Question

If \(s(x)=x^{2}+1\) and \(t(x)=x-3,\) does \(s(t(x))=t(s(x))\) for all values of \(x ?\) Explain.

Short answer

No, \(s(t(x)) eq t(s(x))\) for all values of \(x\).

Step-by-step solution

- Understanding the Functions

The given functions are: - \(s(x)=x^2+1\)- \(t(x)=x-3\)We need to check whether \(s(t(x)) = t(s(x))\) for all values of \(x\).

- Compute s(t(x))

First, find \(s(t(x))\): - \(t(x) = x - 3\)- Substitute \(t(x)\) into \(s(x)\):\[ s(t(x)) = s(x-3)\]- Now substitute \(x-3\) into \(x^2 + 1\):\[ s(x-3) = (x-3)^2 + 1\]- Simplify:\[ s(x-3) = x^2 - 6x + 9 + 1 = x^2 - 6x + 10\]

- Compute t(s(x))

Next, find \(t(s(x))\):- \(s(x) = x^2 + 1\)- Substitute \(s(x)\) into \(t(x)\):\[ t(s(x)) = t(x^2 + 1)\]- Now substitute \(x^2 + 1\) into \(x - 3\):\[ t(x^2 + 1) = (x^2 + 1) - 3\]- Simplify:\[ t(x^2 + 1) = x^2 + 1 - 3 = x^2 - 2\]

- Compare s(t(x)) and t(s(x))

Compare the results from Step 2 and Step 3:\[ s(t(x)) = x^2 - 6x + 10\]\[ t(s(x)) = x^2 - 2\]Since \(s(t(x)) eq t(s(x))\), \(s(t(x))\) does not equal \(t(s(x))\) for all values of \(x\).

Key concepts

Composite Functions

In mathematics, the idea of composite functions comes into play when we use one function inside another function. This is particularly useful for complex mathematical operations, where multiple steps are involved. Composite functions are written as \(f(g(x))\), which means you first apply the function \(g(x)\) and then apply the function \(f(x)\) to the result.

In our exercise, we have two functions: \(s(x) = x^2 + 1\) and \(t(x) = x - 3\). To find the composite function \(s(t(x))\), we substitute \(t(x)\) into \(s(x)\). This action transforms \(t(x) = x - 3\) into the expression \(s(x - 3)\). By substituting and simplifying, you get \(s(x - 3) = (x - 3)^2 + 1 = x^2 - 6x + 10\).

Performing composition the other way around \(t(s(x))\) involves substituting \(s(x)\) into \(t(x)\), and this yields \(t(x^2 + 1) = x^2 + 1 - 3 = x^2 - 2\). Observing that \(s(t(x))\) and \(t(s(x))\) give different results tells us these compositions are not equal for all x.

Polynomial Functions

A polynomial function is a function that involves only non-negative integer powers of \(x\). These are some of the simplest, yet most important, functions in algebra and calculus. Examples include linear functions like \(t(x) = x - 3\), quadratic functions like \(s(x) = x^2 + 1\), and many others.

In the exercise, \(s(x)\) is a polynomial function because it can be expressed in the form \(ax^2 + bx + c \) where \(a = 1, b = 0,\) and \(c = 1\). The function \(s(x)\) is a quadratic polynomial since it has an \(x^2\) term. On the other hand, \(t(x) = x - 3\) is a linear polynomial with \(a = 1 \) and \(c = -3\).

Understanding polynomial functions is crucial for general function manipulation and solving diverse types of algebraic equations. These functions are often involved in modeling real-world phenomena, ranging from simple financial calculations to complex scientific problems.

Function Operations

Operations on functions allow us to create new functions from given ones. Common operations include addition, subtraction, multiplication, division, and composition.

For example, given two functions \(f(x)\) and \(g(x)\), we can create new functions like \(f(x) + g(x)\) or \(f(g(x))\) (the composite function). These operations enable us to explore various functional relationships and behaviors.

In the exercise, we performed the function operation known as composition. This involved substituting one function into another to see how they combine and interact. For \(s(t(x))\), we got \(s(x - 3) = x^2 - 6x + 10\), and for \(t(s(x))\), we got \(t(x^2 + 1) = x^2 - 2\). Comparing these results, we noticed that \(s(t(x))\) is not equal to \(t(s(x))\). This teaches that function operations, specifically composition, do not always yield interchangeable results.

Understanding function operations is fundamental for higher-level math, as it builds the foundation for various topics such as calculus, where functions are manipulated extensively to solve problems and understand systems.