Problem 18 - Calculus for Biology and Medicine

Question

Suppose that \(f(x)=x^{4}, x \geq 3\), and \(g(x)=\sqrt{x+1}, x \geq 3\). Find \((f \circ g)(x)\) together with its domain.

Step-by-Step Solution

Answer

The composite function is \((f \circ g)(x) = (x + 1)^2\) and its domain is \(x \geq 3\).

Step 1: Understanding Composite Functions

Composite functions involve applying one function to the results of another. The notation \((f \circ g)(x)\) means that you first apply \(g(x)\) and then apply \(f\) to the result.

Step 2: Apply the Inner Function

Start by applying the inner function \(g(x)\) to an input \(x\). For this problem, \(g(x) = \sqrt{x + 1}\). So, for any \(x\), \(g(x) = \sqrt{x + 1}\).

Step 3: Apply the Outer Function

Next, apply the outer function \(f(x)\) to the result of the inner function. Here, \(f(x) = x^4\). So, \(f(g(x)) = f(\sqrt{x + 1}) = (\sqrt{x + 1})^4\).

Step 4: Simplify the Expression

Simplify \((\sqrt{x + 1})^4\). This expression can be rewritten as \((x + 1)^{2}\) because \( (\sqrt{x + 1})^4 = (\sqrt{x + 1})^2 \times (\sqrt{x + 1})^2 = (x + 1)^2\).

Step 5: Determining the Domain of the Composite Function

The domain of \((f \circ g)(x)\) is the set of all values of \(x\) where \(g(x)\) is defined and \(f(g(x))\) is also defined. Both conditions must be satisfied. Given \(g(x) = \sqrt{x + 1}\), \(x + 1\) must be non-negative for \(g(x)\) to be real. Thus, \(x \geq -1\). However, from the problem statement, we are told \(x \geq 3\). The domain, therefore, is \(x \geq 3\).

Key Concepts

Domain of Functions

When we talk about the "domain" of a function, we're discussing all the possible input values (x-values) that will work in a function. The domain ensures that the output is a real number under the function's rules. With sets of rules defined by functions, understanding their domain helps avoid mathematical mishaps.

For the two functions given in our exercise, which are:

• \(f(x) = x^4\) and
• \(g(x) = \sqrt{x + 1}\) where both \(x \geq 3\).

Each has specific requirements for what x-values can go into them. For \(g(x)\), the expression inside the square root, \(x + 1\), must be non-negative. Thus, \(x\) must be at least -1. But our problem already requires \(x \geq 3\). For composite functions, both inner and outer functions' domains must be considered.

Since \(g(x)\) requires \(x\geq -1\) and the extra condition from the problem is \(x\geq 3\), the working domain effectively starts at \(x = 3\) to comply with all the necessary conditions.

Function Composition

Function composition is the process where one function is applied to the result of another. It's like a sequence of operations. For our exercise, we had two functions:

• Inner Function: \(g(x) = \sqrt{x+1}\)
• Outer Function: \(f(x) = x^4\)

In the notation \((f \circ g)(x)\), \(g(x)\) is calculated first, and then its output is used as the input for \(f(x)\).

Let's go through it with our specific functions:1. **Apply \(g(x)\):** Take your \(x\) value, modify it as per the rule of \(g\), so \(g(x) = \sqrt{x+1}\).2. **Apply \(f(x)\):** Now, put the result of \(g(x)\) into \(f(x)\), meaning \((f \circ g)(x) = f(g(x)) = (\sqrt{x+1})^4\).

Through function composition, new functions are built, but it’s critical to ensure all function domains intertwine correctly, allowing valid sequences throughout.

Simplifying Expressions

Simplifying expressions is all about making them easier to work with or understand while maintaining their equivalence. In our exercise, after building the expression \((f \circ g)(x) = (\sqrt{x + 1})^4\), there was an opportunity to simplify.

To simplify, unravel \((\sqrt{x+1})^4\):

• First resolve \((\sqrt{x+1})^2\), yielding \((x+1)\).
• Now multiply: \((x+1) \times (x+1) = (x+1)^2\).

The simplification to \((x+1)^2\) makes handling calculations easier, especially for manually working out values.

Remember, while simplifying, the expression should remain equivalent or express the same numerical value as the original. This process should reveal clearer or more manageable operations without altering what the function signifies.