Problem 19 - Calculus for Biology and Medicine

Question

Suppose that \(f(x)=x^{2}, x \geq 0\), and \(g(x)=\sqrt{x}, x \geq 0\). Typically, \(f \circ g \neq g \circ f\), but this is an example in which the order of composition does not matter. Show that \(f \circ g=g \circ f\).

Step-by-Step Solution

Answer

\(f \circ g = g \circ f \) because both result in \( x \).

Step 1: Define the Function Composition

The composition of two functions is represented as \[ f \circ g (x) = f(g(x)) \quad \text{and} \quad g \circ f (x) = g(f(x)) \]We need to compute both of these separately for the given functions to check whether they are equal.

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

Let's substitute the function \( g(x) = \sqrt{x} \) into \( f(x) = x^2 \):\[f \circ g (x) = f(g(x)) = f(\sqrt{x}) = (\sqrt{x})^2 = x.\]

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

Now, let's substitute the function \( f(x) = x^2 \) into \( g(x) = \sqrt{x} \):\[g \circ f (x) = g(f(x)) = g(x^2) = \sqrt{x^2} = x,\]with the understanding that \( x \geq 0 \) ensures \( \sqrt{x^2} = x \).

Step 4: Compare \( f \circ g (x) \) and \( g \circ f (x) \)

From Steps 2 and 3, we find that:\[f \circ g (x) = x \quad \text{and} \quad g \circ f (x) = x.\]Since both compositions result in the same expression \( x \), it follows that:\[f \circ g (x) = g \circ f (x).\]

Key Concepts

Inverse Functions

Inverse functions are like the opposite of regular functions. They help us get back to where we started. If you think of a function as a machine that takes an input and gives an output, an inverse function is like a reverse machine. It takes the output and sends us back to the input. When two functions are inverses of each other, their compositions will return the original input:

• If you have a function \( f(x) \) and its inverse \( f^{-1}(x) \), then \( f(f^{-1}(x)) = x \).
• Similarly, \( f^{-1}(f(x)) = x \).

In the exercise example, although \( f(x)=x^2 \) and \( g(x)=\sqrt{x} \) are not exactly inverse functions due to their domain restrictions, they behave similarly within a restricted domain where \( x \geq 0 \). This restricted domain allows both \((f \circ g)(x)\) and \((g \circ f)(x)\) to simplify to the identity function \( x \), showcasing the essence of inverse-like behavior.

Domain Restrictions

When we talk about domain restrictions, we're focusing on which inputs a function can accept. Not all functions work for all real numbers, and that's totally fine. Some functions need to avoid certain values to ensure they can deliver real, usable results. This is particularly true for operations like division by zero or taking the square root of a negative number. In the context of our problem, both \( f(x) = x^2 \) and \( g(x) = \sqrt{x} \) have specific domain restrictions:

• \( f(x) = x^2 \) is restricted to \( x \geq 0 \) because we are only considering non-negative values to avoid introducing complex numbers when composed with the square root function.
• \( g(x) = \sqrt{x} \) naturally requires \( x \geq 0 \) since square roots of negative numbers are not real.

These domain restrictions ensure the functions remain real and are fundamental to ensure the composition \((f \circ g)(x) = (g \circ f)(x) = x\) holds true.

Square and Square Root Operations

The operations of squaring and taking square roots are closely related. Squaring a number means multiplying it by itself, while taking the square root is finding a number which, when squared, will give the original number back. They're almost like opposite operations, but there's a twist!When you square a positive number or zero, you always get a non-negative result. The square root function essentially reverses this process with one condition:

• It only accepts non-negative numbers as input, precisely because square roots of negative numbers would lead us into the world of imaginary numbers.

In our case: - When you square \( \sqrt{x} \), you simply get \( x \) back, which is demonstrated as \( (f \circ g)(x) = x \).- Similarly, if you take the square root of \( x^2 \), assuming \( x \geq 0 \), you also end up with \( x \), shown in \( (g \circ f)(x) = x \).This pairing of squaring and square rooting works smoothly only under domain restrictions stated, like the ones used in the exercise, ensuring the operations complement each other perfectly.