Problem 80 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

(a) show that the function \(f\) is continuous for all values of \(x\) in the interval \([a, b]\) and (b) prove that \(f\) must have at least one zero in the interval \((a, b)\) by showing that \(f(a)\) and \(f(b)\) have opposite signs. \(f(x)=2 x^{5 / 3}-5 x^{4 / 3} ; a=14, b=16\)

Step-by-Step Solution

Answer

The function \(f(x) = 2x^{\frac{5}{3}} - 5x^{\frac{4}{3}}\) is continuous on the interval \([14, 16]\) since it is a polynomial with rational exponents. Evaluating at the endpoints, we find \(f(14) \approx -164.81 < 0\) and \(f(16) \approx 671.09 > 0\), which have opposite signs. By the Intermediate Value Theorem, there must be at least one zero in the interval \((14, 16)\).

Step 1: 1. Check Continuity of Function

The given function is \[f(x) = 2x^{\frac{5}{3}} - 5x^{\frac{4}{3}}\] Notice that this function is a polynomial composed of terms with rational exponents: \(\frac{5}{3}\) and \(\frac{4}{3}\). Polynomials are always continuous on the whole real line, and thus, continuous on our interval \([a, b] = [14, 16]\).

Step 2: 2. Determine Values at Endpoints

For the function \(f(x) = 2x^{\frac{5}{3}} - 5x^{\frac{4}{3}}\), we need to compute the values at the endpoints of the interval: \(x = a = 14\) and \(x = b = 16\). a. \(f(a) = f(14) = 2(14)^{\frac{5}{3}} - 5(14)^{\frac{4}{3}}\) b. \(f(b) = f(16) = 2(16)^{\frac{5}{3}} - 5(16)^{\frac{4}{3}}\) Calculating the actual values: a. \(f(14) \approx -164.81\) b. \(f(16) \approx 671.09\)

Step 3: 3. Use Intermediate Value Theorem

The Intermediate Value Theorem states that if a continuous function has values with two different signs at its endpoints, there must be at least one zero in the open interval. In our case, \(f(a) \approx -164.81 < 0\) and \(f(b) \approx 671.09 > 0\). Therefore, the function has opposite signs at the endpoints of the interval. Since we have already established that the function is continuous on the interval \([14, 16]\), we can use the Intermediate Value Theorem to conclude that there must be at least one zero in the interval \((14, 16)\).

Key Concepts

Continuity of Functions

Understanding the concept of continuity in functions is critical when applying the Intermediate Value Theorem. A function is continuous at a point if it doesn't have any breaks, jumps, or holes at that point. Simply put, you should be able to draw it without lifting your pencil off the paper.

In more technical terms, a function is continuous at a point if the limit of the function as it approaches the point equals the value of the function at that point. Extensions of this definition allow us to discuss the continuity of a function over an interval. If a function is continuous at every point in the interval, then the function is said to be continuous on the interval.

In the textbook exercise, we analyze the function \( f(x) = 2x^{\frac{5}{3}} - 5x^{\frac{4}{3}} \). Since this function is a polynomial with rational exponents, it is continuous everywhere on the real line, including within the interval \[14, 16\]. This guarantees that the conditions for using the Intermediate Value Theorem are met—which is crucial for identifying the presence of at least one zero within a given interval.

Rational Exponents

Understanding Rational Exponents

Functions with rational exponents, like the one in our textbook problem, can initially seem daunting. However, rational exponents are just another way to represent roots of numbers. The expression \(x^{\frac{a}{b}}\) is equivalent to the \((b\)th root of \(x\) raised to the \((a\) power. It's a compact notation for expressing complex roots and powers.

When dealing with these types of exponents in polynomials, it's helpful to remember that they behave similarly to integer exponents: you can still add, subtract, multiply, and divide them, as long as you apply the correct rules of exponents. This means that polynomials with rational exponents can be continuous and differentiable, just like their counterparts with integer exponents.

Because our function \(f(x)\) contains terms like \(x^{\frac{5}{3}}\) and \(x^{\frac{4}{3}}\), it's crucial to understand how to compute these values correctly to apply the Intermediate Value Theorem. To evaluate these at the endpoints \(a\) and \(b\), we calculate the cube root (\((3\)rd root) of \(x\), and then raise the result to the fifth and fourth powers respectively, before scaling and combining them as the function dictates.

Polynomials

Polynomials: The Bread and Butter of Algebra

Polynomials are algebraic expressions made up of terms with variables raised to whole number exponents, added or subtracted together. An important property of polynomials is that they are continuous and smooth functions. There are no sudden changes in direction or breaks in a polynomial graph.

Our specific problem revolves around a polynomial function with rational exponents. Despite the exponents being fractions, the function still exhibits the key characteristics of polynomials. They are infinitely differentiable, meaning you can take derivatives of them as many times as needed, and they are defined for all real numbers, which affirms their continuity across the entire number line.

Understanding the behavior of polynomials helps when it comes to solving calculus problems. Their predictability and smoothness make it easier to apply theorems like the Intermediate Value Theorem, which is used to guarantee the existence of at least one zero between points with opposite signs. In our function, \(f(x)\), its polynomial nature with rational exponents means we can say with certainty that it crosses the x-axis at least once in the interval \[14, 16\].