Problem 19 - A Graphical Approach to Precalculus with Limits

Question

Suppose that a polynomial function \(P\) is defined in such a way that \(P(2)=-4\) and \(P(2.5)=2\) What conclusion does the intermediate value theorem allow you to make?

Step-by-Step Solution

Answer

There is at least one root of \( P \) in the interval \((2, 2.5)\).

Step 1: Understand the Intermediate Value Theorem

The Intermediate Value Theorem states that if a function \( f \) is continuous on a closed interval \([a, b]\), and \( N \) is any number between \( f(a) \) and \( f(b) \), then there exists at least one \( c \) in the interval \((a, b)\) such that \( f(c) = N \).

Step 2: Identify the Given Values

We are given that for the polynomial function \( P \), \( P(2) = -4 \) and \( P(2.5) = 2 \). This means that at \( x = 2 \), the function value is \(-4\), and at \( x = 2.5 \), the value is \(2\).

Step 3: Establish the Interval

The values \( x = 2 \) and \( x = 2.5 \) establish the interval \([2, 2.5]\).

Step 4: Check for Continuity

Polynomial functions, such as \( P \), are continuous everywhere. Thus, the function \( P \) is continuous on the interval \([2, 2.5]\).

Step 5: Apply the Intermediate Value Theorem

Since \( P \) is continuous on \([2, 2.5]\) and \( P(2) = -4 \) while \( P(2.5) = 2 \), by the Intermediate Value Theorem, \( P(x) \) must equal \(0\) for at least one \( x \) in the interval \( (2, 2.5) \).

Step 6: Conclusion

There exists at least one \( x \) in \((2, 2.5)\) such that \( P(x) = 0 \).

Key Concepts

Polynomial Functions

Polynomial functions are an interesting and important type of mathematical function. They manifest as expressions involving variables raised to whole number exponents, with coefficients. For instance, a simple polynomial could look like this: \[ P(x) = ax^n + bx^{n-1} + ... + zx + c \] Here, each term consists of a coefficient (like \(a, b, \) or \(z\)) and a variable, raised to a power (like \(n\)). You might already recognize common polynomials, such as linear functions (straight lines) and quadratic functions (parabolas). Some key features of polynomial functions include:

• They are defined for all real numbers \(x\).
• They have no breaks, jumps, or holes.
• Their graphs are smooth and continuous.

Thanks to these properties, polynomial functions are excellent candidates for the Intermediate Value Theorem, as they meet the critical requirement of continuity over any interval.

Continuity

Continuity in mathematical terms implies that a function has a smooth connection for every point on its domain. A function is said to be continuous if, when graphed, it can be drawn without lifting the pencil from the paper. The most important thing about continuity is this:

• Within any given interval, the function does not jump or break.

Continuity is a must for applying the Intermediate Value Theorem. If you look at a polynomial function, you'll see it is inherently continuous. This means that our function \(P(x)\) in the exercise smoothly transitions from one value to another without any jarring jumps. For real-world scenarios and mathematical problems, continuity ensures that you can find and count on intermediate values, like knowing there will be a point \(c\) where \(P(x) = 0\) in an interval if \(P(a)\) and \(P(b)\) are of opposite signs.

Closed Intervals

A closed interval is a set of real numbers lying between two endpoints, inclusive. It is denoted with square brackets, like \([a, b]\). The use of square brackets means that the endpoints \(a\) and \(b\) are part of the interval itself. Closed intervals are crucial in many theorems, including the Intermediate Value Theorem, because they ensure that both ends of the described interval are considered, maintaining continuity. When we say a function is continuous on \([a, b]\), we're considering every single point from \(a\) to \(b\), including \(a\) and \(b\) themselves. In our example, we've worked with the closed interval \([2, 2.5]\) when examining \(P(x)\), meaning every value from 2 up to 2.5 is part of the set we're discussing.