Problem 6 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

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

Step-by-Step Solution

Answer

Functions identified for each part: (a) \(\sqrt{1-\sqrt{x}}\), (b) \((1+2^x)^3\), (c) \(\log(2x^2+1)\), (d) \(\sqrt{x^3+1}\), (e) \((1-2^x)^3\), (f) \(\log_2(1+2^x)\).

Step 1: Identify Inner Function of Part a

For the function \( K(x) = \sqrt{1-\sqrt{x}} \), start by identifying the innermost function. The inner function here is \( H(x) = \sqrt{x} \).

Step 2: Identify Middle Function of Part a

Next, apply the function \(G(v)\) on \(H(x)\). Since we have \(1-\) before the square root in \(K(x)\), let the middle function be \(G(v) = 1 - v\), where \( v = \sqrt{x} \).

Step 3: Identify Outer Function of Part a

Lastly, apply the outer function \(F(u)\) which is initially applied to \(G(v)\). The outer function must be \(F(u) = \sqrt{u}\) to match \(K(x)\).

Step 4: Define Functions for Part a

Combine the identified functions: \(H(x) = \sqrt{x}\), \(G(v) = 1 - v\), and \(F(u) = \sqrt{u}\).

Step 5: Identify Inner Function of Part b

For the function \( K(x) = (1+2^x)^3 \), the innermost function is \( H(x) = 2^x \).

Step 6: Identify Middle Function of Part b

Determine the middle function as the next application: \( G(v) = 1 + v \), where \( v = 2^x\).

Step 7: Identify Outer Function of Part b

The outer function is applied next and raises something to the third power: \( F(u) = u^3 \).

Step 8: Define Functions for Part b

The functions are: \(H(x) = 2^x \), \(G(v) = 1 + v\), and \(F(u) = u^3\).

Step 9: Identify Inner Function of Part c

For \( K(x) = \log(2x^2+1) \), start with the innermost function, which is \(H(x) = x^2 \).

Step 10: Identify Middle Function of Part c

The middle function should be the multiplication by a factor and addition: \(G(v) = 2v + 1\).

Step 11: Identify Outer Function of Part c

The outer function is the logarithm: \(F(u) = \log(u) \).

Step 12: Define Functions for Part c

The functions are: \(H(x) = x^2\), \(G(v) = 2v + 1\), and \(F(u) = \log(u)\).

Step 13: Identify Functions for Part d

For \(K(x) = \sqrt{x^3+1}\), we can directly say \(H(x) = x \), \(G(v) = v^3 + 1\), and \(F(u) = \sqrt{u} \).

Step 14: Identify Inner Function of Part e

For \( K(x) = (1-2^x)^3 \), start with the innermost function: \( H(x) = 2^x \).

Step 15: Identify Middle Function of Part e

The middle function is the subtraction: \(G(v) = 1 - v\), where \(v = 2^x \).

Step 16: Identify Outer Function of Part e

The outer function applies the cube: \(F(u) = u^3\).

Step 17: Define Functions for Part e

The functions are: \(H(x) = 2^x\), \(G(v) = 1 - v\), \(F(u) = u^3\).

Step 18: Identify Inner Function of Part f

For \( K(x) = \log_2(1+2^x) \), the innermost function is \(H(x) = 2^x \).

Step 19: Identify Middle Function of Part f

The middle function should be addition: \(G(v) = 1 + v\).

Step 20: Identify Outer Function of Part f

The outer function utilizes the base-2 logarithm: \(F(u) = \log_2(u) \).

Step 21: Define Functions for Part f

The functions are: \(H(x) = 2^x\), \(G(v) = 1 + v\), and \(F(u) = \log_2(u)\).

Key Concepts

Function Composition

Function composition is like stacking functions one inside another, letting one function connect into another. When you compose functions, you take the output of one function and plug it into another function. Think of it like making a sandwich where each part of the sandwich represents a different function.
For instance, if you have functions \( f(x) \) and \( g(x) \), the composition \( f(g(x)) \) means you apply \( g(x) \) first, and then apply \( f(x) \) to the result of \( g(x) \). This is exactly what's happening in the exercises above when functions like \( F(u) \), \( G(v) \), and \( H(x) \) are combined to form \( K(x) \) as \( F(G(H(x))) \).
This concept is fundamental in calculus and analysis as it helps to break down complex operations into simpler, more manageable pieces. The output of one function becomes the input to the next, flowing smoothly like a sequence.

Inner Function

The inner function is like the core or starting point in the process of function composition. It is the function you begin with, whose output feeds into the next function, the middle function. In the language of sandwiches, it might be the first layer of bread.
For example, looking at function \( K(x) = \sqrt{1-\sqrt{x}} \) from the exercise, the innermost part is \( H(x) = \sqrt{x} \). This initial transformation sets the stage for subsequent operations.
The inner function often involves the most basic element of the process, sometimes a simple mathematical operation like exponentiation or taking a square root. In mathematics, choosing the right inner function is crucial as it directly affects how the rest of the composition builds up.

Outer Function

The outer function is like the final shell or the last wrapper in the sequence of transformations in function composition. It acts on the result of the middle function, encompassing the entire operation to arrive at the final expression for \( K(x) \).
Consider \( K(x) = (1+2^x)^3 \) from the examples. Here, the outermost function is \( F(u) = u^3 \), which means that after computing the intermediate result with the middle function, it is raised to the third power.
This ultimate layer is where the completed result emerges in the composite function. Choosing this function involves ensuring that, when applied, it modifies the previous result in a way that aligns perfectly with the overall desired outcome.

Function Decomposition

Function decomposition is the process of breaking down a complex function into simpler, understandable parts. It's like reverse-engineering a task to see each small step in detail. This helps better manage and analyze the function.
For instance, decomposing \( K(x) = \log(2x^2+1) \) involves identifying \( H(x) = x^2 \), \( G(v) = 2v + 1 \), and \( F(u) = \log(u) \). Here, each component serves a distinct purpose, contributing to the final function's outcome.
Decomposition works as a visualization tool, helping you see what transforms are applied in what order. This clarity is invaluable for solving mathematical problems and allows deeper insights into the function's behavior.