Problem 38 - Algebra and Trigonometry

Question

In Exercises \(33-40,\) use the Intermediate Value Theorem to show that each polynomial has a real zero between the given integers. $$f(x)=x^{5}-x^{3}-1 ; \text { between } 1 \text { and } 2$$

Step-by-Step Solution

Answer

The function \(f(x)=x^{5}-x^{3}-1\) has a real zero between 1 and 2 according to the Intermediate Value Theorem.

Step 1: Function Evaluation at Bounds

Evaluate the function \(f(x)=x^{5}-x^{3}-1\) at \(x=1\) and \(x=2\). \[f(1)=1^{5}-1^{3}-1=1-1-1=-1\] \[f(2)=2^{5}-2^{3}-1=32-8-1=23\]

Step 2: Application of the Intermediate Value Theorem

According to the Intermediate Value Theorem, if \(f(a)<0\) and \(f(b)>0\) for a continuous function \(f\) on the closed interval \([a, b]\), then \(f\) must have at least one real zero in the interval \((a, b)\). Here, \(f(1)<0\) and \(f(2)>0\), so \(f\) must possess a real zero between 1 and 2.

Key Concepts

Real Zeros of Polynomials

Understanding the concept of real zeros in polynomials is critical in algebra and calculus. A real zero of a polynomial is a solution to the equation where the polynomial equals zero, visually represented by the x-intercept on a graph. In other words, it's the value of 'x' that makes the polynomial's output zero.

When dealing with polynomials like the one in our exercise, it's particularly useful to know that if a polynomial has real coefficients, any real zeros it has are the x-values where the graph of the polynomial crosses the x-axis. This function behavior ties closely with the Intermediate Value Theorem, where finding a sign change over an interval indicates a zero lying between the interval's bounds. As seen in the provided solution, evaluating the polynomial at two points to reveal a sign change guarantees us the existence of at least one real zero within that interval.

Continuous Functions

The term 'continuous functions' refers to functions that have no interruptions, jumps, or breaks in their graphs. A polynomial function is a classic example of a continuous function over the real numbers, meaning that its graph can be drawn without lifting the pen from paper.

This continuity is a key part of the Intermediate Value Theorem. The theorem relies on the premise that for any value 'c' between 'f(a)' and 'f(b)', where 'f' is continuous over the interval [a, b], there must be at least one 'x' value in the interval where 'f(x) = c'. This core concept is what allowed us to identify the real zero of the polynomial in the interval between 1 and 2 in the example exercise, because the function's continuous nature means it must pass through zero, our 'c' value, at some point in the interval.

Polynomial Evaluation

Polynomial evaluation is simply the process of calculating the value of a polynomial at a given value of 'x'. This procedure is straightforward: you substitute the value of 'x' into the polynomial and carry out the arithmetic operations according to the order of operations.

To understand the significance of this in the example exercise, let's look back at the solution. We evaluated the given polynomial at two different x-values, 1 and 2. By computing these, we could determine the function's signs at these points, which directly informed us whether a real zero exists between our evaluated points. This concept of evaluation not only helps in finding zeros, but also in graphing polynomials and understanding their behavior across different intervals.