circ f)(x)\), first compute \(f(x)\), which is \(x^3\). Then substitute \(x^3\) into \(g(x)\):\[(g \circ f)(x) = g(f(x)) = g(x^3) = 1-x^3\]

Step 5: Verify \((g \circ f)(x) = 1-x^3\)

From Step 4, \((g \circ f)(x) = 1 - x^3\). This matches the expression given in part (b), confirming our solution is correct.

Key Concepts

Understanding Mathematical Functions

Mathematical functions are fundamental tools in mathematics that map every input to a single output. Think of a function as a machine: you give it an input, and it spits out an output. Functions are widely used across various fields of mathematics to describe real-world phenomena and abstract concepts. Every function has:

• A domain: the set of possible input values.
• A range: the set of possible outputs.For instance, in our example, both functions \(f(x) = x^3\) and \(g(x) = 1 - x\) have domains and ranges within the real numbers \(\mathbf{R}\).

Functions are expressed in the form \(f(x)\), where \(x\) is the input variable. These mathematical entities aren't just limited to numbers; they can take more complex forms like polynomials or even other functions as inputs. This versatility makes functions critical in modeling a variety of situations.

Exploring Polynomial Functions

Polynomial functions are a specific type of function that has variables raised to whole number exponents. The general form of a polynomial function is:\[ a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \]where \(a_n, a_{n-1}, \ldots, a_0\) are constants, and \(n\) is a non-negative integer representing the highest degree of the polynomial.
In our exercise:

• Function \(f(x) = x^3\) is a cubic polynomial with a degree of 3, meaning it's of the form \(x^n\), where \(n = 3\).
• This function grows rapidly as \(x\) moves away from zero, either in the positive or negative direction.

Polynomial functions have continuous, smooth curves, which makes them incredibly useful for approximation and interpolation tasks. The coefficients in the polynomial dictate the shape and direction of the curve.

Understanding Function Operations with Compositions

Function operations include several ways of combining functions to create new ones. Function composition is one of the most popular methods. It involves applying one function to the result of another function. In this exercise, we have two compositions:

• For \((f \circ g)(x)\), we first apply \(g(x) = 1 - x\) to an input \(x\), and then apply \(f(x) = x^3\) to the result \(1 - x\).
• This results in \((1-x)^3\), demonstrating how functions are nested through composition.
• For \((g \circ f)(x)\), \(f(x) = x^3\) is applied first, followed by \(g\), leading to \(1 - x^3\).

Compositions like these play a vital role in complex system modeling where multiple processes need to be handled in a specific sequence. Understanding how to manipulate and combine functions through operations enriches our capacity to tackle diverse mathematical problems.