Problem 5 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Find functions, \(F(z)\) and \(G(x)\) so that the following functions, \(H,\) may be written as \(F(G(x))\). $$ \begin{array}{lll} \text { a. } H(x) & =\left(1+x^{2}\right)^{3} & \text { b. } H(x)=10^{\sqrt{x}} & \text { c. } H(x)=\log \left(2 x^{2}+1\right) \\ \text { d. } H(x) & =\sqrt{x^{3}+1} & \text { e. } H(x)=\frac{1-x^{2}}{1+x^{2}} & \text { f. } H(x)=\log _{2}\left(2^{x}\right) \end{array} $$

Step-by-Step Solution

Answer

a) \(F(z) = z^3\), \(G(x) = 1 + x^2\); b) \(F(z) = 10^z\), \(G(x) = \sqrt{x}\); c) \(F(z) = \log(z)\), \(G(x) = 2x^2 + 1\); d) \(F(z) = \sqrt{z}\), \(G(x) = x^3 + 1\); e) \(F(z) = \frac{z_1}{z_2}\), \(G(x) = (1-x^2, 1+x^2)\); f) \(F(z) = \log_2(z)\), \(G(x) = 2^x\).

Step 1: Identify Composition

To express each function as a composition of two functions, we need to think of a natural way to decompose each given function into two separate functions such that one function can be applied to the result of the other.

Step 1: Decompose Function (a)

For the function \( H(x) = (1 + x^2)^3 \), we can set \( G(x) = 1 + x^2 \) and \( F(z) = z^3 \). Thus, \( H(x) = F(G(x)) = (G(x))^3 = (1 + x^2)^3 \).

Step 2: Decompose Function (b)

For \( H(x) = 10^{\sqrt{x}} \), we choose \( G(x) = \sqrt{x} \) and \( F(z) = 10^z \). Then, \( H(x) = F(G(x)) = 10^{G(x)} = 10^{\sqrt{x}} \).

Step 3: Decompose Function (c)

For \( H(x) = \log(2x^2 + 1) \), set \( G(x) = 2x^2 + 1 \) and \( F(z) = \log(z) \). Thus, \( H(x) = F(G(x)) = \log(G(x)) = \log(2x^2 + 1) \).

Step 4: Decompose Function (d)

The function \( H(x) = \sqrt{x^3 + 1} \) can be decomposed with \( G(x) = x^3 + 1 \) and \( F(z) = \sqrt{z} \). So, \( H(x) = F(G(x)) = \sqrt{G(x)} = \sqrt{x^3 + 1} \).

Step 5: Decompose Function (e)

For \( H(x) = \frac{1-x^2}{1+x^2} \), take \( G(x) = (1-x^2, 1+x^2) \) and \( F(z) = \frac{z_1}{z_2} \). Here, \( H(x) = F(G(x)) = \frac{1-x^2}{1+x^2} \).

Step 6: Decompose Function (f)

For the function \( H(x) = \log_2(2^x) \), set \( G(x) = 2^x \) and \( F(z) = \log_2(z) \). Therefore, \( H(x) = F(G(x)) = \log_2(G(x)) = \log_2(2^x) \).

Key Concepts

composition of functions

In calculus, the composition of functions refers to the process of combining two functions where the output of one function becomes the input of another. This method is crucial in math because it allows us to create new functions from existing ones.
Consider function composition as putting one machine inside another: the result of the first machine (or function) becomes what you insert into the second. For example, given functions \( F(z) \) and \( G(x) \), the composition \( F(G(x)) \) involves taking \( G(x) \) and plugging it into \( F \).
Let's see some examples to illustrate this concept:

• For the function \( H(x) = (1 + x^2)^3 \), consider \( G(x) = 1 + x^2 \) and \( F(z) = z^3 \). Here, \( H(x) \) becomes \( F(G(x)) = (G(x))^3 \).
• With \( H(x) = 10^{\sqrt{x}} \), we identify \( G(x) = \sqrt{x} \) and \( F(z) = 10^z \), resulting in \( H(x) = 10^{G(x)} \).

This strategy simplifies the manipulation of complex functions and makes solving calculus problems more manageable.

precalculus concepts

Precalculus is foundational for understanding calculus. It includes concepts like functions, algebra, and trigonometry. These basics help in grasping advanced topics like differentiation and integration.
Understanding the form and behavior of different types of functions is essential in precalculus. For example, polynomial functions, exponential functions, and logarithmic functions all have distinct forms and characteristics.

• Polynomial functions, like \( H(x) = (1 + x^2)^3 \), involve variables raised to whole number powers.
• Exponential functions, such as \( H(x) = 10^{\sqrt{x}} \), have a constant as the base raised to a variable power.
• Logarithmic functions reverse the operation of exponentiation, seen in functions like \( H(x) = \log(2x^2 + 1) \).

Learning how to decompose and compose these functions helps in reading calculus problems better and understanding their solutions.

mathematical modeling

Mathematical modeling involves using mathematical language and techniques to represent real-world systems. It helps us predict and understand complex phenomena by expressing them through equations and functions.
Function decomposition, as seen in the exercise, is a technique often utilized in modeling. By breaking down a complicated part, such as a system or process, into simpler functions, we can analyze each part more easily.
For example, when considering a biological ecosystem's growth, an exponential function might represent the population increase, while a logarithmic function could indicate resource limitations.

• In the case of \( H(x) = \frac{1-x^2}{1+x^2} \), function decomposition allows analyzing differences and ratios separately.
• When using \( H(x) = \log_2(2^x) \), exploring the inverse relationship helps in monitoring changes over time.

By mastering these concepts, mathematical modeling becomes an effective tool for solving real-life problems in fields like physics, economics, and engineering.