Question

Use the Intermediate Value Theorem to show that each polynomial function has a real zero in the given interval. $$ f(x)=x^{5}-x^{4}+7 x^{3}-7 x^{2}-18 x+18 ;[1.4,1.5] $$

Short answer

The Intermediate Value Theorem does not show a zero in [1.4, 1.5] for this polynomial since there is no sign change between the endpoints.

Step-by-step solution

- Understand the Intermediate Value Theorem

The Intermediate Value Theorem states that if a continuous function, say f(x), takes on values f(a) and f(b) at two points a and b, and if N is any number between f(a) and f(b), then there is at least one c in the interval (a, b) such that f(c) = N. This theorem can be used to show that a function has a real zero in an interval if it crosses the x-axis within that interval.

- Evaluate the polynomial at the endpoints of the interval

Evaluate the function at the endpoints of the given interval [1.4, 1.5]. First, calculate f(1.4): \[ f(1.4) = (1.4)^5 - (1.4)^4 + 7(1.4)^3 - 7(1.4)^2 - 18(1.4) + 18\] Then, simplify: \[ f(1.4) = 5.37824 - 3.8416 + 19.3344 - 13.72 - 25.2 + 18 \] \[ f(1.4) = 0.95004 \] Next, calculate f(1.5): \[ f(1.5) = (1.5)^5 - (1.5)^4 + 7(1.5)^3 - 7(1.5)^2 - 18(1.5) + 18 \] Then, simplify: \[ f(1.5) = 7.59375 - 5.0625 + 23.625 - 15.75 - 27 + 18 \] \[ f(1.5) = 1.40625 \]

- Analyze the sign change

Observe the results from Step 2. The values of the function at the endpoints of the interval are: f(1.4) = 0.95004 f(1.5) = 1.40625 Both evaluated values are positive, meaning no sign change occurs between f(1.4) and f(1.5). This suggests that there is no zero in the interval of [1.4, 1.5].

Key concepts

real zeros of polynomials

A real zero of a polynomial is a real number that makes the polynomial equal to zero. In other words, it's a value of x for which the polynomial f(x) = 0. These zeros are also known as the roots of the polynomial. When we plot a polynomial function on a graph, the real zeros are the x-coordinates where the curve intersects the x-axis. Finding real zeros is crucial for understanding the behavior and characteristics of polynomial functions.

For example, if we have the polynomial function f(x) = x^2 - 4, the real zeros would be the solutions to the equation x^2 - 4 = 0. This simplifies to x^2 = 4, which gives us the solutions x = 2 and x = -2. Therefore, the real zeros are 2 and -2.

Finding real zeros often involves various methods such as factoring, using the quadratic formula for quadratic polynomials, or numerical methods for higher degree polynomials. The Intermediate Value Theorem is one such mathematical tool used to show the existence of real zeros within a specific interval.

continuous functions

A function is continuous if you can draw its graph without lifting your pen from the paper. In technical terms, a function f(x) is continuous at a point c if the following three conditions are met: f(c) is defined, the limit of f(x) as x approaches c exists, and the limit of f(x) as x approaches c is equal to f(c).

For polynomial functions, which consist of terms that are sums of variables raised to positive integer powers, continuity is inherent. Polynomials are smooth and unbroken curves, hence they are continuous for all real numbers. This continuity is essential when applying the Intermediate Value Theorem because the theorem relies on the function being continuous over the interval in question.

Imagine we are looking at the polynomial f(x) = x^3 - 3x + 2. This polynomial is continuous everywhere. If we evaluate this polynomial at two points, say a = 0 and b = 2, and find that the function values, f(0) and f(2), have opposite signs, then by the Intermediate Value Theorem, there must be at least one real zero between 0 and 2.

evaluating polynomial functions

Evaluating a polynomial function involves plugging a value into the polynomial and simplifying to find the result. This process is fundamental to applying the Intermediate Value Theorem and understanding the function's behavior over an interval.

Let's say we have the polynomial f(x) = x^3 - 2x^2 + x - 1, and we want to evaluate it at x = 2. We substitute 2 for x and perform the following calculations:

• f(2) = 2^3 - 2(2^2) + 2 - 1
• f(2) = 8 - 8 + 2 - 1
• f(2) = 1

Here, we've evaluated the polynomial at x = 2 and found that f(2) = 1. This evaluation is crucial when we work within an interval to determine if the function might cross the x-axis and thus have a zero in between.

In our original exercise, we evaluated the polynomial f(x) = x^5 - x^4 + 7x^3 - 7x^2 - 18x + 18 at the endpoints of the interval [1.4, 1.5]. The next step was to check if the function's values at these points indicate a zero in between. However, since both f(1.4) and f(1.5) were positive, it suggested no zero in that interval. The act of evaluating at these precise points ensures we have accurate and useful data for further analysis.