Problem 249 - Precalculus

Question

For the following exercises, use the Intermediate Value Theorem to confirm that the given polynomial has at least one zero within the given interval. $$ f(x)=-x^{4}+4, \text { between } x=1 \text { and } x=3 $$

Step-by-Step Solution

Answer

There is at least one zero between \( x=1 \) and \( x=3 \).

Step 1: Understand the Intermediate Value Theorem

The Intermediate Value Theorem (IVT) states that if a continuous function \( f \) takes different signs at two points \( a \) and \( b \), there is at least one zero between \( a \) and \( b \). This means within the interval \([a, b]\), \( f(x) \) will take every value between \( f(a) \) and \( f(b) \).

Step 2: Evaluate f(x) at given points

Calculate the values of \( f(x) \) at \( x=1 \) and \( x=3 \). The given polynomial is \( f(x) = -x^4 + 4 \). First, find \( f(1) \): \[ f(1) = -(1)^4 + 4 = -1 + 4 = 3 \] Next, find \( f(3) \): \[ f(3) = -(3)^4 + 4 = -81 + 4 = -77 \].

Step 3: Check for sign change

Observe the results from Step 2: \( f(1) = 3 \) and \( f(3) = -77 \). Since \( f(1) > 0 \) and \( f(3) < 0 \), there is a sign change between \( x=1 \) and \( x=3 \).

Step 4: Apply the Intermediate Value Theorem

Since \( f(x) \) is continuous for all \( x \), and there is a sign change between \( x=1 \) and \( x=3 \), by the Intermediate Value Theorem, there must be at least one zero in the interval \((1, 3)\).

Key Concepts

Understanding Continuous Functions

In mathematics, a function is called continuous when its graph is a connected curve without any breaks, jumps, or holes. This is a fundamental concept that allows us to predict behavior within an interval. A continuous function ensures that for every small change in the input, there is a small change in the output. This smoothness means that you can draw the entire graph of the function without lifting your pencil from the paper.

The Intermediate Value Theorem heavily relies on the nature of continuous functions. It states that if a continuous function changes sign over an interval, it must cross the x-axis at least once in that interval. This property is crucial in real-world applications and mathematical proofs where predicting the existence of solutions is essential. Therefore, when dealing with continuous functions, especially polynomial expressions, the Intermediate Value Theorem becomes a powerful tool for confirming the existence of zeros.

Exploring Polynomial Functions

Polynomial functions are mathematical expressions involving variables raised to whole number exponents. They take the form:

• Constant: \(c\)
• Linear: \(ax + b\)
• Quadratic: \(ax^2 + bx + c\)
• Cubic: \(ax^3 + bx^2 + cx + d\)

Higher-degree polynomials can go on with more terms and higher powers. In this case, the polynomial given is a fourth-degree: \(-x^4 + 4\).

What makes polynomial functions special is that they are always continuous and smooth. This means there are no sharp corners or gaps, making them ideal candidates for applying the Intermediate Value Theorem. For the exercise mentioned, by evaluating the polynomial at specific points and checking the sign change, we determine the existence of zeros. Because polynomial functions are continuous, any change in sign over an interval guarantees, by the Intermediate Value Theorem, the presence of at least one zero in that interval.

Finding Zeros of a Function

A zero of a function, also known as a root, is the point where the function equals zero. In other words, it's the value of the variable that makes the function's output \(f(x)\) reset to zero. Finding zeros is often the goal in many mathematical problems, as it implies solving the equation \(f(x) = 0\).

To determine if a given interval contains a zero, we use the Intermediate Value Theorem, like in the exercise provided. First, evaluate the function at the start and end of the interval. If one result is positive and the other is negative, there's a sign change, implying at least one zero exists in the interval. This works well for continuous functions, like polynomials. In our exercise, evaluating \(-x^4 + 4\) at \(x = 1\) and \(x = 3\) yielded \(f(1) = 3\) and \(f(3) = -77\), indicating a sign change and thus verifying the presence of a zero between 1 and 3.