Problem 43 - Contemporary Precalculus

Question

Write the given function as the composite of two functions, neither of which is the identity function, as in Examples 6 and 7 . (There may be more than one way to do this.) $$ \begin{array}{l} \text { If } f(x)=x+1 \text { and } g(t)=t^{2} \text { , then } \\ \qquad \begin{aligned} (g \circ f)(x) &=g(f(x))=g(x+1)=(x+1)^{2} \\\ &=x^{2}+2 x+1 \end{aligned} \end{array} $$ Find two other functions \(h(x)\) and \(k(t)\) such that \((k \circ h)(x)=x^{2}+2 x+1\)

Step-by-Step Solution

Answer

Question: Find two new functions, h(x) and k(t), such that their composite function is equal to the given function x^2 + 2x + 1 and neither of them are the identity function. Solution: The functions h(x) = x+1 and k(t) = t^2 satisfy the requirements. Their composite function is (k ⚬ h)(x) = k(x+1) = (x+1)^2 = x^2 + 2x + 1.

Step 1: Choose a function for h(x)

We can choose an initial function for \(h(x)\) such that it involves some form of \(x\). Let's choose a basic linear function for simplicity. In this case, we choose \(h(x) = x+1\).

Step 2: Determine the function for k(t)

Now, we need to find a function \(k(t)\) such that \((k \circ h)(x)=k(h(x))=x^{2}+2x+1\). We have already chosen our \(h(x)\) as \(x+1\), so let's write the composite function in terms of \(t\) and \(h(x)\): $$k(h(x))=k(x+1)=x^{2}+2x+1$$ Now, we need to find an expression for \(k(t)\) that satisfies the above equation. We can choose \(k(t)=t^{2}\) as our function, because when we substitute \(t=x+1\) we get, $$k(x+1)=(x+1)^2=x^2+2x+1$$

Step 3: Confirm the functions satisfy the requirements

Now let's make sure the functions \(h(x)=x+1\) and \(k(t)=t^2\) satisfy the given requirements: 1. Neither function is the identity function: This requirement is satisfied as neither \(h(x)\) nor \(k(t)\) are \(I(x) = x\). 2. The composite function should be equal to the given function: As shown in Step 2, when we substituted \(t=x+1\) in our function \(k(t)\), we got the given function \(x^2+2x+1\). Thus, the requirement is satisfied. So, the functions we found are \(h(x) = x+1\) and \(k(t) = t^2\). Their composite function is \((k \circ h)(x) = k(x+1) = (x+1)^2 = x^2 + 2x + 1\).

Key Concepts

Function Composition

Understanding function composition is one of the key concepts in precalculus and mathematics in general. It involves combining two functions to create a new one, where the output of one function becomes the input of another. In essence, if we have two functions, say h(x) and k(t), to compose them we would calculate k(h(x)).

Imagine function composition as a two-step process in a factory. The first machine (function h) takes raw materials (the original input x) and transforms them to an intermediate product. The second machine (function k) takes this intermediate product and further processes it to create the final product, which is the result of the composite function. In symbol terms, the composite function (k \( circ \) h)(x) represents the entire manufacturing process from start to finish.

To visualize this, consider our exercise problem where we found functions h(x) = x+1 and k(t) = t^2. The composition (k \( circ \) h)(x) means we first apply h(x), so x becomes x+1, and then apply k(t) to (x+1), which results in the final expression (x+1)^2. This practical method ensures that students can piece together the workings of composite functions step by step.

Precalculus

Precalculus is a course that helps bridge the gap between algebra and calculus, providing students with a firmer grasp of functions, complex numbers, matrices, and other pre-calculation concepts. Function composition, as discussed, is a topic within precalculus that has far-reaching applications in calculus, since understanding how to combine functions is fundamental to understanding how more complex mathematical models behave.

For students, mastering precalculus is like gathering all the essential tools you'll need for a calculus journey. It includes not just function composition, but also other key tools like polynomial functions, which are functions made up of terms that are powers of the variable, usually with whole number exponents. An example from our exercise was the quadratic function x^2+2x+1, which is a part of a larger family of polynomial functions. Ensuring a solid understanding of these precalculus concepts is vital for success in higher-level math courses, where exercises become more complex and abstract.

Polynomial Functions

Polynomial functions are the bread and butter of algebra and precalculus. They consist of coefficients and variables raised to whole number exponents. In the context of our exercise, the function f(x) = x^2 + 2x + 1 is a polynomial function, specifically a quadratic function, because the highest power of x is 2.

These functions are smooth, continuous, and take on a wide variety of shapes based on their degree (the highest power of the variable) and their coefficients. A primary reason why polynomial functions are so important is that they are easy to work with and can approximate a vast array of other functions.\(h(x)=x+1\) is also polynomial, albeit a linear one, since it can be written in the form mx + b where m and b are constants. In precalculus, students often manipulate these functions through various operations, including function composition, to understand their properties and how they can model real-world situations.

In essence, familiarity with polynomial functions allows students not only to tackle calculus problems more effectively but also to apply their knowledge to fields like physics, engineering, and economics, where modeling changes over time is crucial.