Problem 13 - Calculus for Biology and Medicine

Question

Suppose that \(f(x)=x^{2}, x \in \mathbf{R}\) and \(g(x)=3+x, x \in \mathbf{R}\). (a) Show that \((f \circ g)(x)=(3+x)^{2}, x \in \mathbf{R}\). (b) Show that \((g \circ f)(x)=3+x^{2}, x \in \mathbf{R}\).

Step-by-Step Solution

Answer

(a) \((f \circ g)(x) = (3+x)^2\); (b) \((g \circ f)(x) = 3+x^2\).

Step 1: Understanding Composition

We need to compute two compositions of functions: \((f \circ g)(x)\), and \((g \circ f)(x)\). The symbol \(\circ\) represents function composition, meaning you substitute the result of one function \(g(x)\) into another function \(f(x)\), or vice versa.

Step 1: Compute \((f \circ g)(x)\)

To find \((f \circ g)(x)\), substitute \(g(x)\) into \(f(x)\). Given that \(f(x) = x^2\) and \(g(x) = 3 + x\), we have:\[(f \circ g)(x) = f(g(x)) = f(3 + x) = (3 + x)^2.\]This matches with what we needed to show for part (a).

Step 2: Expand \((3+x)^2\)

To verify the expression \((3 + x)^2\), we expand it using the formula \((a + b)^2 = a^2 + 2ab + b^2\):\[(3 + x)^2 = 3^2 + 2\cdot3\cdot x + x^2 = 9 + 6x + x^2.\]This confirms that \((3 + x)^2\) is indeed expanded correctly.

Step 3: Compute \((g \circ f)(x)\)

To find \((g \circ f)(x)\), substitute \(f(x)\) into \(g(x)\). Using \(f(x) = x^2\) and \(g(x) = 3 + x\), we have:\[(g \circ f)(x) = g(f(x)) = g(x^2) = 3 + x^2.\]This expression matches the requirement for part (b).

Key Concepts

Composite Function

When dealing with functions, composition is the process of combining two functions to create a new function. Essentially, a composite function is created when the output of one function becomes the input for another. Let's break it down:

1. **Notation**: The composition of the function \( f \) with the function \( g \) is written as \( (f \circ g)(x) \). It translates to "\( f \) of \( g \) of \( x \)."2. **How It Works**: To compute \( (f \circ g)(x) \), we start by evaluating \( g(x) \). The result is then plugged into the function \( f(x) \).

Suppose we have two functions: \( f(x) = x^2 \) and \( g(x) = 3 + x \). To find \( (f \circ g)(x) \):- First, calculate \( g(x) = 3 + x \).- Next, substitute into \( f(x) = x^2 \):- Hence, \( (f \circ g)(x) = (3 + x)^2 \).In essence, the composite function combines these two steps into one smooth operation. It's akin to a two-step process compressed into a single function!

Composite functions are useful because they allow us to build more complex functions from simpler ones. This concept is a powerful tool in calculus and algebra, simplifying the expression of multi-step problems.

Function Substitution

Function substitution is a technique where we take the entire output of one function and substitute it into another function. This is essential for finding composite functions.

For example, if you have \( f(x) = x^2 \) and \( g(x) = 3 + x \), and you need to find \( (g \circ f)(x) \):- You first consider \( f(x) \) which gives you \( x^2 \).- Then, take this output \( x^2 \) and plug it into \( g \) to replace the \( x \) in \( g(x) = 3 + x \).- Consequently, \( (g \circ f)(x) = g(x^2) = 3 + x^2 \).This method can be visualized as one function feeding into the other, with the output of the first becoming the input for the second.

Substituting one function into another allows mathematicians and students to transform problems into more manageable forms. It simplifies calculations and helps in solving equations where functions overlap. This approach ensures that operations on functions are systematic and consistent.

Expansion of Binomials

Expanding binomials involves expanding expressions raised to a power and is often necessary when dealing with compositions of functions in algebra.When you have an expression like \((3 + x)^2\), you apply the binomial expansion formula, which is:\[(a + b)^2 = a^2 + 2ab + b^2\]For our case with the binomial \(3 + x\):- Here, \(a = 3\) and \(b = x\).- Plug these into the expansion formula:\[(3 + x)^2 = 3^2 + 2 \cdot 3 \cdot x + x^2 = 9 + 6x + x^2\].

This expanded form is essential for further simplifications or solving equations. Binomial expansion comes into play not only in composite functions but across many areas of mathematics where polynomial expressions are prevalent.

Mastering the technique of expanding binomials broadens the scope of analyzing and transforming algebraic expressions efficiently, making problem-solving simpler and more intuitive.