Problem 8 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

For each part, find two pairs, \(F\) and \(G,\) so that \(F \circ G\) is \(H\). a \(H(x)=\sqrt{1-\sqrt{x}}\) b \(H(x)=\frac{1}{1-\sqrt{x}} \quad\) c \(H(x)=\left(1+x^{2}\right)^{3}\) \(\mathrm{d} \quad H(x)=\left(x^{\left(x^{2}\right)}\right)^{3}\) e \(H(x)=2^{\left(x^{2}\right)}\) \(\mathrm{f} \quad H(x)=\left(2^{x}\right) 2\)

Step-by-Step Solution

Answer

Part a: \(F(x)=\sqrt{1-x}\), \(G(x)=\sqrt{x}\); Part b: \(F(x)=\frac{1}{1-x}\), \(G(x)=\sqrt{x}\); Part c: \(F(x)=x^3\), \(G(x)=1+x^2\); Part d: \(F(x)=x^3\), \(G(x)=x^{(x^2)}\); Part e: \(F(x)=2^x\), \(G(x)=x^2\); Part f: \(F(x)=x \cdot 2\), \(G(x)=2^x\).

Step 1: Identify Inner and Outer Functions for Part a

Given: \(H(x) = \sqrt{1-\sqrt{x}}\). To write this as \(F(G(x))\), we consider an inner function \(G(x)\) and an outer function \(F(x)\). Let \(G(x) = \sqrt{x}\), then \(F(x) = \sqrt{1-x}\). Thus, \(F(G(x)) = \sqrt{1-\sqrt{x}} = H(x)\).

Step 2: Identify Inner and Outer Functions for Part b

Given: \(H(x) = \frac{1}{1-\sqrt{x}}\). Choose \(G(x) = \sqrt{x}\) and \(F(x) = \frac{1}{1-x}\). Consequently, \(F(G(x)) = \frac{1}{1-\sqrt{x}} = H(x)\).

Step 3: Identify Inner and Outer Functions for Part c

Given: \(H(x) = (1+x^2)^3\). Let's choose \(G(x) = 1+x^2\) and \(F(x) = x^3\). This gives \(F(G(x)) = (1+x^2)^3 = H(x)\).

Step 4: Identify Inner and Outer Functions for Part d

Given: \(H(x) = \left(x^{(x^2)}\right)^3\). Choose \(G(x) = x^{(x^2)}\) and \(F(x) = x^3\). Thus, \(F(G(x)) = \left(x^{(x^2)}\right)^3 = H(x)\).

Step 5: Identify Inner and Outer Functions for Part e

Given: \(H(x) = 2^{(x^2)}\). Choose \(G(x) = x^2\) and \(F(x) = 2^x\). Hence, \(F(G(x)) = 2^{(x^2)} = H(x)\).

Step 6: Identify Inner and Outer Functions for Part f

Given: \(H(x) = (2^x) \cdot 2\). Set \(G(x) = 2^x\) and \(F(x) = x \cdot 2\). Therefore, \(F(G(x)) = (2^x) \cdot 2 = H(x)\).

Key Concepts

Inner and Outer Functions

Understanding the concept of inner and outer functions is a core part of mastering function composition. Function composition is when you apply one function to the result of another. To solve a function composition problem, we need to identify both the inner and outer functions in the expression. For example, if given the composite function \(H(x) = \sqrt{1-\sqrt{x}}\), we determine:

• The inner function \(G(x)\) could be \(\sqrt{x}\).
• The outer function \(F(x)\) would be \(\sqrt{1-x}\).

The reason we call them "inner" and "outer" is due to how they are applied. The inner function \(G(x)\) is evaluated first, and then \(F(x)\) is applied to the result. As a result, the overall computation follows a clear order. By identifying these, students can better deconstruct complex expressions into manageable parts.

Mathematical Modeling

Mathematical modeling involves using mathematics to represent, analyze, and solve real-world problems. Function composition is a valuable tool in this process. When setting up a model, it is essential to identify the relationships between different variables. In the case of function composition, the relationship between variables \(x\) is mapped through \(G(x)\) before applying \(F(x)\).

• Consider \(H(x) = \frac{1}{1-\sqrt{x}}\): this could model scenarios where you need to process one input and use the result in another calculation.
• First, process \(G(x) = \sqrt{x}\) to prepare the input.
• Then, apply \(F(x) = \frac{1}{1-x}\) to the outcome of \(G(x)\) to obtain \(H(x)\).

Thus, function composition helps in breaking down problems into smaller steps, making complex scenarios easier to manage and solve. Such methods are instrumental in fields such as physics, engineering, and economics.

Calculus Problem Solving

Calculus problem solving often utilizes function composition to simplify or solve problems. By understanding function composition, we can efficiently handle complicated expressions or derivations. In calculus, composite functions are pivotal when performing operations such as differentiation or integration. Let's take the example \(H(x) = (1+x^2)^3\):

• The inner function is \(G(x) = 1+x^2\).
• The outer function is \(F(x) = x^3\).

By defining these, calculus problems like finding the derivative \(H'(x)\) become more structured. For example, apply the chain rule on \(H(x)\): differentiate \(F(x)\) as \(3(1+x^2)^2\), then multiply by the derivative of \(G(x)\), which is \(2x\). This simplification aids learners in tackling calculus tasks and understanding compound relationships within functions more deeply. As skills develop, handling multi-step problems becomes more intuitive.